task_url stringlengths 30 116 | task_name stringlengths 2 86 | task_description stringlengths 0 14.4k | language_url stringlengths 2 53 | language_name stringlengths 1 52 | code stringlengths 0 61.9k |
|---|---|---|---|---|---|
http://rosettacode.org/wiki/Non-decimal_radices/Convert | Non-decimal radices/Convert | Number base conversion is when you express a stored integer in an integer base, such as in octal (base 8) or binary (base 2). It also is involved when you take a string representing a number in a given base and convert it to the stored integer form. Normally, a stored integer is in binary, but that's typically invisible to the user, who normally enters or sees stored integers as decimal.
Task
Write a function (or identify the built-in function) which is passed a non-negative integer to convert, and another integer representing the base.
It should return a string containing the digits of the resulting number, without leading zeros except for the number 0 itself.
For the digits beyond 9, one should use the lowercase English alphabet, where the digit a = 9+1, b = a+1, etc.
For example: the decimal number 26 expressed in base 16 would be 1a.
Write a second function which is passed a string and an integer base, and it returns an integer representing that string interpreted in that base.
The programs may be limited by the word size or other such constraint of a given language. There is no need to do error checking for negatives, bases less than 2, or inappropriate digits.
| #Standard_ML | Standard ML | fun toBase b v = let
fun toBase' (a, 0) = a
| toBase' (a, v) = toBase' (v mod b :: a, v div b)
in
toBase' ([], v)
end
fun fromBase b ds =
foldl (fn (k, n) => n * b + k) 0 ds
val toAlphaDigits = let
fun convert n = if n < 10 then chr (n + ord #"0")
else chr (n + ord #"a" - 10)
in
implode o map convert
end
val fromAlphaDigits = let
fun convert c = if Char.isDigit c then ord c - ord #"0"
else if Char.isUpper c then ord c - ord #"A" + 10
else if Char.isLower c then ord c - ord #"a" + 10
else raise Match
in
map convert o explode
end |
http://rosettacode.org/wiki/Narcissistic_decimal_number | Narcissistic decimal number | A Narcissistic decimal number is a non-negative integer,
n
{\displaystyle n}
, that is equal to the sum of the
m
{\displaystyle m}
-th powers of each of the digits in the decimal representation of
n
{\displaystyle n}
, where
m
{\displaystyle m}
is the number of digits in the decimal representation of
n
{\displaystyle n}
.
Narcissistic (decimal) numbers are sometimes called Armstrong numbers, named after Michael F. Armstrong.
They are also known as Plus Perfect numbers.
An example
if
n
{\displaystyle n}
is 153
then
m
{\displaystyle m}
, (the number of decimal digits) is 3
we have 13 + 53 + 33 = 1 + 125 + 27 = 153
and so 153 is a narcissistic decimal number
Task
Generate and show here the first 25 narcissistic decimal numbers.
Note:
0
1
=
0
{\displaystyle 0^{1}=0}
, the first in the series.
See also
the OEIS entry: Armstrong (or Plus Perfect, or narcissistic) numbers.
MathWorld entry: Narcissistic Number.
Wikipedia entry: Narcissistic number.
| #Quackery | Quackery | [ [] swap
[ 10 /mod
rot join swap
dup 0 = until ]
drop ] is digits ( n --> [ )
[ dup digits
0 over size rot
witheach
[ over ** rot + swap ]
drop = ] is narcissistic ( n --> b )
[] 0
[ dup narcissistic if
[ tuck join swap ]
1+ over size 25 = until ]
drop echo |
http://rosettacode.org/wiki/Narcissistic_decimal_number | Narcissistic decimal number | A Narcissistic decimal number is a non-negative integer,
n
{\displaystyle n}
, that is equal to the sum of the
m
{\displaystyle m}
-th powers of each of the digits in the decimal representation of
n
{\displaystyle n}
, where
m
{\displaystyle m}
is the number of digits in the decimal representation of
n
{\displaystyle n}
.
Narcissistic (decimal) numbers are sometimes called Armstrong numbers, named after Michael F. Armstrong.
They are also known as Plus Perfect numbers.
An example
if
n
{\displaystyle n}
is 153
then
m
{\displaystyle m}
, (the number of decimal digits) is 3
we have 13 + 53 + 33 = 1 + 125 + 27 = 153
and so 153 is a narcissistic decimal number
Task
Generate and show here the first 25 narcissistic decimal numbers.
Note:
0
1
=
0
{\displaystyle 0^{1}=0}
, the first in the series.
See also
the OEIS entry: Armstrong (or Plus Perfect, or narcissistic) numbers.
MathWorld entry: Narcissistic Number.
Wikipedia entry: Narcissistic number.
| #R | R | for (u in 1:10000000) {
j <- nchar(u)
set2 <- c()
for (i in 1:j) {
set2[i] <- as.numeric(substr(u, i, i))
}
control <- c()
for (k in 1:j) {
control[k] <- set2[k]^(j)
}
if (sum(control) == u) print(u)
} |
http://rosettacode.org/wiki/Munching_squares | Munching squares | Render a graphical pattern where each pixel is colored by the value of 'x xor y' from an arbitrary color table.
| #Wren | Wren | import "graphics" for Canvas, Color
import "dome" for Window
class Game {
static init() {
Window.title = "Munching squares"
var w = 512
var h = 512
Window.resize(w, h)
Canvas.resize(w, h)
for (x in 0...w) {
for (y in 0...h) {
var c = (x ^ y) & 255
Canvas.pset(x, y, Color.rgb(255-c, (c/2).floor, c))
}
}
}
static update() {}
static draw(alpha) {}
} |
http://rosettacode.org/wiki/Munchausen_numbers | Munchausen numbers | A Munchausen number is a natural number n the sum of whose digits (in base 10), each raised to the power of itself, equals n.
(Munchausen is also spelled: Münchhausen.)
For instance: 3435 = 33 + 44 + 33 + 55
Task
Find all Munchausen numbers between 1 and 5000.
Also see
The OEIS entry: A046253
The Wikipedia entry: Perfect digit-to-digit invariant, redirected from Munchausen Number
| #FreeBASIC | FreeBASIC | ' FB 1.05.0 Win64
' Cache n ^ n for the digits 1 to 9
' Note than 0 ^ 0 specially treated as 0 (not 1) for this purpose
Dim Shared powers(1 To 9) As UInteger
For i As UInteger = 1 To 9
Dim power As UInteger = i
For j As UInteger = 2 To i
power *= i
Next j
powers(i) = power
Next i
Function isMunchausen(n As UInteger) As Boolean
Dim p As UInteger = n
Dim As UInteger digit, sum
While p > 0
digit = p Mod 10
If digit > 0 Then sum += powers(digit)
p \= 10
Wend
Return n = sum
End Function
Print "The Munchausen numbers between 0 and 500000000 are : "
For i As UInteger = 0 To 500000000
If isMunchausen(i) Then Print i
Next
Print
Print "Press any key to quit"
Sleep |
http://rosettacode.org/wiki/Mutual_recursion | Mutual recursion | Two functions are said to be mutually recursive if the first calls the second,
and in turn the second calls the first.
Write two mutually recursive functions that compute members of the Hofstadter Female and Male sequences defined as:
F
(
0
)
=
1
;
M
(
0
)
=
0
F
(
n
)
=
n
−
M
(
F
(
n
−
1
)
)
,
n
>
0
M
(
n
)
=
n
−
F
(
M
(
n
−
1
)
)
,
n
>
0.
{\displaystyle {\begin{aligned}F(0)&=1\ ;\ M(0)=0\\F(n)&=n-M(F(n-1)),\quad n>0\\M(n)&=n-F(M(n-1)),\quad n>0.\end{aligned}}}
(If a language does not allow for a solution using mutually recursive functions
then state this rather than give a solution by other means).
| #E | E | def [F, M] := [
fn n { if (n <=> 0) { 1 } else { n - M(F(n - 1)) } },
fn n { if (n <=> 0) { 0 } else { n - F(M(n - 1)) } },
] |
http://rosettacode.org/wiki/Multisplit | Multisplit | It is often necessary to split a string into pieces
based on several different (potentially multi-character) separator strings,
while still retaining the information about which separators were present in the input.
This is particularly useful when doing small parsing tasks.
The task is to write code to demonstrate this.
The function (or procedure or method, as appropriate) should
take an input string and an ordered collection of separators.
The order of the separators is significant:
The delimiter order represents priority in matching, with the first defined delimiter having the highest priority.
In cases where there would be an ambiguity as to
which separator to use at a particular point
(e.g., because one separator is a prefix of another)
the separator with the highest priority should be used.
Delimiters can be reused and the output from the function should be an ordered sequence of substrings.
Test your code using the input string “a!===b=!=c” and the separators “==”, “!=” and “=”.
For these inputs the string should be parsed as "a" (!=) "" (==) "b" (=) "" (!=) "c", where matched delimiters are shown in parentheses, and separated strings are quoted, so our resulting output is "a", empty string, "b", empty string, "c".
Note that the quotation marks are shown for clarity and do not form part of the output.
Extra Credit: provide information that indicates which separator was matched at each separation point and where in the input string that separator was matched.
| #REXX | REXX | /*REXX program splits a (character) string based on different separator delimiters.*/
parse arg $ /*obtain optional string from the C.L. */
if $='' then $= "a!===b=!=c" /*None specified? Then use the default*/
say 'old string:' $ /*display the old string to the screen.*/
null= '0'x /*null char. It can be most anything.*/
seps= '== != =' /*list of separator strings to be used.*/
/* [↓] process the tokens in SEPS. */
do j=1 for words(seps) /*parse the string with all the seps. */
sep=word(seps,j) /*pick a separator to use in below code*/
/* [↓] process characters in the sep.*/
do k=1 for length(sep) /*parse for various separator versions.*/
sep=strip(insert(null, sep, k), , null) /*allow imbedded "nulls" in separator, */
$=changestr(sep, $, null) /* ··· but not trailing "nulls". */
/* [↓] process strings in the input. */
do until $==old; old=$ /*keep changing until no more changes. */
$=changestr(null || null, $, null) /*reduce replicated "nulls" in string. */
end /*until*/
/* [↓] use BIF or external program.*/
sep=changestr(null, sep, '') /*remove true nulls from the separator.*/
end /*k*/
end /*j*/
showNull= ' {} ' /*just one more thing, display the ··· */
$=changestr(null, $, showNull) /* ··· showing of "null" chars. */
say 'new string:' $ /*now, display the new string to term. */
/*stick a fork in it, we're all done. */ |
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #AutoHotkey | AutoHotkey | a := []
Loop, %n%
a[A_Index] := new Foo() |
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #BBC_BASIC | BBC BASIC | REM Determine object count at runtime:
n% = RND(1000)
REM Declare an array of structures; all members are initialised to zero:
DIM objects{(n%) a%, b$}
REM Initialise the objects to distinct values:
FOR i% = 0 TO DIM(objects{()},1)
objects{(i%)}.a% = i%
objects{(i%)}.b$ = STR$(i%)
NEXT
REM This is how to create an array of pointers to the same object:
DIM objects%(n%), object{a%, b$}
FOR i% = 0 TO DIM(objects%(),1)
objects%(i%) = object{}
NEXT |
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #Brat | Brat | n.of foo.new |
http://rosettacode.org/wiki/Multifactorial | Multifactorial | The factorial of a number, written as
n
!
{\displaystyle n!}
, is defined as
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
.
Multifactorials generalize factorials as follows:
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
n
!
!
=
n
(
n
−
2
)
(
n
−
4
)
.
.
.
{\displaystyle n!!=n(n-2)(n-4)...}
n
!
!
!
=
n
(
n
−
3
)
(
n
−
6
)
.
.
.
{\displaystyle n!!!=n(n-3)(n-6)...}
n
!
!
!
!
=
n
(
n
−
4
)
(
n
−
8
)
.
.
.
{\displaystyle n!!!!=n(n-4)(n-8)...}
n
!
!
!
!
!
=
n
(
n
−
5
)
(
n
−
10
)
.
.
.
{\displaystyle n!!!!!=n(n-5)(n-10)...}
In all cases, the terms in the products are positive integers.
If we define the degree of the multifactorial as the difference in successive terms that are multiplied together for a multifactorial (the number of exclamation marks), then the task is twofold:
Write a function that given n and the degree, calculates the multifactorial.
Use the function to generate and display here a table of the first ten members (1 to 10) of the first five degrees of multifactorial.
Note: The wikipedia entry on multifactorials gives a different formula. This task uses the Wolfram mathworld definition.
| #11l | 11l | F multifact(n, d)
R product((n .< 1).step(-d))
L(d) 1..5
print(‘Degree ’d‘: ’(1..10).map(n -> multifact(n, @d))) |
http://rosettacode.org/wiki/N-queens_problem | N-queens problem |
Solve the eight queens puzzle.
You can extend the problem to solve the puzzle with a board of size NxN.
For the number of solutions for small values of N, see OEIS: A000170.
Related tasks
A* search algorithm
Solve a Hidato puzzle
Solve a Holy Knight's tour
Knight's tour
Peaceful chess queen armies
Solve a Hopido puzzle
Solve a Numbrix puzzle
Solve the no connection puzzle
| #Common_Lisp | Common Lisp | (defun queens (n &optional (m n))
(if (zerop n)
(list nil)
(loop for solution in (queens (1- n) m)
nconc (loop for new-col from 1 to m
when (loop for row from 1 to n
for col in solution
always (/= new-col col (+ col row) (- col row)))
collect (cons new-col solution)))))
(defun print-solution (solution)
(loop for queen-col in solution
do (loop for col from 1 to (length solution)
do (write-char (if (= col queen-col) #\Q #\.)))
(terpri))
(terpri))
(defun print-queens (n)
(mapc #'print-solution (queens n))) |
http://rosettacode.org/wiki/Multiplicative_order | Multiplicative order | The multiplicative order of a relative to m is the least positive integer n such that a^n is 1 (modulo m).
Example
The multiplicative order of 37 relative to 1000 is 100 because 37^100 is 1 (modulo 1000), and no number smaller than 100 would do.
One possible algorithm that is efficient also for large numbers is the following: By the Chinese Remainder Theorem, it's enough to calculate the multiplicative order for each prime exponent p^k of m, and
combine the results with the least common multiple operation.
Now the order of a with regard to p^k must divide Φ(p^k). Call this number t, and determine it's factors q^e. Since each multiple of the order will also yield 1 when used as exponent for a, it's enough to find the least d such that (q^d)*(t/(q^e)) yields 1 when used as exponent.
Task
Implement a routine to calculate the multiplicative order along these lines. You may assume that routines to determine the factorization into prime powers are available in some library.
An algorithm for the multiplicative order can be found in Bach & Shallit, Algorithmic Number Theory, Volume I: Efficient Algorithms, The MIT Press, 1996:
Exercise 5.8, page 115:
Suppose you are given a prime p and a complete factorization
of p-1. Show how to compute the order of an
element a in (Z/(p))* using O((lg p)4/(lg lg p)) bit
operations.
Solution, page 337:
Let the prime factorization of p-1 be q1e1q2e2...qkek . We use the following observation:
if x^((p-1)/qifi) = 1 (mod p) ,
and fi=ei or x^((p-1)/qifi+1) != 1 (mod p) , then qiei-fi||ordp x. (This follows by combining Exercises 5.1 and 2.10.)
Hence it suffices to find, for each i , the exponent fi such that the condition above holds.
This can be done as follows: first compute q1e1, q2e2, ... ,
qkek . This can be done using O((lg p)2) bit operations. Next, compute y1=(p-1)/q1e1, ... , yk=(p-1)/qkek .
This can be done using O((lg p)2) bit operations. Now, using the binary method,
compute x1=ay1(mod p), ... , xk=ayk(mod p) .
This can be done using O(k(lg p)3) bit operations, and k=O((lg p)/(lg lg p)) by Theorem 8.8.10.
Finally, for each i , repeatedly raise xi to the qi-th power (mod p) (as many as ei-1 times), checking to see when 1 is obtained.
This can be done using O((lg p)3) steps.
The total cost is dominated by O(k(lg p)3) , which is O((lg p)4/(lg lg p)).
| #EchoLisp | EchoLisp |
(require 'bigint)
;; factor-exp returns a list ((p k) ..) : a = p1^k1 * p2^k2 ..
(define (factor-exp a)
(map (lambda (g) (list (first g) (length g)))
(group* (prime-factors a))))
;; copied from Ruby
(define (_mult_order a p k (x))
(define pk (expt p k))
(define t (* (1- p) (expt p (1- k))))
(define r 1)
(for [((q e) (factor-exp t))]
(set! x (powmod a (/ t (expt q e)) pk))
(while (!= x 1)
(*= r q)
(set! x (powmod x q pk))))
r)
(define (order a m)
"multiplicative order : (order a m) → n : a^n = 1 (mod m)"
(assert (= 1 (gcd a m)) "a and m must be coprimes")
(define mopks (for/list [((p k) (factor-exp m))] (_mult_order a p k)))
(for/fold (n 1) ((mopk mopks)) (lcm n mopk)))
;; results
order 37 1000)
→ 100
(order (+ (expt 10 100) 1) 7919)
→ 3959
(order (+ (expt 10 1000) 1) 15485863)
→ 15485862
|
http://rosettacode.org/wiki/Nth_root | Nth root | Task
Implement the algorithm to compute the principal nth root
A
n
{\displaystyle {\sqrt[{n}]{A}}}
of a positive real number A, as explained at the Wikipedia page.
| #Phix | Phix | with javascript_semantics
function pow_(atom x, integer e)
atom r = 1
for i=1 to e do
r *= x
end for
return r
end function
function nth_root(atom y, n)
atom eps = 1e-15, -- relative accuracy
x = 1
while 1 do
-- atom d = ( y / power(x,n-1) - x ) / n
atom d = ( y / pow_(x,n-1) - x ) / n
x += d
atom e = eps*x -- absolute accuracy
if d > -e and d < e then exit end if
end while
return x
end function
procedure test(sequence yn)
atom {y,n} = yn
printf(1,"nth_root(%d,%d) = %.10g, builtin = %.10g\n",{y,n,nth_root(y,n),power(y,1/n)})
end procedure
papply({{1024,10},{27,3},{2,2},{5642,125},{4913,3},{8,3},{16,2},{16,4},{125,3},{1000000000,3},{1000000000,9}},test)
|
http://rosettacode.org/wiki/N%27th | N'th | Write a function/method/subroutine/... that when given an integer greater than or equal to zero returns a string of the number followed by an apostrophe then the ordinal suffix.
Example
Returns would include 1'st 2'nd 3'rd 11'th 111'th 1001'st 1012'th
Task
Use your routine to show here the output for at least the following (inclusive) ranges of integer inputs:
0..25, 250..265, 1000..1025
Note: apostrophes are now optional to allow correct apostrophe-less English.
| #Haskell | Haskell | import Data.Array
ordSuffs :: Array Integer String
ordSuffs = listArray (0,9) ["th", "st", "nd", "rd", "th",
"th", "th", "th", "th", "th"]
ordSuff :: Integer -> String
ordSuff n = show n ++ suff n
where suff m | (m `rem` 100) >= 11 && (m `rem` 100) <= 13 = "th"
| otherwise = ordSuffs ! (m `rem` 10)
printOrdSuffs :: [Integer] -> IO ()
printOrdSuffs = putStrLn . unwords . map ordSuff
main :: IO ()
main = do
printOrdSuffs [ 0.. 25]
printOrdSuffs [ 250.. 265]
printOrdSuffs [1000..1025] |
http://rosettacode.org/wiki/Non-decimal_radices/Convert | Non-decimal radices/Convert | Number base conversion is when you express a stored integer in an integer base, such as in octal (base 8) or binary (base 2). It also is involved when you take a string representing a number in a given base and convert it to the stored integer form. Normally, a stored integer is in binary, but that's typically invisible to the user, who normally enters or sees stored integers as decimal.
Task
Write a function (or identify the built-in function) which is passed a non-negative integer to convert, and another integer representing the base.
It should return a string containing the digits of the resulting number, without leading zeros except for the number 0 itself.
For the digits beyond 9, one should use the lowercase English alphabet, where the digit a = 9+1, b = a+1, etc.
For example: the decimal number 26 expressed in base 16 would be 1a.
Write a second function which is passed a string and an integer base, and it returns an integer representing that string interpreted in that base.
The programs may be limited by the word size or other such constraint of a given language. There is no need to do error checking for negatives, bases less than 2, or inappropriate digits.
| #Swift | Swift | println(String(26, radix: 16)) // prints "1a" |
http://rosettacode.org/wiki/Narcissistic_decimal_number | Narcissistic decimal number | A Narcissistic decimal number is a non-negative integer,
n
{\displaystyle n}
, that is equal to the sum of the
m
{\displaystyle m}
-th powers of each of the digits in the decimal representation of
n
{\displaystyle n}
, where
m
{\displaystyle m}
is the number of digits in the decimal representation of
n
{\displaystyle n}
.
Narcissistic (decimal) numbers are sometimes called Armstrong numbers, named after Michael F. Armstrong.
They are also known as Plus Perfect numbers.
An example
if
n
{\displaystyle n}
is 153
then
m
{\displaystyle m}
, (the number of decimal digits) is 3
we have 13 + 53 + 33 = 1 + 125 + 27 = 153
and so 153 is a narcissistic decimal number
Task
Generate and show here the first 25 narcissistic decimal numbers.
Note:
0
1
=
0
{\displaystyle 0^{1}=0}
, the first in the series.
See also
the OEIS entry: Armstrong (or Plus Perfect, or narcissistic) numbers.
MathWorld entry: Narcissistic Number.
Wikipedia entry: Narcissistic number.
| #Racket | Racket | ;; OEIS: A005188 defines these as positive numbers, so I will follow that definition in the function
;; definitions.
;;
;; 0: assuming it is represented as the single digit 0 (and not an empty string, which is not the
;; usual convention for 0 in decimal), is not: sum(0^0), which is 1. 0^0 is a strange one,
;; wolfram alpha calls returns 0^0 as indeterminate -- so I will defer to the brains behind OEIS
;; on the definition here, rather than copy what I'm seeing in some of the results here
#lang racket
;; Included for the serious efficientcy gains we get from fxvectors vs. general vectors.
;;
;; We also use fx+/fx- etc. As it stands, they do a check for fixnumness, for safety.
;; We can link them in as "unsafe" operations (see the documentation on racket/fixnum);
;; but we get a result from this program quickly enough for my tastes.
(require racket/fixnum)
; uses a precalculated (fx)vector of powers -- caller provided, please.
(define (sub-narcissitic? N powered-digits)
(let loop ((n N) (target N))
(cond
[(fx> 0 target) #f]
[(fx= 0 target) (fx= 0 n)]
[(fx= 0 n) #f]
[else (loop (fxquotient n 10)
(fx- target (fxvector-ref powered-digits (fxremainder n 10))))])))
; Can be used as standalone, since it doesn't require caller to care about things like order of
; magnitude etc. However, it *is* slow, since it regenerates the powered-digits vector every time.
(define (narcissitic? n) ; n is +ve
(define oom+1 (fx+ 1 (order-of-magnitude n)))
(define powered-digits (for/fxvector ((i 10)) (expt i oom+1)))
(sub-narcissitic? n powered-digits))
;; next m primes > z
(define (next-narcissitics z m) ; naming convention following math/number-theory's next-primes
(let-values
([(i l)
(for*/fold ((i (fx+ 1 z)) (l empty))
((oom (in-naturals))
(dgts^oom (in-value (for/fxvector ((i 10)) (expt i (add1 oom)))))
(n (in-range (expt 10 oom) (expt 10 (add1 oom))))
#:when (sub-narcissitic? n dgts^oom)
; everyone else uses ^C to break...
; that's a bit of a manual process, don't you think?
#:final (= (fx+ 1 (length l)) m))
(values (+ i 1) (append l (list n))))])
l)) ; we only want the list
(module+ main
(next-narcissitics 0 25)
; here's another list... depending on whether you believe sloane or wolfram :-)
(cons 0 (next-narcissitics 0 25)))
(module+ test
(require rackunit)
; example given at head of task
(check-true (narcissitic? 153))
; rip off the first 12 (and 0, since Armstrong numbers seem to be postivie) from
; http://oeis.org/A005188 for testing
(check-equal?
(for/list ((i (in-range 12))
(n (sequence-filter narcissitic? (in-naturals 1)))) n)
'(1 2 3 4 5 6 7 8 9 153 370 371))
(check-equal? (next-narcissitics 0 12) '(1 2 3 4 5 6 7 8 9 153 370 371))) |
http://rosettacode.org/wiki/Munching_squares | Munching squares | Render a graphical pattern where each pixel is colored by the value of 'x xor y' from an arbitrary color table.
| #XPL0 | XPL0 | include c:\cxpl\codes; \intrinsic 'code' declarations
int X, Y;
[SetVid($101); \set 640x480 graphics with 8-bit color
port($3C8):= 0; \set color registers with beautiful shades
for X:= 0 to 256-1 do
[port($3C9):= X>>1; \red
port($3C9):= X>>3; \green
port($3C9):= X; \blue
];
for Y:= 0 to 256-1 do \"color table" is array of 256 registers
for X:= 0 to 256-1 do
Point(X, Y, X|Y); \"|" = XOR, not OR which is "!"
X:= ChIn(1); \wait for keystroke
SetVid(3); \restore normal text mode
] |
http://rosettacode.org/wiki/Munching_squares | Munching squares | Render a graphical pattern where each pixel is colored by the value of 'x xor y' from an arbitrary color table.
| #Yabasic | Yabasic | w = 256
open window w, w
For x = 0 To w-1
For y = 0 To w-1
r =and(xor(x, y), 255)
color r, and(r*2, 255), and(r*3, 255)
dot x, y
Next
Next |
http://rosettacode.org/wiki/Munchausen_numbers | Munchausen numbers | A Munchausen number is a natural number n the sum of whose digits (in base 10), each raised to the power of itself, equals n.
(Munchausen is also spelled: Münchhausen.)
For instance: 3435 = 33 + 44 + 33 + 55
Task
Find all Munchausen numbers between 1 and 5000.
Also see
The OEIS entry: A046253
The Wikipedia entry: Perfect digit-to-digit invariant, redirected from Munchausen Number
| #F.C5.8Drmul.C3.A6 | Fōrmulæ | package main
import(
"fmt"
"math"
)
var powers [10]int
func isMunchausen(n int) bool {
if n < 0 { return false }
n64 := int64(n)
nn := n64
var sum int64 = 0
for nn > 0 {
sum += int64(powers[nn % 10])
if sum > n64 { return false }
nn /= 10
}
return sum == n64
}
func main() {
// cache n ^ n for n in 0..9, defining 0 ^ 0 = 0 for this purpose
for i := 1; i <= 9; i++ {
d := float64(i)
powers[i] = int(math.Pow(d, d))
}
// check numbers 0 to 500 million
fmt.Println("The Munchausen numbers between 0 and 500 million are:")
for i := 0; i <= 500000000; i++ {
if isMunchausen(i) { fmt.Printf("%d ", i) }
}
fmt.Println()
} |
http://rosettacode.org/wiki/Mutual_recursion | Mutual recursion | Two functions are said to be mutually recursive if the first calls the second,
and in turn the second calls the first.
Write two mutually recursive functions that compute members of the Hofstadter Female and Male sequences defined as:
F
(
0
)
=
1
;
M
(
0
)
=
0
F
(
n
)
=
n
−
M
(
F
(
n
−
1
)
)
,
n
>
0
M
(
n
)
=
n
−
F
(
M
(
n
−
1
)
)
,
n
>
0.
{\displaystyle {\begin{aligned}F(0)&=1\ ;\ M(0)=0\\F(n)&=n-M(F(n-1)),\quad n>0\\M(n)&=n-F(M(n-1)),\quad n>0.\end{aligned}}}
(If a language does not allow for a solution using mutually recursive functions
then state this rather than give a solution by other means).
| #Eiffel | Eiffel |
class
APPLICATION
create
make
feature
make
-- Test of the mutually recursive functions Female and Male.
do
across
0 |..| 19 as c
loop
io.put_string (Female (c.item).out + " ")
end
io.new_line
across
0 |..| 19 as c
loop
io.put_string (Male (c.item).out + " ")
end
end
Female (n: INTEGER): INTEGER
-- Female sequence of the Hofstadter Female and Male sequences.
require
n_not_negative: n >= 0
do
Result := 1
if n /= 0 then
Result := n - Male (Female (n - 1))
end
end
Male (n: INTEGER): INTEGER
-- Male sequence of the Hofstadter Female and Male sequences.
require
n_not_negative: n >= 0
do
Result := 0
if n /= 0 then
Result := n - Female (Male (n - 1))
end
end
end
|
http://rosettacode.org/wiki/Multisplit | Multisplit | It is often necessary to split a string into pieces
based on several different (potentially multi-character) separator strings,
while still retaining the information about which separators were present in the input.
This is particularly useful when doing small parsing tasks.
The task is to write code to demonstrate this.
The function (or procedure or method, as appropriate) should
take an input string and an ordered collection of separators.
The order of the separators is significant:
The delimiter order represents priority in matching, with the first defined delimiter having the highest priority.
In cases where there would be an ambiguity as to
which separator to use at a particular point
(e.g., because one separator is a prefix of another)
the separator with the highest priority should be used.
Delimiters can be reused and the output from the function should be an ordered sequence of substrings.
Test your code using the input string “a!===b=!=c” and the separators “==”, “!=” and “=”.
For these inputs the string should be parsed as "a" (!=) "" (==) "b" (=) "" (!=) "c", where matched delimiters are shown in parentheses, and separated strings are quoted, so our resulting output is "a", empty string, "b", empty string, "c".
Note that the quotation marks are shown for clarity and do not form part of the output.
Extra Credit: provide information that indicates which separator was matched at each separation point and where in the input string that separator was matched.
| #Ring | Ring |
# Project : Multisplit
str = "a!===b=!=c"
sep = "=== != =! b =!="
sep = str2list(substr(sep, " ", nl))
for n = 1 to len(sep)
pos = substr(str, sep[n])
see "" + n + ": " + substr(str, 1, pos-1) + " Sep By: " + sep[n] + nl
next
|
http://rosettacode.org/wiki/Multisplit | Multisplit | It is often necessary to split a string into pieces
based on several different (potentially multi-character) separator strings,
while still retaining the information about which separators were present in the input.
This is particularly useful when doing small parsing tasks.
The task is to write code to demonstrate this.
The function (or procedure or method, as appropriate) should
take an input string and an ordered collection of separators.
The order of the separators is significant:
The delimiter order represents priority in matching, with the first defined delimiter having the highest priority.
In cases where there would be an ambiguity as to
which separator to use at a particular point
(e.g., because one separator is a prefix of another)
the separator with the highest priority should be used.
Delimiters can be reused and the output from the function should be an ordered sequence of substrings.
Test your code using the input string “a!===b=!=c” and the separators “==”, “!=” and “=”.
For these inputs the string should be parsed as "a" (!=) "" (==) "b" (=) "" (!=) "c", where matched delimiters are shown in parentheses, and separated strings are quoted, so our resulting output is "a", empty string, "b", empty string, "c".
Note that the quotation marks are shown for clarity and do not form part of the output.
Extra Credit: provide information that indicates which separator was matched at each separation point and where in the input string that separator was matched.
| #Ruby | Ruby | text = 'a!===b=!=c'
separators = ['==', '!=', '=']
def multisplit_simple(text, separators)
text.split(Regexp.union(separators))
end
p multisplit_simple(text, separators) # => ["a", "", "b", "", "c"]
|
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #C | C | foo *foos = malloc(n * sizeof(*foos));
for (int i = 0; i < n; i++)
init_foo(&foos[i]); |
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #C.23 | C# | using System;
using System.Linq;
using System.Collections.Generic;
List<Foo> foos = Enumerable.Range(1, n).Select(x => new Foo()).ToList(); |
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #C.2B.2B | C++ | // this assumes T is a default-constructible type (all built-in types are)
T* p = new T[n]; // if T is POD, the objects are uninitialized, otherwise they are default-initialized
//If default initialisation is not what you want, or if T is a POD type which will be uninitialized
for(size_t i = 0; i != n; ++i)
p[i] = make_a_T(); //or some other expression of type T
// when you don't need the objects any more, get rid of them
delete[] p; |
http://rosettacode.org/wiki/Multifactorial | Multifactorial | The factorial of a number, written as
n
!
{\displaystyle n!}
, is defined as
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
.
Multifactorials generalize factorials as follows:
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
n
!
!
=
n
(
n
−
2
)
(
n
−
4
)
.
.
.
{\displaystyle n!!=n(n-2)(n-4)...}
n
!
!
!
=
n
(
n
−
3
)
(
n
−
6
)
.
.
.
{\displaystyle n!!!=n(n-3)(n-6)...}
n
!
!
!
!
=
n
(
n
−
4
)
(
n
−
8
)
.
.
.
{\displaystyle n!!!!=n(n-4)(n-8)...}
n
!
!
!
!
!
=
n
(
n
−
5
)
(
n
−
10
)
.
.
.
{\displaystyle n!!!!!=n(n-5)(n-10)...}
In all cases, the terms in the products are positive integers.
If we define the degree of the multifactorial as the difference in successive terms that are multiplied together for a multifactorial (the number of exclamation marks), then the task is twofold:
Write a function that given n and the degree, calculates the multifactorial.
Use the function to generate and display here a table of the first ten members (1 to 10) of the first five degrees of multifactorial.
Note: The wikipedia entry on multifactorials gives a different formula. This task uses the Wolfram mathworld definition.
| #360_Assembly | 360 Assembly | * Multifactorial 09/05/2016
MULFACR CSECT
USING MULFACR,13
SAVEAR B STM-SAVEAR(15)
DC 17F'0'
STM STM 14,12,12(13) prolog
ST 13,4(15) "
ST 15,8(13) "
LR 13,15 "
LA I,1 i=1
LOOPI C I,D do i=1 to deg
BH ELOOPI leave i
LA L,W+4 l=@p
LA J,1 j=1
LOOPJ C J,N do j=1 to num
BH ELOOPJ leave j
LA R,1 r=1
LCR S,I s=-i
LR K,J k=j
LOOPK C K,=F'2' do k=j to 2 by s
BL ELOOPK leave k
MR RR,K r=r*k
AR K,S k=k+s
B LOOPK next k
ELOOPK CVD R,Y pack r
MVC X,=XL12'402020202020202020202120' ed mask
ED X,Y+2 edit r
MVC 0(8,L),X+4 output r
LA L,8(L) l=l+8
LA J,1(J) j=j+1
B LOOPJ next j
ELOOPJ WTO MF=(E,W)
LA I,1(I) i=i+1
B LOOPI next i
ELOOPI L 13,4(0,13) epilog
LM 14,12,12(13) "
XR 15,15 "
BR 14 "
N DC F'10' number
D DC F'5' degree
W DC 0F,H'84',H'0',CL80' ' length,zero,text
X DS CL12 temp
Y DS D packed PL8
I EQU 6
J EQU 7
K EQU 8
S EQU 9
RR EQU 10 even reg of R for MR opcode
R EQU 11
L EQU 12
END MULFACR |
http://rosettacode.org/wiki/Multifactorial | Multifactorial | The factorial of a number, written as
n
!
{\displaystyle n!}
, is defined as
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
.
Multifactorials generalize factorials as follows:
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
n
!
!
=
n
(
n
−
2
)
(
n
−
4
)
.
.
.
{\displaystyle n!!=n(n-2)(n-4)...}
n
!
!
!
=
n
(
n
−
3
)
(
n
−
6
)
.
.
.
{\displaystyle n!!!=n(n-3)(n-6)...}
n
!
!
!
!
=
n
(
n
−
4
)
(
n
−
8
)
.
.
.
{\displaystyle n!!!!=n(n-4)(n-8)...}
n
!
!
!
!
!
=
n
(
n
−
5
)
(
n
−
10
)
.
.
.
{\displaystyle n!!!!!=n(n-5)(n-10)...}
In all cases, the terms in the products are positive integers.
If we define the degree of the multifactorial as the difference in successive terms that are multiplied together for a multifactorial (the number of exclamation marks), then the task is twofold:
Write a function that given n and the degree, calculates the multifactorial.
Use the function to generate and display here a table of the first ten members (1 to 10) of the first five degrees of multifactorial.
Note: The wikipedia entry on multifactorials gives a different formula. This task uses the Wolfram mathworld definition.
| #Action.21 | Action! | INCLUDE "D2:REAL.ACT" ;from the Action! Tool Kit
PROC Multifactorial(INT n,d REAL POINTER res)
REAL r
IntToReal(1,res)
WHILE n>1
DO
IntToReal(n,r)
RealMult(res,r,res)
n==-d
OD
RETURN
PROC Main()
BYTE n,d
REAL r
Put(125) PutE() ;clear the screen
FOR d=1 TO 5
DO
PrintF("Degree %B:",d)
FOR n=1 TO 10
DO
Multifactorial(n,d,r)
Put(32) PrintR(r)
OD
PutE()
OD
RETURN |
http://rosettacode.org/wiki/Multi-base_primes | Multi-base primes | Prime numbers are prime no matter what base they are represented in.
A prime number in a base other than 10 may not look prime at first glance.
For instance: 19 base 10 is 25 in base 7.
Several different prime numbers may be expressed as the "same" string when converted to a different base.
107 base 10 converted to base 6 == 255
173 base 10 converted to base 8 == 255
353 base 10 converted to base 12 == 255
467 base 10 converted to base 14 == 255
743 base 10 converted to base 18 == 255
1277 base 10 converted to base 24 == 255
1487 base 10 converted to base 26 == 255
2213 base 10 converted to base 32 == 255
Task
Restricted to bases 2 through 36; find the strings that have the most different bases that evaluate to that string when converting prime numbers to a base.
Find the conversion string, the amount of bases that evaluate a prime to that string and the enumeration of bases that evaluate a prime to that string.
Display here, on this page, the string, the count and the list for all of the: 1 character, 2 character, 3 character, and 4 character strings that have the maximum base count that evaluate to that string.
Should be no surprise, the string '2' has the largest base count for single character strings.
Stretch goal
Do the same for the maximum 5 character string.
| #C.2B.2B | C++ | #include <algorithm>
#include <cmath>
#include <cstdint>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <string>
#include <vector>
#include <primesieve.hpp>
class prime_sieve {
public:
explicit prime_sieve(uint64_t limit);
bool is_prime(uint64_t n) const {
return n == 2 || ((n & 1) == 1 && sieve[n >> 1]);
}
private:
std::vector<bool> sieve;
};
prime_sieve::prime_sieve(uint64_t limit) : sieve((limit + 1) / 2, false) {
primesieve::iterator iter;
uint64_t prime = iter.next_prime(); // consume 2
while ((prime = iter.next_prime()) <= limit) {
sieve[prime >> 1] = true;
}
}
template <typename T> void print(std::ostream& out, const std::vector<T>& v) {
if (!v.empty()) {
out << '[';
auto i = v.begin();
out << *i++;
for (; i != v.end(); ++i)
out << ", " << *i;
out << ']';
}
}
std::string to_string(const std::vector<unsigned int>& v) {
static constexpr char digits[] =
"0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
std::string str;
for (auto i : v)
str += digits[i];
return str;
}
bool increment(std::vector<unsigned int>& digits, unsigned int max_base) {
for (auto i = digits.rbegin(); i != digits.rend(); ++i) {
if (*i + 1 != max_base) {
++*i;
return true;
}
*i = 0;
}
return false;
}
void multi_base_primes(unsigned int max_base, unsigned int max_length) {
prime_sieve sieve(static_cast<uint64_t>(std::pow(max_base, max_length)));
for (unsigned int length = 1; length <= max_length; ++length) {
std::cout << length
<< "-character strings which are prime in most bases: ";
unsigned int most_bases = 0;
std::vector<
std::pair<std::vector<unsigned int>, std::vector<unsigned int>>>
max_strings;
std::vector<unsigned int> digits(length, 0);
digits[0] = 1;
std::vector<unsigned int> bases;
do {
auto max = std::max_element(digits.begin(), digits.end());
unsigned int min_base = 2;
if (max != digits.end())
min_base = std::max(min_base, *max + 1);
if (most_bases > max_base - min_base + 1)
continue;
bases.clear();
for (unsigned int b = min_base; b <= max_base; ++b) {
if (max_base - b + 1 + bases.size() < most_bases)
break;
uint64_t n = 0;
for (auto d : digits)
n = n * b + d;
if (sieve.is_prime(n))
bases.push_back(b);
}
if (bases.size() > most_bases) {
most_bases = bases.size();
max_strings.clear();
}
if (bases.size() == most_bases)
max_strings.emplace_back(digits, bases);
} while (increment(digits, max_base));
std::cout << most_bases << '\n';
for (const auto& m : max_strings) {
std::cout << to_string(m.first) << " -> ";
print(std::cout, m.second);
std::cout << '\n';
}
std::cout << '\n';
}
}
int main(int argc, char** argv) {
unsigned int max_base = 36;
unsigned int max_length = 4;
for (int arg = 1; arg + 1 < argc; ++arg) {
if (strcmp(argv[arg], "-max_base") == 0)
max_base = strtoul(argv[++arg], nullptr, 10);
else if (strcmp(argv[arg], "-max_length") == 0)
max_length = strtoul(argv[++arg], nullptr, 10);
}
if (max_base > 62) {
std::cerr << "Max base cannot be greater than 62.\n";
return EXIT_FAILURE;
}
if (max_base < 2) {
std::cerr << "Max base cannot be less than 2.\n";
return EXIT_FAILURE;
}
multi_base_primes(max_base, max_length);
return EXIT_SUCCESS;
} |
http://rosettacode.org/wiki/N-queens_problem | N-queens problem |
Solve the eight queens puzzle.
You can extend the problem to solve the puzzle with a board of size NxN.
For the number of solutions for small values of N, see OEIS: A000170.
Related tasks
A* search algorithm
Solve a Hidato puzzle
Solve a Holy Knight's tour
Knight's tour
Peaceful chess queen armies
Solve a Hopido puzzle
Solve a Numbrix puzzle
Solve the no connection puzzle
| #Curry | Curry |
-- 8-queens implementation with the Constrained Constructor pattern
-- Sergio Antoy
-- Fri Jul 13 07:05:32 PDT 2001
-- Place 8 queens on a chessboard so that no queen can capture
-- (and be captured by) any other queen.
-- Non-deterministic choice operator
infixl 0 !
X ! _ = X
_ ! Y = Y
-- A solution is represented by a list of integers.
-- The i-th integer in the list is the column of the board
-- in which the queen in the i-th row is placed.
-- Rows and columns are numbered from 1 to 8.
-- For example, [4,2,7,3,6,8,5,1] is a solution where the
-- the queen in row 1 is in column 4, etc.
-- Any solution must be a permutation of [1,2,...,8].
-- The state of a queen is its position, row and column, on the board.
-- Operation column is a particularly simple instance
-- of a Constrained Constructor pattern.
-- When it is invoked, it produces only valid states.
column = 1 ! 2 ! 3 ! 4 ! 5 ! 6 ! 7 ! 8
-- A path of the puzzle is a sequence of successive placements of
-- queens on the board. It is not explicitly defined as a type.
-- A path is a potential solution in the making.
-- Constrained Constructor on a path
-- Any path must be valid, i.e., any column must be in the range 1..8
-- and different from any other column in the path.
-- Furthermore, the path must be safe for the queens.
-- No queen in a path may capture any other queen in the path.
-- Operation makePath add column n to path c or fails.
makePath c n | valid c && safe c 1 = n:c
where valid c | n =:= column = uniq c
where uniq [] = True
uniq (c:cs) = n /= c && uniq cs
safe [] _ = True
safe (c:cs) k = abs (n-c) /= k && safe cs (k+1)
where abs x = if x < 0 then -x else x
-- extend the path argument till all the queens are on the board
-- see the Incremental Solution pattern
extend p = if (length p == 8)
then p
else extend (makePath p x)
where x free
-- solve the puzzle
main = extend []
|
http://rosettacode.org/wiki/Multiplicative_order | Multiplicative order | The multiplicative order of a relative to m is the least positive integer n such that a^n is 1 (modulo m).
Example
The multiplicative order of 37 relative to 1000 is 100 because 37^100 is 1 (modulo 1000), and no number smaller than 100 would do.
One possible algorithm that is efficient also for large numbers is the following: By the Chinese Remainder Theorem, it's enough to calculate the multiplicative order for each prime exponent p^k of m, and
combine the results with the least common multiple operation.
Now the order of a with regard to p^k must divide Φ(p^k). Call this number t, and determine it's factors q^e. Since each multiple of the order will also yield 1 when used as exponent for a, it's enough to find the least d such that (q^d)*(t/(q^e)) yields 1 when used as exponent.
Task
Implement a routine to calculate the multiplicative order along these lines. You may assume that routines to determine the factorization into prime powers are available in some library.
An algorithm for the multiplicative order can be found in Bach & Shallit, Algorithmic Number Theory, Volume I: Efficient Algorithms, The MIT Press, 1996:
Exercise 5.8, page 115:
Suppose you are given a prime p and a complete factorization
of p-1. Show how to compute the order of an
element a in (Z/(p))* using O((lg p)4/(lg lg p)) bit
operations.
Solution, page 337:
Let the prime factorization of p-1 be q1e1q2e2...qkek . We use the following observation:
if x^((p-1)/qifi) = 1 (mod p) ,
and fi=ei or x^((p-1)/qifi+1) != 1 (mod p) , then qiei-fi||ordp x. (This follows by combining Exercises 5.1 and 2.10.)
Hence it suffices to find, for each i , the exponent fi such that the condition above holds.
This can be done as follows: first compute q1e1, q2e2, ... ,
qkek . This can be done using O((lg p)2) bit operations. Next, compute y1=(p-1)/q1e1, ... , yk=(p-1)/qkek .
This can be done using O((lg p)2) bit operations. Now, using the binary method,
compute x1=ay1(mod p), ... , xk=ayk(mod p) .
This can be done using O(k(lg p)3) bit operations, and k=O((lg p)/(lg lg p)) by Theorem 8.8.10.
Finally, for each i , repeatedly raise xi to the qi-th power (mod p) (as many as ei-1 times), checking to see when 1 is obtained.
This can be done using O((lg p)3) steps.
The total cost is dominated by O(k(lg p)3) , which is O((lg p)4/(lg lg p)).
| #Factor | Factor | USING: kernel math math.functions math.primes.factors sequences ;
: (ord) ( a pair -- n )
first2 dupd ^ swap dupd [ /i ] keep 1 - * divisors
[ swap ^mod 1 = ] 2with find nip ;
: ord ( a n -- m )
2dup gcd nip 1 =
[ group-factors [ (ord) ] with [ lcm ] map-reduce ]
[ 2drop 0/0. ] if ; |
http://rosettacode.org/wiki/Multiplicative_order | Multiplicative order | The multiplicative order of a relative to m is the least positive integer n such that a^n is 1 (modulo m).
Example
The multiplicative order of 37 relative to 1000 is 100 because 37^100 is 1 (modulo 1000), and no number smaller than 100 would do.
One possible algorithm that is efficient also for large numbers is the following: By the Chinese Remainder Theorem, it's enough to calculate the multiplicative order for each prime exponent p^k of m, and
combine the results with the least common multiple operation.
Now the order of a with regard to p^k must divide Φ(p^k). Call this number t, and determine it's factors q^e. Since each multiple of the order will also yield 1 when used as exponent for a, it's enough to find the least d such that (q^d)*(t/(q^e)) yields 1 when used as exponent.
Task
Implement a routine to calculate the multiplicative order along these lines. You may assume that routines to determine the factorization into prime powers are available in some library.
An algorithm for the multiplicative order can be found in Bach & Shallit, Algorithmic Number Theory, Volume I: Efficient Algorithms, The MIT Press, 1996:
Exercise 5.8, page 115:
Suppose you are given a prime p and a complete factorization
of p-1. Show how to compute the order of an
element a in (Z/(p))* using O((lg p)4/(lg lg p)) bit
operations.
Solution, page 337:
Let the prime factorization of p-1 be q1e1q2e2...qkek . We use the following observation:
if x^((p-1)/qifi) = 1 (mod p) ,
and fi=ei or x^((p-1)/qifi+1) != 1 (mod p) , then qiei-fi||ordp x. (This follows by combining Exercises 5.1 and 2.10.)
Hence it suffices to find, for each i , the exponent fi such that the condition above holds.
This can be done as follows: first compute q1e1, q2e2, ... ,
qkek . This can be done using O((lg p)2) bit operations. Next, compute y1=(p-1)/q1e1, ... , yk=(p-1)/qkek .
This can be done using O((lg p)2) bit operations. Now, using the binary method,
compute x1=ay1(mod p), ... , xk=ayk(mod p) .
This can be done using O(k(lg p)3) bit operations, and k=O((lg p)/(lg lg p)) by Theorem 8.8.10.
Finally, for each i , repeatedly raise xi to the qi-th power (mod p) (as many as ei-1 times), checking to see when 1 is obtained.
This can be done using O((lg p)3) steps.
The total cost is dominated by O(k(lg p)3) , which is O((lg p)4/(lg lg p)).
| #Go | Go | package main
import (
"fmt"
"math/big"
)
func main() {
moTest(big.NewInt(37), big.NewInt(3343))
b := big.NewInt(100)
moTest(b.Add(b.Exp(ten, b, nil), one), big.NewInt(7919))
moTest(b.Add(b.Exp(ten, b.SetInt64(1000), nil), one), big.NewInt(15485863))
moTest(b.Sub(b.Exp(ten, b.SetInt64(10000), nil), one),
big.NewInt(22801763489))
moTest(big.NewInt(1511678068), big.NewInt(7379191741))
moTest(big.NewInt(3047753288), big.NewInt(2257683301))
}
func moTest(a, n *big.Int) {
if a.BitLen() < 100 {
fmt.Printf("ord(%v)", a)
} else {
fmt.Print("ord([big])")
}
if n.BitLen() < 100 {
fmt.Printf(" mod %v ", n)
} else {
fmt.Print(" mod [big] ")
}
if !n.ProbablyPrime(20) {
fmt.Println("not computed. modulus must be prime for this algorithm.")
return
}
fmt.Println("=", moBachShallit58(a, n, factor(new(big.Int).Sub(n, one))))
}
var one = big.NewInt(1)
var two = big.NewInt(2)
var ten = big.NewInt(10)
func moBachShallit58(a, n *big.Int, pf []pExp) *big.Int {
n1 := new(big.Int).Sub(n, one)
var x, y, o1, g big.Int
mo := big.NewInt(1)
for _, pe := range pf {
y.Quo(n1, y.Exp(pe.prime, big.NewInt(pe.exp), nil))
var o int64
for x.Exp(a, &y, n); x.Cmp(one) > 0; o++ {
x.Exp(&x, pe.prime, n)
}
o1.Exp(pe.prime, o1.SetInt64(o), nil)
mo.Mul(mo, o1.Quo(&o1, g.GCD(nil, nil, mo, &o1)))
}
return mo
}
type pExp struct {
prime *big.Int
exp int64
}
func factor(n *big.Int) (pf []pExp) {
var e int64
for ; n.Bit(int(e)) == 0; e++ {
}
if e > 0 {
n.Rsh(n, uint(e))
pf = []pExp{{big.NewInt(2), e}}
}
s := sqrt(n)
q, r := new(big.Int), new(big.Int)
for d := big.NewInt(3); n.Cmp(one) > 0; d.Add(d, two) {
if d.Cmp(s) > 0 {
d.Set(n)
}
for e = 0; ; e++ {
q.QuoRem(n, d, r)
if r.BitLen() > 0 {
break
}
n.Set(q)
}
if e > 0 {
pf = append(pf, pExp{new(big.Int).Set(d), e})
s = sqrt(n)
}
}
return
}
func sqrt(n *big.Int) *big.Int {
a := new(big.Int)
for b := new(big.Int).Set(n); ; {
a.Set(b)
b.Rsh(b.Add(b.Quo(n, a), a), 1)
if b.Cmp(a) >= 0 {
return a
}
}
return a.SetInt64(0)
} |
http://rosettacode.org/wiki/Nth_root | Nth root | Task
Implement the algorithm to compute the principal nth root
A
n
{\displaystyle {\sqrt[{n}]{A}}}
of a positive real number A, as explained at the Wikipedia page.
| #Phixmonti | Phixmonti | def nthroot
var n var y
1e-15 var eps /# relative accuracy #/
1 var x
true
while
y x n 1 - power / x - n / var d
x d + var x
eps x * var e /# absolute accuracy #/
d 0 e - < d e > or
endwhile
x
enddef
def printList
len for get print endfor
enddef
10 1024 3 27 2 2 125 5642 4 16 stklen tolist
len 1 swap 2 3 tolist
for
var i
i get swap i 1 + get rot var e var b
"The " e "th root of " b " is " b 1 e / power " (" b e nthroot ")" 9 tolist
printList drop nl
endfor |
http://rosettacode.org/wiki/Nth_root | Nth root | Task
Implement the algorithm to compute the principal nth root
A
n
{\displaystyle {\sqrt[{n}]{A}}}
of a positive real number A, as explained at the Wikipedia page.
| #PHP | PHP | function nthroot($number, $root, $p = P)
{
$x[0] = $number;
$x[1] = $number/$root;
while(abs($x[1]-$x[0]) > $p)
{
$x[0] = $x[1];
$x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root;
}
return $x[1];
} |
http://rosettacode.org/wiki/N%27th | N'th | Write a function/method/subroutine/... that when given an integer greater than or equal to zero returns a string of the number followed by an apostrophe then the ordinal suffix.
Example
Returns would include 1'st 2'nd 3'rd 11'th 111'th 1001'st 1012'th
Task
Use your routine to show here the output for at least the following (inclusive) ranges of integer inputs:
0..25, 250..265, 1000..1025
Note: apostrophes are now optional to allow correct apostrophe-less English.
| #Icon_and_Unicon | Icon and Unicon | procedure main(A)
every writes(" ",nth(0 to 25) | "\n")
every writes(" ",nth(250 to 265) | "\n")
every writes(" ",nth(1000 to 1025) | "\n")
end
procedure nth(n)
return n || ((n%10 = 1, n%100 ~= 11, "st") |
(n%10 = 2, n%100 ~= 12, "nd") |
(n%10 = 3, n%100 ~= 13, "rd") | "th")
end |
http://rosettacode.org/wiki/Non-decimal_radices/Convert | Non-decimal radices/Convert | Number base conversion is when you express a stored integer in an integer base, such as in octal (base 8) or binary (base 2). It also is involved when you take a string representing a number in a given base and convert it to the stored integer form. Normally, a stored integer is in binary, but that's typically invisible to the user, who normally enters or sees stored integers as decimal.
Task
Write a function (or identify the built-in function) which is passed a non-negative integer to convert, and another integer representing the base.
It should return a string containing the digits of the resulting number, without leading zeros except for the number 0 itself.
For the digits beyond 9, one should use the lowercase English alphabet, where the digit a = 9+1, b = a+1, etc.
For example: the decimal number 26 expressed in base 16 would be 1a.
Write a second function which is passed a string and an integer base, and it returns an integer representing that string interpreted in that base.
The programs may be limited by the word size or other such constraint of a given language. There is no need to do error checking for negatives, bases less than 2, or inappropriate digits.
| #Tcl | Tcl | namespace eval baseconvert {
variable chars "0123456789abcdefghijklmnopqrstuvwxyz"
namespace export baseconvert
}
proc baseconvert::dec2base {n b} {
variable chars
expr {$n == 0 ? 0
: "[string trimleft [dec2base [expr {$n/$b}] $b] 0][string index $chars [expr {$n%$b}]]"
}
}
proc baseconvert::base2dec {n b} {
variable chars
set sum 0
foreach char [split $n ""] {
set d [string first $char [string range $chars 0 [expr {$b - 1}]]]
if {$d == -1} {error "invalid base-$b digit '$char' in $n"}
set sum [expr {$sum * $b + $d}]
}
return $sum
}
proc baseconvert::baseconvert {n basefrom baseto} {
dec2base [base2dec $n $basefrom] $baseto
}
namespace import baseconvert::baseconvert
baseconvert 12345 10 23 ;# ==> 107h
baseconvert 107h 23 7 ;# ==> 50664
baseconvert 50664 7 10 ;# ==> 12345 |
http://rosettacode.org/wiki/Narcissistic_decimal_number | Narcissistic decimal number | A Narcissistic decimal number is a non-negative integer,
n
{\displaystyle n}
, that is equal to the sum of the
m
{\displaystyle m}
-th powers of each of the digits in the decimal representation of
n
{\displaystyle n}
, where
m
{\displaystyle m}
is the number of digits in the decimal representation of
n
{\displaystyle n}
.
Narcissistic (decimal) numbers are sometimes called Armstrong numbers, named after Michael F. Armstrong.
They are also known as Plus Perfect numbers.
An example
if
n
{\displaystyle n}
is 153
then
m
{\displaystyle m}
, (the number of decimal digits) is 3
we have 13 + 53 + 33 = 1 + 125 + 27 = 153
and so 153 is a narcissistic decimal number
Task
Generate and show here the first 25 narcissistic decimal numbers.
Note:
0
1
=
0
{\displaystyle 0^{1}=0}
, the first in the series.
See also
the OEIS entry: Armstrong (or Plus Perfect, or narcissistic) numbers.
MathWorld entry: Narcissistic Number.
Wikipedia entry: Narcissistic number.
| #Raku | Raku | sub is-narcissistic(Int $n) { $n == [+] $n.comb »**» $n.chars }
for 0 .. * {
if .&is-narcissistic {
.say;
last if ++state$ >= 25;
}
} |
http://rosettacode.org/wiki/Munching_squares | Munching squares | Render a graphical pattern where each pixel is colored by the value of 'x xor y' from an arbitrary color table.
| #zkl | zkl | fcn muncher{
bitmap:=PPM(256,256);
coolness:=(1).random(0x10000); // 55379, 18180, 40, 51950, 57619, 43514, 65465
foreach y,x in ([0 .. 255],[0 .. 255]){
b:=x.bitXor(y); // shades of blue
// rgb:=b*coolness; // kaleidoscopic image
// rgb:=(b*coolness + b)*coolness + b; // more coolness
rgb:=(b*0x10000 + b)*0x10000 + b; // copy ADA image
bitmap[x,y]=rgb;
}
bitmap.write(File("foo.ppm","wb"));
}(); |
http://rosettacode.org/wiki/Munchausen_numbers | Munchausen numbers | A Munchausen number is a natural number n the sum of whose digits (in base 10), each raised to the power of itself, equals n.
(Munchausen is also spelled: Münchhausen.)
For instance: 3435 = 33 + 44 + 33 + 55
Task
Find all Munchausen numbers between 1 and 5000.
Also see
The OEIS entry: A046253
The Wikipedia entry: Perfect digit-to-digit invariant, redirected from Munchausen Number
| #Go | Go | package main
import(
"fmt"
"math"
)
var powers [10]int
func isMunchausen(n int) bool {
if n < 0 { return false }
n64 := int64(n)
nn := n64
var sum int64 = 0
for nn > 0 {
sum += int64(powers[nn % 10])
if sum > n64 { return false }
nn /= 10
}
return sum == n64
}
func main() {
// cache n ^ n for n in 0..9, defining 0 ^ 0 = 0 for this purpose
for i := 1; i <= 9; i++ {
d := float64(i)
powers[i] = int(math.Pow(d, d))
}
// check numbers 0 to 500 million
fmt.Println("The Munchausen numbers between 0 and 500 million are:")
for i := 0; i <= 500000000; i++ {
if isMunchausen(i) { fmt.Printf("%d ", i) }
}
fmt.Println()
} |
http://rosettacode.org/wiki/Mutual_recursion | Mutual recursion | Two functions are said to be mutually recursive if the first calls the second,
and in turn the second calls the first.
Write two mutually recursive functions that compute members of the Hofstadter Female and Male sequences defined as:
F
(
0
)
=
1
;
M
(
0
)
=
0
F
(
n
)
=
n
−
M
(
F
(
n
−
1
)
)
,
n
>
0
M
(
n
)
=
n
−
F
(
M
(
n
−
1
)
)
,
n
>
0.
{\displaystyle {\begin{aligned}F(0)&=1\ ;\ M(0)=0\\F(n)&=n-M(F(n-1)),\quad n>0\\M(n)&=n-F(M(n-1)),\quad n>0.\end{aligned}}}
(If a language does not allow for a solution using mutually recursive functions
then state this rather than give a solution by other means).
| #Elena | Elena | import extensions;
import system'collections;
F = (n => (n == 0) ? 1 : (n - M(F(n-1))) );
M = (n => (n == 0) ? 0 : (n - F(M(n-1))) );
public program()
{
var ra := new ArrayList();
var rb := new ArrayList();
for(int i := 0, i <= 19, i += 1)
{
ra.append(F(i));
rb.append(M(i))
};
console.printLine(ra.asEnumerable());
console.printLine(rb.asEnumerable())
} |
http://rosettacode.org/wiki/Multisplit | Multisplit | It is often necessary to split a string into pieces
based on several different (potentially multi-character) separator strings,
while still retaining the information about which separators were present in the input.
This is particularly useful when doing small parsing tasks.
The task is to write code to demonstrate this.
The function (or procedure or method, as appropriate) should
take an input string and an ordered collection of separators.
The order of the separators is significant:
The delimiter order represents priority in matching, with the first defined delimiter having the highest priority.
In cases where there would be an ambiguity as to
which separator to use at a particular point
(e.g., because one separator is a prefix of another)
the separator with the highest priority should be used.
Delimiters can be reused and the output from the function should be an ordered sequence of substrings.
Test your code using the input string “a!===b=!=c” and the separators “==”, “!=” and “=”.
For these inputs the string should be parsed as "a" (!=) "" (==) "b" (=) "" (!=) "c", where matched delimiters are shown in parentheses, and separated strings are quoted, so our resulting output is "a", empty string, "b", empty string, "c".
Note that the quotation marks are shown for clarity and do not form part of the output.
Extra Credit: provide information that indicates which separator was matched at each separation point and where in the input string that separator was matched.
| #Run_BASIC | Run BASIC | str$ = "a!===b=!=c"
sep$ = "=== != =! b =!="
while word$(sep$,i+1," ") <> ""
i = i + 1
theSep$ = word$(sep$,i," ")
split$ = word$(str$,1,theSep$)
print i;" ";split$;" Sep By: ";theSep$
wend |
http://rosettacode.org/wiki/Multisplit | Multisplit | It is often necessary to split a string into pieces
based on several different (potentially multi-character) separator strings,
while still retaining the information about which separators were present in the input.
This is particularly useful when doing small parsing tasks.
The task is to write code to demonstrate this.
The function (or procedure or method, as appropriate) should
take an input string and an ordered collection of separators.
The order of the separators is significant:
The delimiter order represents priority in matching, with the first defined delimiter having the highest priority.
In cases where there would be an ambiguity as to
which separator to use at a particular point
(e.g., because one separator is a prefix of another)
the separator with the highest priority should be used.
Delimiters can be reused and the output from the function should be an ordered sequence of substrings.
Test your code using the input string “a!===b=!=c” and the separators “==”, “!=” and “=”.
For these inputs the string should be parsed as "a" (!=) "" (==) "b" (=) "" (!=) "c", where matched delimiters are shown in parentheses, and separated strings are quoted, so our resulting output is "a", empty string, "b", empty string, "c".
Note that the quotation marks are shown for clarity and do not form part of the output.
Extra Credit: provide information that indicates which separator was matched at each separation point and where in the input string that separator was matched.
| #Scala | Scala | import scala.annotation.tailrec
def multiSplit(str:String, sep:Seq[String])={
def findSep(index:Int)=sep find (str startsWith (_, index))
@tailrec def nextSep(index:Int):(Int,Int)=
if(index>str.size) (index, 0) else findSep(index) match {
case Some(sep) => (index, sep.size)
case _ => nextSep(index + 1)
}
def getParts(start:Int, pos:(Int,Int)):List[String]={
val part=str slice (start, pos._1)
if(pos._2==0) List(part) else part :: getParts(pos._1+pos._2, nextSep(pos._1+pos._2))
}
getParts(0, nextSep(0))
}
println(multiSplit("a!===b=!=c", Seq("!=", "==", "="))) |
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #Clojure | Clojure | user> (take 3 (repeat (rand))) ; repeating the same random number three times
(0.2787011365537204 0.2787011365537204 0.2787011365537204)
user> (take 3 (repeatedly rand)) ; creating three different random number
(0.8334795669220695 0.08405601245793926 0.5795448744634744)
user> |
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #Common_Lisp | Common Lisp | (make-list n :initial-element (make-the-distinct-thing))
(make-array n :initial-element (make-the-distinct-thing)) |
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #D | D | auto fooArray = new Foo[n];
foreach (ref item; fooArray)
item = new Foo();
|
http://rosettacode.org/wiki/Multifactorial | Multifactorial | The factorial of a number, written as
n
!
{\displaystyle n!}
, is defined as
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
.
Multifactorials generalize factorials as follows:
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
n
!
!
=
n
(
n
−
2
)
(
n
−
4
)
.
.
.
{\displaystyle n!!=n(n-2)(n-4)...}
n
!
!
!
=
n
(
n
−
3
)
(
n
−
6
)
.
.
.
{\displaystyle n!!!=n(n-3)(n-6)...}
n
!
!
!
!
=
n
(
n
−
4
)
(
n
−
8
)
.
.
.
{\displaystyle n!!!!=n(n-4)(n-8)...}
n
!
!
!
!
!
=
n
(
n
−
5
)
(
n
−
10
)
.
.
.
{\displaystyle n!!!!!=n(n-5)(n-10)...}
In all cases, the terms in the products are positive integers.
If we define the degree of the multifactorial as the difference in successive terms that are multiplied together for a multifactorial (the number of exclamation marks), then the task is twofold:
Write a function that given n and the degree, calculates the multifactorial.
Use the function to generate and display here a table of the first ten members (1 to 10) of the first five degrees of multifactorial.
Note: The wikipedia entry on multifactorials gives a different formula. This task uses the Wolfram mathworld definition.
| #Ada | Ada | with Ada.Text_IO; use Ada.Text_IO;
procedure Mfact is
function MultiFact (num : Natural; deg : Positive) return Natural is
Result, N : Integer := num;
begin
if N = 0 then return 1; end if;
loop
N := N - deg; exit when N <= 0; Result := Result * N;
end loop; return Result;
end MultiFact;
begin
for deg in 1..5 loop
Put("Degree"& Integer'Image(deg) &":");
for num in 1..10 loop Put(Integer'Image(MultiFact(num,deg))); end loop;
New_line;
end loop;
end Mfact; |
http://rosettacode.org/wiki/Multi-base_primes | Multi-base primes | Prime numbers are prime no matter what base they are represented in.
A prime number in a base other than 10 may not look prime at first glance.
For instance: 19 base 10 is 25 in base 7.
Several different prime numbers may be expressed as the "same" string when converted to a different base.
107 base 10 converted to base 6 == 255
173 base 10 converted to base 8 == 255
353 base 10 converted to base 12 == 255
467 base 10 converted to base 14 == 255
743 base 10 converted to base 18 == 255
1277 base 10 converted to base 24 == 255
1487 base 10 converted to base 26 == 255
2213 base 10 converted to base 32 == 255
Task
Restricted to bases 2 through 36; find the strings that have the most different bases that evaluate to that string when converting prime numbers to a base.
Find the conversion string, the amount of bases that evaluate a prime to that string and the enumeration of bases that evaluate a prime to that string.
Display here, on this page, the string, the count and the list for all of the: 1 character, 2 character, 3 character, and 4 character strings that have the maximum base count that evaluate to that string.
Should be no surprise, the string '2' has the largest base count for single character strings.
Stretch goal
Do the same for the maximum 5 character string.
| #F.23 | F# |
// Multi-base primes. Nigel Galloway: July 4th., 2021
let digits="0123456789abcdefghijklmnopqrstuvwxyz"
let fG n g=let rec fN g=function i when i<n->i::g |i->fN((i%n)::g)(i/n) in primes32()|>Seq.skipWhile((>)(pown n (g-1)))|>Seq.takeWhile((>)(pown n g))|>Seq.map(fun g->(n,fN [] g))
let fN g={2..36}|>Seq.collect(fun n->fG n g)|>Seq.groupBy snd|>Seq.groupBy(snd>>(Seq.length))|>Seq.maxBy fst
{1..4}|>Seq.iter(fun g->let n,i=fN g in printfn "The following strings of length %d represent primes in the maximum number of bases(%d):" g n
i|>Seq.iter(fun(n,g)->printf " %s->" (n|>List.map(fun n->digits.[n])|>Array.ofList|>System.String)
g|>Seq.iter(fun(g,_)->printf "%d " g); printfn ""); printfn "")
|
http://rosettacode.org/wiki/N-queens_problem | N-queens problem |
Solve the eight queens puzzle.
You can extend the problem to solve the puzzle with a board of size NxN.
For the number of solutions for small values of N, see OEIS: A000170.
Related tasks
A* search algorithm
Solve a Hidato puzzle
Solve a Holy Knight's tour
Knight's tour
Peaceful chess queen armies
Solve a Hopido puzzle
Solve a Numbrix puzzle
Solve the no connection puzzle
| #D | D | void main() {
import std.stdio, std.algorithm, std.range, permutations2;
enum n = 8;
n.iota.array.permutations.filter!(p =>
n.iota.map!(i => p[i] + i).array.sort().uniq.count == n &&
n.iota.map!(i => p[i] - i).array.sort().uniq.count == n)
.count.writeln;
} |
http://rosettacode.org/wiki/Multiplicative_order | Multiplicative order | The multiplicative order of a relative to m is the least positive integer n such that a^n is 1 (modulo m).
Example
The multiplicative order of 37 relative to 1000 is 100 because 37^100 is 1 (modulo 1000), and no number smaller than 100 would do.
One possible algorithm that is efficient also for large numbers is the following: By the Chinese Remainder Theorem, it's enough to calculate the multiplicative order for each prime exponent p^k of m, and
combine the results with the least common multiple operation.
Now the order of a with regard to p^k must divide Φ(p^k). Call this number t, and determine it's factors q^e. Since each multiple of the order will also yield 1 when used as exponent for a, it's enough to find the least d such that (q^d)*(t/(q^e)) yields 1 when used as exponent.
Task
Implement a routine to calculate the multiplicative order along these lines. You may assume that routines to determine the factorization into prime powers are available in some library.
An algorithm for the multiplicative order can be found in Bach & Shallit, Algorithmic Number Theory, Volume I: Efficient Algorithms, The MIT Press, 1996:
Exercise 5.8, page 115:
Suppose you are given a prime p and a complete factorization
of p-1. Show how to compute the order of an
element a in (Z/(p))* using O((lg p)4/(lg lg p)) bit
operations.
Solution, page 337:
Let the prime factorization of p-1 be q1e1q2e2...qkek . We use the following observation:
if x^((p-1)/qifi) = 1 (mod p) ,
and fi=ei or x^((p-1)/qifi+1) != 1 (mod p) , then qiei-fi||ordp x. (This follows by combining Exercises 5.1 and 2.10.)
Hence it suffices to find, for each i , the exponent fi such that the condition above holds.
This can be done as follows: first compute q1e1, q2e2, ... ,
qkek . This can be done using O((lg p)2) bit operations. Next, compute y1=(p-1)/q1e1, ... , yk=(p-1)/qkek .
This can be done using O((lg p)2) bit operations. Now, using the binary method,
compute x1=ay1(mod p), ... , xk=ayk(mod p) .
This can be done using O(k(lg p)3) bit operations, and k=O((lg p)/(lg lg p)) by Theorem 8.8.10.
Finally, for each i , repeatedly raise xi to the qi-th power (mod p) (as many as ei-1 times), checking to see when 1 is obtained.
This can be done using O((lg p)3) steps.
The total cost is dominated by O(k(lg p)3) , which is O((lg p)4/(lg lg p)).
| #Haskell | Haskell | powerMod
:: (Integral a, Integral b)
=> a -> a -> b -> a
powerMod m _ 0 = 1
powerMod m x n
| n > 0 = f x_ (n - 1) x_
where
x_ = x `rem` m
f _ 0 y = y
f a d y = g a d
where
g b i
| even i = g (b * b `rem` m) (i `quot` 2)
| otherwise = f b (i - 1) (b * y `rem` m)
powerMod m _ _ = error "powerMod: negative exponent" |
http://rosettacode.org/wiki/Nth_root | Nth root | Task
Implement the algorithm to compute the principal nth root
A
n
{\displaystyle {\sqrt[{n}]{A}}}
of a positive real number A, as explained at the Wikipedia page.
| #Picat | Picat | go =>
L = [[2,2],
[34,5],
[34**5,5],
[7131.5**10],
[7,0.5],
[1024,10],
[5642, 125]
],
foreach([A,N] in L)
R = nthroot(A,N),
printf("nthroot(%8w,%8w) %20w (check: %w)\n",A,N,R,A**(1/N))
end,
nl.
%
% x^n = a
%
% Given a and n, find x (to Precision)
%
nthroot(A,N) = nthroot(A,N,0.000001).
nthroot(A,N,Precision) = X1 =>
NF = N * 1.0, % float version of N
X0 = A / NF,
X1 = 1.0,
do
X0 := X1,
X1 := (1.0 / NF)*((NF - 1.0)*X0 + (A / (X0 ** (NF - 1))))
while( abs(X0-X1) > Precision). |
http://rosettacode.org/wiki/N%27th | N'th | Write a function/method/subroutine/... that when given an integer greater than or equal to zero returns a string of the number followed by an apostrophe then the ordinal suffix.
Example
Returns would include 1'st 2'nd 3'rd 11'th 111'th 1001'st 1012'th
Task
Use your routine to show here the output for at least the following (inclusive) ranges of integer inputs:
0..25, 250..265, 1000..1025
Note: apostrophes are now optional to allow correct apostrophe-less English.
| #J | J | suf=: (;:'th st nd rd th') {::~ 4 <. 10 10 (* 1&~:)~/@#: ]
nth=: [: ;:inv (": , suf)each |
http://rosettacode.org/wiki/Non-decimal_radices/Convert | Non-decimal radices/Convert | Number base conversion is when you express a stored integer in an integer base, such as in octal (base 8) or binary (base 2). It also is involved when you take a string representing a number in a given base and convert it to the stored integer form. Normally, a stored integer is in binary, but that's typically invisible to the user, who normally enters or sees stored integers as decimal.
Task
Write a function (or identify the built-in function) which is passed a non-negative integer to convert, and another integer representing the base.
It should return a string containing the digits of the resulting number, without leading zeros except for the number 0 itself.
For the digits beyond 9, one should use the lowercase English alphabet, where the digit a = 9+1, b = a+1, etc.
For example: the decimal number 26 expressed in base 16 would be 1a.
Write a second function which is passed a string and an integer base, and it returns an integer representing that string interpreted in that base.
The programs may be limited by the word size or other such constraint of a given language. There is no need to do error checking for negatives, bases less than 2, or inappropriate digits.
| #Ursala | Ursala | #import std
#import nat
num_to_string "b" = ||'0'! (-: num digits--letters)*+ @NiX ~&r->l ^|rrPlCrlPX/~& division\"b"
string_to_num "b" = @x =>0 sum^|/(-:@rlXS num digits--letters) product/"b" |
http://rosettacode.org/wiki/Non-decimal_radices/Convert | Non-decimal radices/Convert | Number base conversion is when you express a stored integer in an integer base, such as in octal (base 8) or binary (base 2). It also is involved when you take a string representing a number in a given base and convert it to the stored integer form. Normally, a stored integer is in binary, but that's typically invisible to the user, who normally enters or sees stored integers as decimal.
Task
Write a function (or identify the built-in function) which is passed a non-negative integer to convert, and another integer representing the base.
It should return a string containing the digits of the resulting number, without leading zeros except for the number 0 itself.
For the digits beyond 9, one should use the lowercase English alphabet, where the digit a = 9+1, b = a+1, etc.
For example: the decimal number 26 expressed in base 16 would be 1a.
Write a second function which is passed a string and an integer base, and it returns an integer representing that string interpreted in that base.
The programs may be limited by the word size or other such constraint of a given language. There is no need to do error checking for negatives, bases less than 2, or inappropriate digits.
| #VBA | VBA | Private Function to_base(ByVal number As Long, base As Integer) As String
Dim digits As String, result As String
Dim i As Integer, digit As Integer
digits = "0123456789abcdefghijklmnopqrstuvwxyz"
Do While number > 0
digit = number Mod base
result = Mid(digits, digit + 1, 1) & result
number = number \ base
Loop
to_base = result
End Function
Private Function from_base(number As String, base As Integer) As Long
Dim digits As String, result As Long
Dim i As Integer
digits = "0123456789abcdefghijklmnopqrstuvwxyz"
result = Val(InStr(1, digits, Mid(number, 1, 1), vbTextCompare) - 1)
For i = 2 To Len(number)
result = result * base + Val(InStr(1, digits, Mid(number, i, 1), vbTextCompare) - 1)
Next i
from_base = result
End Function
Public Sub Non_decimal_radices_Convert()
Debug.Print "26 decimal in base 16 is: "; to_base(26, 16); ". Conversely, hexadecimal 1a in decimal is: "; from_base("1a", 16)
End Sub |
http://rosettacode.org/wiki/Narcissistic_decimal_number | Narcissistic decimal number | A Narcissistic decimal number is a non-negative integer,
n
{\displaystyle n}
, that is equal to the sum of the
m
{\displaystyle m}
-th powers of each of the digits in the decimal representation of
n
{\displaystyle n}
, where
m
{\displaystyle m}
is the number of digits in the decimal representation of
n
{\displaystyle n}
.
Narcissistic (decimal) numbers are sometimes called Armstrong numbers, named after Michael F. Armstrong.
They are also known as Plus Perfect numbers.
An example
if
n
{\displaystyle n}
is 153
then
m
{\displaystyle m}
, (the number of decimal digits) is 3
we have 13 + 53 + 33 = 1 + 125 + 27 = 153
and so 153 is a narcissistic decimal number
Task
Generate and show here the first 25 narcissistic decimal numbers.
Note:
0
1
=
0
{\displaystyle 0^{1}=0}
, the first in the series.
See also
the OEIS entry: Armstrong (or Plus Perfect, or narcissistic) numbers.
MathWorld entry: Narcissistic Number.
Wikipedia entry: Narcissistic number.
| #REXX | REXX | /*REXX program generates and displays a number of narcissistic (Armstrong) numbers. */
numeric digits 39 /*be able to handle largest Armstrong #*/
parse arg N . /*obtain optional argument from the CL.*/
if N=='' | N=="," then N=25 /*Not specified? Then use the default.*/
N=min(N, 89) /*there are only 89 narcissistic #s. */
#=0 /*number of narcissistic numbers so far*/
do j=0 until #==N; L=length(j) /*get length of the J decimal number.*/
$=left(j, 1) **L /*1st digit in J raised to the L pow.*/
do k=2 for L-1 until $>j /*perform for each decimal digit in J.*/
$=$ + substr(j, k, 1) ** L /*add digit raised to power to the sum.*/
end /*k*/ /* [↑] calculate the rest of the sum. */
if $\==j then iterate /*does the sum equal to J? No, skip it*/
#=# + 1 /*bump count of narcissistic numbers. */
say right(#, 9) ' narcissistic:' j /*display index and narcissistic number*/
end /*j*/ /*stick a fork in it, we're all done. */ |
http://rosettacode.org/wiki/Munchausen_numbers | Munchausen numbers | A Munchausen number is a natural number n the sum of whose digits (in base 10), each raised to the power of itself, equals n.
(Munchausen is also spelled: Münchhausen.)
For instance: 3435 = 33 + 44 + 33 + 55
Task
Find all Munchausen numbers between 1 and 5000.
Also see
The OEIS entry: A046253
The Wikipedia entry: Perfect digit-to-digit invariant, redirected from Munchausen Number
| #Haskell | Haskell | import Control.Monad (join)
import Data.List (unfoldr)
isMunchausen :: Integer -> Bool
isMunchausen =
(==)
<*> (sum . map (join (^)) . unfoldr digit)
digit 0 = Nothing
digit n = Just (r, q) where (q, r) = n `divMod` 10
main :: IO ()
main = print $ filter isMunchausen [1 .. 5000] |
http://rosettacode.org/wiki/Mutual_recursion | Mutual recursion | Two functions are said to be mutually recursive if the first calls the second,
and in turn the second calls the first.
Write two mutually recursive functions that compute members of the Hofstadter Female and Male sequences defined as:
F
(
0
)
=
1
;
M
(
0
)
=
0
F
(
n
)
=
n
−
M
(
F
(
n
−
1
)
)
,
n
>
0
M
(
n
)
=
n
−
F
(
M
(
n
−
1
)
)
,
n
>
0.
{\displaystyle {\begin{aligned}F(0)&=1\ ;\ M(0)=0\\F(n)&=n-M(F(n-1)),\quad n>0\\M(n)&=n-F(M(n-1)),\quad n>0.\end{aligned}}}
(If a language does not allow for a solution using mutually recursive functions
then state this rather than give a solution by other means).
| #Elixir | Elixir | defmodule MutualRecursion do
def f(0), do: 1
def f(n), do: n - m(f(n - 1))
def m(0), do: 0
def m(n), do: n - f(m(n - 1))
end
IO.inspect Enum.map(0..19, fn n -> MutualRecursion.f(n) end)
IO.inspect Enum.map(0..19, fn n -> MutualRecursion.m(n) end) |
http://rosettacode.org/wiki/Multisplit | Multisplit | It is often necessary to split a string into pieces
based on several different (potentially multi-character) separator strings,
while still retaining the information about which separators were present in the input.
This is particularly useful when doing small parsing tasks.
The task is to write code to demonstrate this.
The function (or procedure or method, as appropriate) should
take an input string and an ordered collection of separators.
The order of the separators is significant:
The delimiter order represents priority in matching, with the first defined delimiter having the highest priority.
In cases where there would be an ambiguity as to
which separator to use at a particular point
(e.g., because one separator is a prefix of another)
the separator with the highest priority should be used.
Delimiters can be reused and the output from the function should be an ordered sequence of substrings.
Test your code using the input string “a!===b=!=c” and the separators “==”, “!=” and “=”.
For these inputs the string should be parsed as "a" (!=) "" (==) "b" (=) "" (!=) "c", where matched delimiters are shown in parentheses, and separated strings are quoted, so our resulting output is "a", empty string, "b", empty string, "c".
Note that the quotation marks are shown for clarity and do not form part of the output.
Extra Credit: provide information that indicates which separator was matched at each separation point and where in the input string that separator was matched.
| #Scheme | Scheme | (use srfi-13)
(use srfi-42)
(define (shatter separators the-string)
(let loop ((str the-string) (tmp ""))
(if (string=? "" str)
(list tmp)
(if-let1 sep (find (^s (string-prefix? s str)) separators)
(cons* tmp sep
(loop (string-drop str (string-length sep)) ""))
(loop (string-drop str 1) (string-append tmp (string-take str 1)))))))
(define (glean shards)
(list-ec (: x (index i) shards)
(if (even? i)) x)) |
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #Delphi | Delphi | var
i: Integer;
lObject: TMyObject;
lList: TObjectList<TMyObject>;
begin
lList := TObjectList<TMyObject>.Create;
lObject := TMyObject.Create;
for i := 1 to 10 do
lList.Add(lObject);
// ... |
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #E | E | pragma.enable("accumulator")
...
accum [] for _ in 1..n { _.with(makeWhatever()) } |
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #EchoLisp | EchoLisp |
;; wrong - make-vector is evaluated one time - same vector
(define L (make-list 3 (make-vector 4)))
L → (#(0 0 0 0) #(0 0 0 0) #(0 0 0 0))
(vector-set! (first L ) 1 '🔴) ;; sets the 'first' vector
L → (#(0 🔴 0 0) #(0 🔴 0 0) #(0 🔴 0 0))
;; right - three different vectors
(define L(map make-vector (make-list 3 4)))
L → (#(0 0 0 0) #(0 0 0 0) #(0 0 0 0))
(vector-set! (first L ) 1 '🔵) ;; sets the first vector
L → (#(0 🔵 0 0) #(0 0 0 0) #(0 0 0 0)) ;; OK
|
http://rosettacode.org/wiki/Move-to-front_algorithm | Move-to-front algorithm | Given a symbol table of a zero-indexed array of all possible input symbols
this algorithm reversibly transforms a sequence
of input symbols into an array of output numbers (indices).
The transform in many cases acts to give frequently repeated input symbols
lower indices which is useful in some compression algorithms.
Encoding algorithm
for each symbol of the input sequence:
output the index of the symbol in the symbol table
move that symbol to the front of the symbol table
Decoding algorithm
# Using the same starting symbol table
for each index of the input sequence:
output the symbol at that index of the symbol table
move that symbol to the front of the symbol table
Example
Encoding the string of character symbols 'broood' using a symbol table of the lowercase characters a-to-z
Input
Output
SymbolTable
broood
1
'abcdefghijklmnopqrstuvwxyz'
broood
1 17
'bacdefghijklmnopqrstuvwxyz'
broood
1 17 15
'rbacdefghijklmnopqstuvwxyz'
broood
1 17 15 0
'orbacdefghijklmnpqstuvwxyz'
broood
1 17 15 0 0
'orbacdefghijklmnpqstuvwxyz'
broood
1 17 15 0 0 5
'orbacdefghijklmnpqstuvwxyz'
Decoding the indices back to the original symbol order:
Input
Output
SymbolTable
1 17 15 0 0 5
b
'abcdefghijklmnopqrstuvwxyz'
1 17 15 0 0 5
br
'bacdefghijklmnopqrstuvwxyz'
1 17 15 0 0 5
bro
'rbacdefghijklmnopqstuvwxyz'
1 17 15 0 0 5
broo
'orbacdefghijklmnpqstuvwxyz'
1 17 15 0 0 5
brooo
'orbacdefghijklmnpqstuvwxyz'
1 17 15 0 0 5
broood
'orbacdefghijklmnpqstuvwxyz'
Task
Encode and decode the following three strings of characters using the symbol table of the lowercase characters a-to-z as above.
Show the strings and their encoding here.
Add a check to ensure that the decoded string is the same as the original.
The strings are:
broood
bananaaa
hiphophiphop
(Note the misspellings in the above strings.)
| #11l | 11l | V symboltable = Array(‘a’..‘z’)
F move2front_encode(strng)
[Int] sequence
V pad = copy(:symboltable)
L(char) strng
V indx = pad.index(char)
sequence.append(indx)
pad = [pad.pop(indx)] [+] pad
R sequence
F move2front_decode(sequence)
[Char] chars
V pad = copy(:symboltable)
L(indx) sequence
V char = pad[indx]
chars.append(char)
pad = [pad.pop(indx)] [+] pad
R chars.join(‘’)
L(s) [‘broood’, ‘bananaaa’, ‘hiphophiphop’]
V encode = move2front_encode(s)
print(‘#14 encodes to #.’.format(s, encode), end' ‘, ’)
V decode = move2front_decode(encode)
print(‘which decodes back to #.’.format(decode))
assert(s == decode, ‘Whoops!’) |
http://rosettacode.org/wiki/Multi-dimensional_array | Multi-dimensional array | For the purposes of this task, the actual memory layout or access method of this data structure is not mandated.
It is enough to:
State the number and extent of each index to the array.
Provide specific, ordered, integer indices for all dimensions of the array together with a new value to update the indexed value.
Provide specific, ordered, numeric indices for all dimensions of the array to obtain the arrays value at that indexed position.
Task
State if the language supports multi-dimensional arrays in its syntax and usual implementation.
State whether the language uses row-major or column major order for multi-dimensional array storage, or any other relevant kind of storage.
Show how to create a four dimensional array in your language and set, access, set to another value; and access the new value of an integer-indexed item of the array.
The idiomatic method for the language is preferred.
The array should allow a range of five, four, three and two (or two three four five if convenient), in each of the indices, in order. (For example, if indexing starts at zero for the first index then a range of 0..4 inclusive would suffice).
State if memory allocation is optimised for the array - especially if contiguous memory is likely to be allocated.
If the language has exceptional native multi-dimensional array support such as optional bounds checking, reshaping, or being able to state both the lower and upper bounds of index ranges, then this is the task to mention them.
Show all output here, (but you may judiciously use ellipses to shorten repetitive output text).
| #11l | 11l | V arr = [
[
[1,2,3],
[4,5,6]
],
[
[11,22,33],
[44,55,66]
],
[
[111,222,333],
[444,555,666]
],
[
[1111,2222,3333],
[4444,5555,6666]
]
] |
http://rosettacode.org/wiki/Multiple_regression | Multiple regression | Task
Given a set of data vectors in the following format:
y
=
{
y
1
,
y
2
,
.
.
.
,
y
n
}
{\displaystyle y=\{y_{1},y_{2},...,y_{n}\}\,}
X
i
=
{
x
i
1
,
x
i
2
,
.
.
.
,
x
i
n
}
,
i
∈
1..
k
{\displaystyle X_{i}=\{x_{i1},x_{i2},...,x_{in}\},i\in 1..k\,}
Compute the vector
β
=
{
β
1
,
β
2
,
.
.
.
,
β
k
}
{\displaystyle \beta =\{\beta _{1},\beta _{2},...,\beta _{k}\}}
using ordinary least squares regression using the following equation:
y
j
=
Σ
i
β
i
⋅
x
i
j
,
j
∈
1..
n
{\displaystyle y_{j}=\Sigma _{i}\beta _{i}\cdot x_{ij},j\in 1..n}
You can assume y is given to you as a vector (a one-dimensional array), and X is given to you as a two-dimensional array (i.e. matrix).
| #Ada | Ada | generic
type Element_Type is private;
Zero : Element_Type;
One : Element_Type;
with function "+" (Left, Right : Element_Type) return Element_Type is <>;
with function "-" (Left, Right : Element_Type) return Element_Type is <>;
with function "*" (Left, Right : Element_Type) return Element_Type is <>;
with function "/" (Left, Right : Element_Type) return Element_Type is <>;
package Matrices is
type Vector is array (Positive range <>) of Element_Type;
type Matrix is
array (Positive range <>, Positive range <>) of Element_Type;
function "*" (Left, Right : Matrix) return Matrix;
function Invert (Source : Matrix) return Matrix;
function Reduced_Row_Echelon_Form (Source : Matrix) return Matrix;
function Regression_Coefficients
(Source : Vector;
Regressors : Matrix)
return Vector;
function To_Column_Vector
(Source : Matrix;
Row : Positive := 1)
return Vector;
function To_Matrix
(Source : Vector;
Column_Vector : Boolean := True)
return Matrix;
function To_Row_Vector
(Source : Matrix;
Column : Positive := 1)
return Vector;
function Transpose (Source : Matrix) return Matrix;
Size_Mismatch : exception;
Not_Square_Matrix : exception;
Not_Invertible : exception;
end Matrices; |
http://rosettacode.org/wiki/Multifactorial | Multifactorial | The factorial of a number, written as
n
!
{\displaystyle n!}
, is defined as
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
.
Multifactorials generalize factorials as follows:
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
n
!
!
=
n
(
n
−
2
)
(
n
−
4
)
.
.
.
{\displaystyle n!!=n(n-2)(n-4)...}
n
!
!
!
=
n
(
n
−
3
)
(
n
−
6
)
.
.
.
{\displaystyle n!!!=n(n-3)(n-6)...}
n
!
!
!
!
=
n
(
n
−
4
)
(
n
−
8
)
.
.
.
{\displaystyle n!!!!=n(n-4)(n-8)...}
n
!
!
!
!
!
=
n
(
n
−
5
)
(
n
−
10
)
.
.
.
{\displaystyle n!!!!!=n(n-5)(n-10)...}
In all cases, the terms in the products are positive integers.
If we define the degree of the multifactorial as the difference in successive terms that are multiplied together for a multifactorial (the number of exclamation marks), then the task is twofold:
Write a function that given n and the degree, calculates the multifactorial.
Use the function to generate and display here a table of the first ten members (1 to 10) of the first five degrees of multifactorial.
Note: The wikipedia entry on multifactorials gives a different formula. This task uses the Wolfram mathworld definition.
| #Aime | Aime | mf(integer a, n)
{
integer o;
o = 1;
do {
o *= a;
} while (0 < (a -= n));
o;
}
main(void)
{
integer i, j;
i = 0;
while ((i += 1) <= 5) {
o_("degree ", i, ":");
j = 0;
while ((j += 1) <= 10) {
o_("\t", mf(j, i));
}
o_("\n");
}
0;
} |
http://rosettacode.org/wiki/Multi-base_primes | Multi-base primes | Prime numbers are prime no matter what base they are represented in.
A prime number in a base other than 10 may not look prime at first glance.
For instance: 19 base 10 is 25 in base 7.
Several different prime numbers may be expressed as the "same" string when converted to a different base.
107 base 10 converted to base 6 == 255
173 base 10 converted to base 8 == 255
353 base 10 converted to base 12 == 255
467 base 10 converted to base 14 == 255
743 base 10 converted to base 18 == 255
1277 base 10 converted to base 24 == 255
1487 base 10 converted to base 26 == 255
2213 base 10 converted to base 32 == 255
Task
Restricted to bases 2 through 36; find the strings that have the most different bases that evaluate to that string when converting prime numbers to a base.
Find the conversion string, the amount of bases that evaluate a prime to that string and the enumeration of bases that evaluate a prime to that string.
Display here, on this page, the string, the count and the list for all of the: 1 character, 2 character, 3 character, and 4 character strings that have the maximum base count that evaluate to that string.
Should be no surprise, the string '2' has the largest base count for single character strings.
Stretch goal
Do the same for the maximum 5 character string.
| #Factor | Factor | USING: assocs assocs.extras formatting io kernel math
math.functions math.parser math.primes math.ranges present
sequences ;
: prime?* ( n -- ? ) [ prime? ] [ f ] if* ; inline
: (bases) ( n -- range quot )
present 2 36 [a,b] [ base> prime?* ] with ; inline
: <digits> ( n -- range ) [ 1 - ] keep [ 10^ ] bi@ [a,b) ;
: multibase ( n -- assoc )
<digits> [ (bases) count ] zip-with assoc-invert
expand-keys-push-at >alist [ first ] supremum-by ;
: multibase. ( n -- )
dup multibase first2
[ "%d-digit numbers that are prime in the most bases: %d\n" printf ] dip
[ dup (bases) filter "%d => %[%d, %]\n" printf ] each ;
4 [1,b] [ multibase. nl ] each |
http://rosettacode.org/wiki/Multi-base_primes | Multi-base primes | Prime numbers are prime no matter what base they are represented in.
A prime number in a base other than 10 may not look prime at first glance.
For instance: 19 base 10 is 25 in base 7.
Several different prime numbers may be expressed as the "same" string when converted to a different base.
107 base 10 converted to base 6 == 255
173 base 10 converted to base 8 == 255
353 base 10 converted to base 12 == 255
467 base 10 converted to base 14 == 255
743 base 10 converted to base 18 == 255
1277 base 10 converted to base 24 == 255
1487 base 10 converted to base 26 == 255
2213 base 10 converted to base 32 == 255
Task
Restricted to bases 2 through 36; find the strings that have the most different bases that evaluate to that string when converting prime numbers to a base.
Find the conversion string, the amount of bases that evaluate a prime to that string and the enumeration of bases that evaluate a prime to that string.
Display here, on this page, the string, the count and the list for all of the: 1 character, 2 character, 3 character, and 4 character strings that have the maximum base count that evaluate to that string.
Should be no surprise, the string '2' has the largest base count for single character strings.
Stretch goal
Do the same for the maximum 5 character string.
| #Go | Go | package main
import (
"fmt"
"math"
"rcu"
)
var maxDepth = 6
var maxBase = 36
var c = rcu.PrimeSieve(int(math.Pow(float64(maxBase), float64(maxDepth))), true)
var digits = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"
var maxStrings [][][]int
var mostBases = -1
func maxSlice(a []int) int {
max := 0
for _, e := range a {
if e > max {
max = e
}
}
return max
}
func maxInt(a, b int) int {
if a > b {
return a
}
return b
}
func process(indices []int) {
minBase := maxInt(2, maxSlice(indices)+1)
if maxBase - minBase + 1 < mostBases {
return // can't affect results so return
}
var bases []int
for b := minBase; b <= maxBase; b++ {
n := 0
for _, i := range indices {
n = n*b + i
}
if !c[n] {
bases = append(bases, b)
}
}
count := len(bases)
if count > mostBases {
mostBases = count
indices2 := make([]int, len(indices))
copy(indices2, indices)
maxStrings = [][][]int{[][]int{indices2, bases}}
} else if count == mostBases {
indices2 := make([]int, len(indices))
copy(indices2, indices)
maxStrings = append(maxStrings, [][]int{indices2, bases})
}
}
func printResults() {
fmt.Printf("%d\n", len(maxStrings[0][1]))
for _, m := range maxStrings {
s := ""
for _, i := range m[0] {
s = s + string(digits[i])
}
fmt.Printf("%s -> %v\n", s, m[1])
}
}
func nestedFor(indices []int, length, level int) {
if level == len(indices) {
process(indices)
} else {
indices[level] = 0
if level == 0 {
indices[level] = 1
}
for indices[level] < length {
nestedFor(indices, length, level+1)
indices[level]++
}
}
}
func main() {
for depth := 1; depth <= maxDepth; depth++ {
fmt.Print(depth, " character strings which are prime in most bases: ")
maxStrings = maxStrings[:0]
mostBases = -1
indices := make([]int, depth)
nestedFor(indices, maxBase, 0)
printResults()
fmt.Println()
}
} |
http://rosettacode.org/wiki/N-queens_problem | N-queens problem |
Solve the eight queens puzzle.
You can extend the problem to solve the puzzle with a board of size NxN.
For the number of solutions for small values of N, see OEIS: A000170.
Related tasks
A* search algorithm
Solve a Hidato puzzle
Solve a Holy Knight's tour
Knight's tour
Peaceful chess queen armies
Solve a Hopido puzzle
Solve a Numbrix puzzle
Solve the no connection puzzle
| #Dart | Dart | /**
Return true if queen placement q[n] does not conflict with
other queens q[0] through q[n-1]
*/
isConsistent(List q, int n) {
for (int i=0; i<n; i++) {
if (q[i] == q[n]) {
return false; // Same column
}
if ((q[i] - q[n]) == (n - i)) {
return false; // Same major diagonal
}
if ((q[n] - q[i]) == (n - i)) {
return false; // Same minor diagonal
}
}
return true;
}
/**
Print out N-by-N placement of queens from permutation q in ASCII.
*/
printQueens(List q) {
int N = q.length;
for (int i=0; i<N; i++) {
StringBuffer sb = new StringBuffer();
for (int j=0; j<N; j++) {
if (q[i] == j) {
sb.write("Q ");
} else {
sb.write("* ");
}
}
print(sb.toString());
}
print("");
}
/**
Try all permutations using backtracking
*/
enumerate(int N) {
var a = new List(N);
_enumerate(a, 0);
}
_enumerate(List q, int n) {
if (n == q.length) {
printQueens(q);
} else {
for (int i = 0; i < q.length; i++) {
q[n] = i;
if (isConsistent(q, n)){
_enumerate(q, n+1);
}
}
}
}
void main() {
enumerate(4);
} |
http://rosettacode.org/wiki/Multiplicative_order | Multiplicative order | The multiplicative order of a relative to m is the least positive integer n such that a^n is 1 (modulo m).
Example
The multiplicative order of 37 relative to 1000 is 100 because 37^100 is 1 (modulo 1000), and no number smaller than 100 would do.
One possible algorithm that is efficient also for large numbers is the following: By the Chinese Remainder Theorem, it's enough to calculate the multiplicative order for each prime exponent p^k of m, and
combine the results with the least common multiple operation.
Now the order of a with regard to p^k must divide Φ(p^k). Call this number t, and determine it's factors q^e. Since each multiple of the order will also yield 1 when used as exponent for a, it's enough to find the least d such that (q^d)*(t/(q^e)) yields 1 when used as exponent.
Task
Implement a routine to calculate the multiplicative order along these lines. You may assume that routines to determine the factorization into prime powers are available in some library.
An algorithm for the multiplicative order can be found in Bach & Shallit, Algorithmic Number Theory, Volume I: Efficient Algorithms, The MIT Press, 1996:
Exercise 5.8, page 115:
Suppose you are given a prime p and a complete factorization
of p-1. Show how to compute the order of an
element a in (Z/(p))* using O((lg p)4/(lg lg p)) bit
operations.
Solution, page 337:
Let the prime factorization of p-1 be q1e1q2e2...qkek . We use the following observation:
if x^((p-1)/qifi) = 1 (mod p) ,
and fi=ei or x^((p-1)/qifi+1) != 1 (mod p) , then qiei-fi||ordp x. (This follows by combining Exercises 5.1 and 2.10.)
Hence it suffices to find, for each i , the exponent fi such that the condition above holds.
This can be done as follows: first compute q1e1, q2e2, ... ,
qkek . This can be done using O((lg p)2) bit operations. Next, compute y1=(p-1)/q1e1, ... , yk=(p-1)/qkek .
This can be done using O((lg p)2) bit operations. Now, using the binary method,
compute x1=ay1(mod p), ... , xk=ayk(mod p) .
This can be done using O(k(lg p)3) bit operations, and k=O((lg p)/(lg lg p)) by Theorem 8.8.10.
Finally, for each i , repeatedly raise xi to the qi-th power (mod p) (as many as ei-1 times), checking to see when 1 is obtained.
This can be done using O((lg p)3) steps.
The total cost is dominated by O(k(lg p)3) , which is O((lg p)4/(lg lg p)).
| #J | J | mo=: 4 : 0
a=. x: x
m=. x: y
assert. 1=a+.m
*./ a mopk"1 |: __ q: m
) |
http://rosettacode.org/wiki/Multiplicative_order | Multiplicative order | The multiplicative order of a relative to m is the least positive integer n such that a^n is 1 (modulo m).
Example
The multiplicative order of 37 relative to 1000 is 100 because 37^100 is 1 (modulo 1000), and no number smaller than 100 would do.
One possible algorithm that is efficient also for large numbers is the following: By the Chinese Remainder Theorem, it's enough to calculate the multiplicative order for each prime exponent p^k of m, and
combine the results with the least common multiple operation.
Now the order of a with regard to p^k must divide Φ(p^k). Call this number t, and determine it's factors q^e. Since each multiple of the order will also yield 1 when used as exponent for a, it's enough to find the least d such that (q^d)*(t/(q^e)) yields 1 when used as exponent.
Task
Implement a routine to calculate the multiplicative order along these lines. You may assume that routines to determine the factorization into prime powers are available in some library.
An algorithm for the multiplicative order can be found in Bach & Shallit, Algorithmic Number Theory, Volume I: Efficient Algorithms, The MIT Press, 1996:
Exercise 5.8, page 115:
Suppose you are given a prime p and a complete factorization
of p-1. Show how to compute the order of an
element a in (Z/(p))* using O((lg p)4/(lg lg p)) bit
operations.
Solution, page 337:
Let the prime factorization of p-1 be q1e1q2e2...qkek . We use the following observation:
if x^((p-1)/qifi) = 1 (mod p) ,
and fi=ei or x^((p-1)/qifi+1) != 1 (mod p) , then qiei-fi||ordp x. (This follows by combining Exercises 5.1 and 2.10.)
Hence it suffices to find, for each i , the exponent fi such that the condition above holds.
This can be done as follows: first compute q1e1, q2e2, ... ,
qkek . This can be done using O((lg p)2) bit operations. Next, compute y1=(p-1)/q1e1, ... , yk=(p-1)/qkek .
This can be done using O((lg p)2) bit operations. Now, using the binary method,
compute x1=ay1(mod p), ... , xk=ayk(mod p) .
This can be done using O(k(lg p)3) bit operations, and k=O((lg p)/(lg lg p)) by Theorem 8.8.10.
Finally, for each i , repeatedly raise xi to the qi-th power (mod p) (as many as ei-1 times), checking to see when 1 is obtained.
This can be done using O((lg p)3) steps.
The total cost is dominated by O(k(lg p)3) , which is O((lg p)4/(lg lg p)).
| #Java | Java | import java.math.BigInteger;
import java.util.ArrayList;
import java.util.List;
public class MultiplicativeOrder {
private static final BigInteger ONE = BigInteger.ONE;
private static final BigInteger TWO = BigInteger.valueOf(2);
private static final BigInteger THREE = BigInteger.valueOf(3);
private static final BigInteger TEN = BigInteger.TEN;
private static class PExp {
BigInteger prime;
long exp;
PExp(BigInteger prime, long exp) {
this.prime = prime;
this.exp = exp;
}
}
private static void moTest(BigInteger a, BigInteger n) {
if (!n.isProbablePrime(20)) {
System.out.println("Not computed. Modulus must be prime for this algorithm.");
return;
}
if (a.bitLength() < 100) System.out.printf("ord(%s)", a);
else System.out.print("ord([big])");
if (n.bitLength() < 100) System.out.printf(" mod %s ", n);
else System.out.print(" mod [big] ");
BigInteger mob = moBachShallit58(a, n, factor(n.subtract(ONE)));
System.out.println("= " + mob);
}
private static BigInteger moBachShallit58(BigInteger a, BigInteger n, List<PExp> pf) {
BigInteger n1 = n.subtract(ONE);
BigInteger mo = ONE;
for (PExp pe : pf) {
BigInteger y = n1.divide(pe.prime.pow((int) pe.exp));
long o = 0;
BigInteger x = a.modPow(y, n.abs());
while (x.compareTo(ONE) > 0) {
x = x.modPow(pe.prime, n.abs());
o++;
}
BigInteger o1 = BigInteger.valueOf(o);
o1 = pe.prime.pow(o1.intValue());
o1 = o1.divide(mo.gcd(o1));
mo = mo.multiply(o1);
}
return mo;
}
private static List<PExp> factor(BigInteger n) {
List<PExp> pf = new ArrayList<>();
BigInteger nn = n;
Long e = 0L;
while (!nn.testBit(e.intValue())) e++;
if (e > 0L) {
nn = nn.shiftRight(e.intValue());
pf.add(new PExp(TWO, e));
}
BigInteger s = sqrt(nn);
BigInteger d = THREE;
while (nn.compareTo(ONE) > 0) {
if (d.compareTo(s) > 0) d = nn;
e = 0L;
while (true) {
BigInteger[] qr = nn.divideAndRemainder(d);
if (qr[1].bitLength() > 0) break;
nn = qr[0];
e++;
}
if (e > 0L) {
pf.add(new PExp(d, e));
s = sqrt(nn);
}
d = d.add(TWO);
}
return pf;
}
private static BigInteger sqrt(BigInteger n) {
BigInteger b = n;
while (true) {
BigInteger a = b;
b = n.divide(a).add(a).shiftRight(1);
if (b.compareTo(a) >= 0) return a;
}
}
public static void main(String[] args) {
moTest(BigInteger.valueOf(37), BigInteger.valueOf(3343));
BigInteger b = TEN.pow(100).add(ONE);
moTest(b, BigInteger.valueOf(7919));
b = TEN.pow(1000).add(ONE);
moTest(b, BigInteger.valueOf(15485863));
b = TEN.pow(10000).subtract(ONE);
moTest(b, BigInteger.valueOf(22801763489L));
moTest(BigInteger.valueOf(1511678068), BigInteger.valueOf(7379191741L));
moTest(BigInteger.valueOf(3047753288L), BigInteger.valueOf(2257683301L));
}
} |
http://rosettacode.org/wiki/Nth_root | Nth root | Task
Implement the algorithm to compute the principal nth root
A
n
{\displaystyle {\sqrt[{n}]{A}}}
of a positive real number A, as explained at the Wikipedia page.
| #PicoLisp | PicoLisp | (load "@lib/math.l")
(de nthRoot (N A)
(let (X1 A X2 (*/ A N))
(until (= X1 X2)
(setq
X1 X2
X2 (*/
(+
(* X1 (dec N))
(*/ A 1.0 (pow X1 (* (dec N) 1.0))) )
N ) ) )
X2 ) )
(prinl (format (nthRoot 2 2.0) *Scl))
(prinl (format (nthRoot 3 12.3) *Scl))
(prinl (format (nthRoot 4 45.6) *Scl)) |
http://rosettacode.org/wiki/Nth_root | Nth root | Task
Implement the algorithm to compute the principal nth root
A
n
{\displaystyle {\sqrt[{n}]{A}}}
of a positive real number A, as explained at the Wikipedia page.
| #PL.2FI | PL/I | /* Finds the N-th root of the number A */
root: procedure (A, N) returns (float);
declare A float, N fixed binary;
declare (xi, xip1) float;
xi = 1; /* An initial guess */
do forever;
xip1 = ((n-1)*xi + A/xi**(n-1) ) / n;
if abs(xip1-xi) < 1e-5 then leave;
xi = xip1;
end;
return (xi);
end root; |
http://rosettacode.org/wiki/N%27th | N'th | Write a function/method/subroutine/... that when given an integer greater than or equal to zero returns a string of the number followed by an apostrophe then the ordinal suffix.
Example
Returns would include 1'st 2'nd 3'rd 11'th 111'th 1001'st 1012'th
Task
Use your routine to show here the output for at least the following (inclusive) ranges of integer inputs:
0..25, 250..265, 1000..1025
Note: apostrophes are now optional to allow correct apostrophe-less English.
| #Java | Java | public class Nth {
public static String ordinalAbbrev(int n){
String ans = "th"; //most of the time it should be "th"
if(n % 100 / 10 == 1) return ans; //teens are all "th"
switch(n % 10){
case 1: ans = "st"; break;
case 2: ans = "nd"; break;
case 3: ans = "rd"; break;
}
return ans;
}
public static void main(String[] args){
for(int i = 0; i <= 25;i++){
System.out.print(i + ordinalAbbrev(i) + " ");
}
System.out.println();
for(int i = 250; i <= 265;i++){
System.out.print(i + ordinalAbbrev(i) + " ");
}
System.out.println();
for(int i = 1000; i <= 1025;i++){
System.out.print(i + ordinalAbbrev(i) + " ");
}
}
} |
http://rosettacode.org/wiki/Non-decimal_radices/Convert | Non-decimal radices/Convert | Number base conversion is when you express a stored integer in an integer base, such as in octal (base 8) or binary (base 2). It also is involved when you take a string representing a number in a given base and convert it to the stored integer form. Normally, a stored integer is in binary, but that's typically invisible to the user, who normally enters or sees stored integers as decimal.
Task
Write a function (or identify the built-in function) which is passed a non-negative integer to convert, and another integer representing the base.
It should return a string containing the digits of the resulting number, without leading zeros except for the number 0 itself.
For the digits beyond 9, one should use the lowercase English alphabet, where the digit a = 9+1, b = a+1, etc.
For example: the decimal number 26 expressed in base 16 would be 1a.
Write a second function which is passed a string and an integer base, and it returns an integer representing that string interpreted in that base.
The programs may be limited by the word size or other such constraint of a given language. There is no need to do error checking for negatives, bases less than 2, or inappropriate digits.
| #Wolframalpha | Wolframalpha | import "/fmt" for Conv
System.print(Conv.itoa(26, 16))
System.print(Conv.atoi("1a", 16)) |
http://rosettacode.org/wiki/Narcissistic_decimal_number | Narcissistic decimal number | A Narcissistic decimal number is a non-negative integer,
n
{\displaystyle n}
, that is equal to the sum of the
m
{\displaystyle m}
-th powers of each of the digits in the decimal representation of
n
{\displaystyle n}
, where
m
{\displaystyle m}
is the number of digits in the decimal representation of
n
{\displaystyle n}
.
Narcissistic (decimal) numbers are sometimes called Armstrong numbers, named after Michael F. Armstrong.
They are also known as Plus Perfect numbers.
An example
if
n
{\displaystyle n}
is 153
then
m
{\displaystyle m}
, (the number of decimal digits) is 3
we have 13 + 53 + 33 = 1 + 125 + 27 = 153
and so 153 is a narcissistic decimal number
Task
Generate and show here the first 25 narcissistic decimal numbers.
Note:
0
1
=
0
{\displaystyle 0^{1}=0}
, the first in the series.
See also
the OEIS entry: Armstrong (or Plus Perfect, or narcissistic) numbers.
MathWorld entry: Narcissistic Number.
Wikipedia entry: Narcissistic number.
| #Ring | Ring |
n = 0
count = 0
size = 15
while count != size
m = isNarc(n)
if m=1 see "" + n + " is narcisstic" + nl
count = count + 1 ok
n = n + 1
end
func isNarc n
m = len(string(n))
sum = 0
digit = 0
for pos = 1 to m
digit = number(substr(string(n), pos, 1))
sum = sum + pow(digit,m)
next
nr = (sum = n)
return nr
|
http://rosettacode.org/wiki/Narcissistic_decimal_number | Narcissistic decimal number | A Narcissistic decimal number is a non-negative integer,
n
{\displaystyle n}
, that is equal to the sum of the
m
{\displaystyle m}
-th powers of each of the digits in the decimal representation of
n
{\displaystyle n}
, where
m
{\displaystyle m}
is the number of digits in the decimal representation of
n
{\displaystyle n}
.
Narcissistic (decimal) numbers are sometimes called Armstrong numbers, named after Michael F. Armstrong.
They are also known as Plus Perfect numbers.
An example
if
n
{\displaystyle n}
is 153
then
m
{\displaystyle m}
, (the number of decimal digits) is 3
we have 13 + 53 + 33 = 1 + 125 + 27 = 153
and so 153 is a narcissistic decimal number
Task
Generate and show here the first 25 narcissistic decimal numbers.
Note:
0
1
=
0
{\displaystyle 0^{1}=0}
, the first in the series.
See also
the OEIS entry: Armstrong (or Plus Perfect, or narcissistic) numbers.
MathWorld entry: Narcissistic Number.
Wikipedia entry: Narcissistic number.
| #Ruby | Ruby | class Integer
def narcissistic?
return false if negative?
digs = self.digits
m = digs.size
digs.map{|d| d**m}.sum == self
end
end
puts 0.step.lazy.select(&:narcissistic?).first(25) |
http://rosettacode.org/wiki/Munchausen_numbers | Munchausen numbers | A Munchausen number is a natural number n the sum of whose digits (in base 10), each raised to the power of itself, equals n.
(Munchausen is also spelled: Münchhausen.)
For instance: 3435 = 33 + 44 + 33 + 55
Task
Find all Munchausen numbers between 1 and 5000.
Also see
The OEIS entry: A046253
The Wikipedia entry: Perfect digit-to-digit invariant, redirected from Munchausen Number
| #J | J | munch=: +/@(^~@(10&#.inv))
(#~ ] = munch"0) 1+i.5000
1 3435 |
http://rosettacode.org/wiki/Mutual_recursion | Mutual recursion | Two functions are said to be mutually recursive if the first calls the second,
and in turn the second calls the first.
Write two mutually recursive functions that compute members of the Hofstadter Female and Male sequences defined as:
F
(
0
)
=
1
;
M
(
0
)
=
0
F
(
n
)
=
n
−
M
(
F
(
n
−
1
)
)
,
n
>
0
M
(
n
)
=
n
−
F
(
M
(
n
−
1
)
)
,
n
>
0.
{\displaystyle {\begin{aligned}F(0)&=1\ ;\ M(0)=0\\F(n)&=n-M(F(n-1)),\quad n>0\\M(n)&=n-F(M(n-1)),\quad n>0.\end{aligned}}}
(If a language does not allow for a solution using mutually recursive functions
then state this rather than give a solution by other means).
| #Erlang | Erlang | -module(mutrec).
-export([mutrec/0, f/1, m/1]).
f(0) -> 1;
f(N) -> N - m(f(N-1)).
m(0) -> 0;
m(N) -> N - f(m(N-1)).
mutrec() -> lists:map(fun(X) -> io:format("~w ", [f(X)]) end, lists:seq(0,19)),
io:format("~n", []),
lists:map(fun(X) -> io:format("~w ", [m(X)]) end, lists:seq(0,19)),
io:format("~n", []). |
http://rosettacode.org/wiki/Multisplit | Multisplit | It is often necessary to split a string into pieces
based on several different (potentially multi-character) separator strings,
while still retaining the information about which separators were present in the input.
This is particularly useful when doing small parsing tasks.
The task is to write code to demonstrate this.
The function (or procedure or method, as appropriate) should
take an input string and an ordered collection of separators.
The order of the separators is significant:
The delimiter order represents priority in matching, with the first defined delimiter having the highest priority.
In cases where there would be an ambiguity as to
which separator to use at a particular point
(e.g., because one separator is a prefix of another)
the separator with the highest priority should be used.
Delimiters can be reused and the output from the function should be an ordered sequence of substrings.
Test your code using the input string “a!===b=!=c” and the separators “==”, “!=” and “=”.
For these inputs the string should be parsed as "a" (!=) "" (==) "b" (=) "" (!=) "c", where matched delimiters are shown in parentheses, and separated strings are quoted, so our resulting output is "a", empty string, "b", empty string, "c".
Note that the quotation marks are shown for clarity and do not form part of the output.
Extra Credit: provide information that indicates which separator was matched at each separation point and where in the input string that separator was matched.
| #SenseTalk | SenseTalk | set source to "a!===b=!=c"
set separators to ["==", "!=", "="]
put each line delimited by separators of source |
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #Elena | Elena | import system'routines;
import extensions;
class Foo;
// create a list of disting object
fill(n)
= RangeEnumerator.new(1,n).selectBy:(x => new Foo()).toArray();
// testing
public program()
{
var foos := fill(10);
} |
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #Elixir | Elixir | randoms = for _ <- 1..10, do: :rand.uniform(1000) |
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #Erlang | Erlang | Randoms = [random:uniform(1000) || _ <- lists:seq(1,10)].
|
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #F.23 | F# | >List.replicate 3 (System.Guid.NewGuid());;
val it : Guid list =
[485632d7-1fd6-4d9e-8910-7949d7b2b485; 485632d7-1fd6-4d9e-8910-7949d7b2b485;
485632d7-1fd6-4d9e-8910-7949d7b2b485] |
http://rosettacode.org/wiki/Move-to-front_algorithm | Move-to-front algorithm | Given a symbol table of a zero-indexed array of all possible input symbols
this algorithm reversibly transforms a sequence
of input symbols into an array of output numbers (indices).
The transform in many cases acts to give frequently repeated input symbols
lower indices which is useful in some compression algorithms.
Encoding algorithm
for each symbol of the input sequence:
output the index of the symbol in the symbol table
move that symbol to the front of the symbol table
Decoding algorithm
# Using the same starting symbol table
for each index of the input sequence:
output the symbol at that index of the symbol table
move that symbol to the front of the symbol table
Example
Encoding the string of character symbols 'broood' using a symbol table of the lowercase characters a-to-z
Input
Output
SymbolTable
broood
1
'abcdefghijklmnopqrstuvwxyz'
broood
1 17
'bacdefghijklmnopqrstuvwxyz'
broood
1 17 15
'rbacdefghijklmnopqstuvwxyz'
broood
1 17 15 0
'orbacdefghijklmnpqstuvwxyz'
broood
1 17 15 0 0
'orbacdefghijklmnpqstuvwxyz'
broood
1 17 15 0 0 5
'orbacdefghijklmnpqstuvwxyz'
Decoding the indices back to the original symbol order:
Input
Output
SymbolTable
1 17 15 0 0 5
b
'abcdefghijklmnopqrstuvwxyz'
1 17 15 0 0 5
br
'bacdefghijklmnopqrstuvwxyz'
1 17 15 0 0 5
bro
'rbacdefghijklmnopqstuvwxyz'
1 17 15 0 0 5
broo
'orbacdefghijklmnpqstuvwxyz'
1 17 15 0 0 5
brooo
'orbacdefghijklmnpqstuvwxyz'
1 17 15 0 0 5
broood
'orbacdefghijklmnpqstuvwxyz'
Task
Encode and decode the following three strings of characters using the symbol table of the lowercase characters a-to-z as above.
Show the strings and their encoding here.
Add a check to ensure that the decoded string is the same as the original.
The strings are:
broood
bananaaa
hiphophiphop
(Note the misspellings in the above strings.)
| #Action.21 | Action! | DEFINE SYMBOL_TABLE_SIZE="26"
PROC InitSymbolTable(BYTE ARRAY table BYTE len)
BYTE i
FOR i=0 TO len-1
DO
table(i)=i+'a
OD
RETURN
BYTE FUNC Find(BYTE ARRAY table BYTE len,c)
BYTE i
FOR i=0 TO len-1
DO
IF table(i)=c THEN
RETURN (i)
FI
OD
Break()
RETURN (0)
PROC MoveToFront(BYTE ARRAY table BYTE len,pos)
BYTE sym
sym=table(pos)
WHILE pos>0
DO
table(pos)=table(pos-1)
pos==-1
OD
table(pos)=sym
RETURN
PROC Encode(CHAR ARRAY in BYTE ARRAY out BYTE POINTER outLen)
BYTE ARRAY table(SYMBOL_TABLE_SIZE)
BYTE i,pos
outLen^=0
InitSymbolTable(table,SYMBOL_TABLE_SIZE)
FOR i=1 TO in(0)
DO
pos=Find(table,SYMBOL_TABLE_SIZE,in(i))
out(outLen^)=pos
outLen^==+1
MoveToFront(table,SYMBOL_TABLE_SIZE,pos)
OD
RETURN
PROC Decode(BYTE ARRAY in BYTE inLen CHAR ARRAY out)
BYTE ARRAY table(SYMBOL_TABLE_SIZE)
BYTE i,pos,len
len=0
InitSymbolTable(table,SYMBOL_TABLE_SIZE)
FOR i=0 TO inLen-1
DO
pos=in(i)
len==+1
out(len)=table(pos)
MoveToFront(table,SYMBOL_TABLE_SIZE,pos)
OD
out(0)=len
RETURN
PROC Test(CHAR ARRAY s)
BYTE ARRAY encoded(255)
CHAR ARRAY decoded(256)
BYTE encodedLength,i
Print("Source: ")
PrintE(s)
Encode(s,encoded,@encodedLength)
Print("Encoded: ")
FOR i=0 TO encodedLength-1
DO
PrintB(encoded(i)) Put(32)
OD
PutE()
Decode(encoded,encodedLength,decoded)
Print("Decoded: ")
PrintE(decoded)
IF SCompare(s,decoded)=0 THEN
PrintE("Decoded is equal to the source.")
ELSE
PrintE("Decoded is different from the source!")
FI
PutE()
RETURN
PROC Main()
Test("broood")
Test("bananaaa")
Test("hiphophiphop")
RETURN |
http://rosettacode.org/wiki/Multi-dimensional_array | Multi-dimensional array | For the purposes of this task, the actual memory layout or access method of this data structure is not mandated.
It is enough to:
State the number and extent of each index to the array.
Provide specific, ordered, integer indices for all dimensions of the array together with a new value to update the indexed value.
Provide specific, ordered, numeric indices for all dimensions of the array to obtain the arrays value at that indexed position.
Task
State if the language supports multi-dimensional arrays in its syntax and usual implementation.
State whether the language uses row-major or column major order for multi-dimensional array storage, or any other relevant kind of storage.
Show how to create a four dimensional array in your language and set, access, set to another value; and access the new value of an integer-indexed item of the array.
The idiomatic method for the language is preferred.
The array should allow a range of five, four, three and two (or two three four five if convenient), in each of the indices, in order. (For example, if indexing starts at zero for the first index then a range of 0..4 inclusive would suffice).
State if memory allocation is optimised for the array - especially if contiguous memory is likely to be allocated.
If the language has exceptional native multi-dimensional array support such as optional bounds checking, reshaping, or being able to state both the lower and upper bounds of index ranges, then this is the task to mention them.
Show all output here, (but you may judiciously use ellipses to shorten repetitive output text).
| #Ada | Ada |
type Grade is range 0..100;
subtype Lower_Case is Character range 'a'..'z';
subtype Natural is Integer range 0..Integer'Last;
subtype Positive is Integer range 1..Integer'Last;
type Deflection is range -180..180;
|
http://rosettacode.org/wiki/Multi-dimensional_array | Multi-dimensional array | For the purposes of this task, the actual memory layout or access method of this data structure is not mandated.
It is enough to:
State the number and extent of each index to the array.
Provide specific, ordered, integer indices for all dimensions of the array together with a new value to update the indexed value.
Provide specific, ordered, numeric indices for all dimensions of the array to obtain the arrays value at that indexed position.
Task
State if the language supports multi-dimensional arrays in its syntax and usual implementation.
State whether the language uses row-major or column major order for multi-dimensional array storage, or any other relevant kind of storage.
Show how to create a four dimensional array in your language and set, access, set to another value; and access the new value of an integer-indexed item of the array.
The idiomatic method for the language is preferred.
The array should allow a range of five, four, three and two (or two three four five if convenient), in each of the indices, in order. (For example, if indexing starts at zero for the first index then a range of 0..4 inclusive would suffice).
State if memory allocation is optimised for the array - especially if contiguous memory is likely to be allocated.
If the language has exceptional native multi-dimensional array support such as optional bounds checking, reshaping, or being able to state both the lower and upper bounds of index ranges, then this is the task to mention them.
Show all output here, (but you may judiciously use ellipses to shorten repetitive output text).
| #ALGOL_68 | ALGOL 68 | [ 1 : 5, 1 : 4, 1 : 3, 1 : 2 ]INT a; |
http://rosettacode.org/wiki/Multiple_regression | Multiple regression | Task
Given a set of data vectors in the following format:
y
=
{
y
1
,
y
2
,
.
.
.
,
y
n
}
{\displaystyle y=\{y_{1},y_{2},...,y_{n}\}\,}
X
i
=
{
x
i
1
,
x
i
2
,
.
.
.
,
x
i
n
}
,
i
∈
1..
k
{\displaystyle X_{i}=\{x_{i1},x_{i2},...,x_{in}\},i\in 1..k\,}
Compute the vector
β
=
{
β
1
,
β
2
,
.
.
.
,
β
k
}
{\displaystyle \beta =\{\beta _{1},\beta _{2},...,\beta _{k}\}}
using ordinary least squares regression using the following equation:
y
j
=
Σ
i
β
i
⋅
x
i
j
,
j
∈
1..
n
{\displaystyle y_{j}=\Sigma _{i}\beta _{i}\cdot x_{ij},j\in 1..n}
You can assume y is given to you as a vector (a one-dimensional array), and X is given to you as a two-dimensional array (i.e. matrix).
| #BBC_BASIC | BBC BASIC | *FLOAT 64
INSTALL @lib$+"ARRAYLIB"
DIM y(14), x(2,14), c(2)
y() = 52.21, 53.12, 54.48, 55.84, 57.20, 58.57, 59.93, 61.29, \
\ 63.11, 64.47, 66.28, 68.10, 69.92, 72.19, 74.46
x() = 1.47, 1.50, 1.52, 1.55, 1.57, 1.60, 1.63, 1.65, \
\ 1.68, 1.70, 1.73, 1.75, 1.78, 1.80, 1.83
FOR row% = DIM(x(),1) TO 0 STEP -1
FOR col% = 0 TO DIM(x(),2)
x(row%,col%) = x(0,col%) ^ row%
NEXT
NEXT row%
PROCmultipleregression(y(), x(), c())
FOR i% = 0 TO DIM(c(),1) : PRINT c(i%) " "; : NEXT
PRINT
END
DEF PROCmultipleregression(y(), x(), c())
LOCAL m(), t()
DIM m(DIM(x(),1), DIM(x(),1)), t(DIM(x(),2),DIM(x(),1))
PROC_transpose(x(), t())
m() = x().t()
PROC_invert(m())
t() = t().m()
c() = y().t()
ENDPROC |
http://rosettacode.org/wiki/Multifactorial | Multifactorial | The factorial of a number, written as
n
!
{\displaystyle n!}
, is defined as
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
.
Multifactorials generalize factorials as follows:
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
n
!
!
=
n
(
n
−
2
)
(
n
−
4
)
.
.
.
{\displaystyle n!!=n(n-2)(n-4)...}
n
!
!
!
=
n
(
n
−
3
)
(
n
−
6
)
.
.
.
{\displaystyle n!!!=n(n-3)(n-6)...}
n
!
!
!
!
=
n
(
n
−
4
)
(
n
−
8
)
.
.
.
{\displaystyle n!!!!=n(n-4)(n-8)...}
n
!
!
!
!
!
=
n
(
n
−
5
)
(
n
−
10
)
.
.
.
{\displaystyle n!!!!!=n(n-5)(n-10)...}
In all cases, the terms in the products are positive integers.
If we define the degree of the multifactorial as the difference in successive terms that are multiplied together for a multifactorial (the number of exclamation marks), then the task is twofold:
Write a function that given n and the degree, calculates the multifactorial.
Use the function to generate and display here a table of the first ten members (1 to 10) of the first five degrees of multifactorial.
Note: The wikipedia entry on multifactorials gives a different formula. This task uses the Wolfram mathworld definition.
| #ALGOL_68 | ALGOL 68 | BEGIN
INT highest degree = 5;
INT largest number = 10;
CO Recursive implementation of multifactorial function CO
PROC multi fact = (INT n, deg) INT :
(n <= deg | n | n * multi fact(n - deg, deg));
CO Iterative implementation of multifactorial function CO
PROC multi fact i = (INT n, deg) INT :
BEGIN
INT result := n, nn := n;
WHILE (nn >= deg + 1) DO
result TIMESAB nn - deg;
nn MINUSAB deg
OD;
result
END;
CO Print out multifactorials CO
FOR i TO highest degree DO
printf (($l, "Degree ", g(0), ":"$, i));
FOR j TO largest number DO
printf (($xg(0)$, multi fact (j, i)))
OD
OD
END
|
http://rosettacode.org/wiki/Mouse_position | Mouse position | Task
Get the current location of the mouse cursor relative to the active window.
Please specify if the window may be externally created.
| #8086_Assembly | 8086 Assembly | mov ax,3
int 33h |
http://rosettacode.org/wiki/Multi-base_primes | Multi-base primes | Prime numbers are prime no matter what base they are represented in.
A prime number in a base other than 10 may not look prime at first glance.
For instance: 19 base 10 is 25 in base 7.
Several different prime numbers may be expressed as the "same" string when converted to a different base.
107 base 10 converted to base 6 == 255
173 base 10 converted to base 8 == 255
353 base 10 converted to base 12 == 255
467 base 10 converted to base 14 == 255
743 base 10 converted to base 18 == 255
1277 base 10 converted to base 24 == 255
1487 base 10 converted to base 26 == 255
2213 base 10 converted to base 32 == 255
Task
Restricted to bases 2 through 36; find the strings that have the most different bases that evaluate to that string when converting prime numbers to a base.
Find the conversion string, the amount of bases that evaluate a prime to that string and the enumeration of bases that evaluate a prime to that string.
Display here, on this page, the string, the count and the list for all of the: 1 character, 2 character, 3 character, and 4 character strings that have the maximum base count that evaluate to that string.
Should be no surprise, the string '2' has the largest base count for single character strings.
Stretch goal
Do the same for the maximum 5 character string.
| #Julia | Julia | using Primes
function maxprimebases(ndig, maxbase)
maxprimebases = [Int[]]
nwithbases = [0]
maxprime = 10^(ndig) - 1
for p in div(maxprime + 1, 10):maxprime
dig = digits(p)
bases = [b for b in 2:maxbase if (isprime(evalpoly(b, dig)) && all(i -> i < b, dig))]
if length(bases) > length(first(maxprimebases))
maxprimebases = [bases]
nwithbases = [p]
elseif length(bases) == length(first(maxprimebases))
push!(maxprimebases, bases)
push!(nwithbases, p)
end
end
alen, vlen = length(first(maxprimebases)), length(maxprimebases)
println("\nThe maximum number of prime valued bases for base 10 numeric strings of length ",
ndig, " is $alen. The base 10 value list of ", vlen > 1 ? "these" : "this", " is:")
for i in eachindex(maxprimebases)
println(nwithbases[i], " => ", maxprimebases[i])
end
end
@time for n in 1:6
maxprimebases(n, 36)
end
|
http://rosettacode.org/wiki/Multi-base_primes | Multi-base primes | Prime numbers are prime no matter what base they are represented in.
A prime number in a base other than 10 may not look prime at first glance.
For instance: 19 base 10 is 25 in base 7.
Several different prime numbers may be expressed as the "same" string when converted to a different base.
107 base 10 converted to base 6 == 255
173 base 10 converted to base 8 == 255
353 base 10 converted to base 12 == 255
467 base 10 converted to base 14 == 255
743 base 10 converted to base 18 == 255
1277 base 10 converted to base 24 == 255
1487 base 10 converted to base 26 == 255
2213 base 10 converted to base 32 == 255
Task
Restricted to bases 2 through 36; find the strings that have the most different bases that evaluate to that string when converting prime numbers to a base.
Find the conversion string, the amount of bases that evaluate a prime to that string and the enumeration of bases that evaluate a prime to that string.
Display here, on this page, the string, the count and the list for all of the: 1 character, 2 character, 3 character, and 4 character strings that have the maximum base count that evaluate to that string.
Should be no surprise, the string '2' has the largest base count for single character strings.
Stretch goal
Do the same for the maximum 5 character string.
| #Mathematica.2FWolfram_Language | Mathematica/Wolfram Language | ClearAll[OtherBasePrimes, OtherBasePrimesPower]
OtherBasePrimes[n_Integer] := Module[{digs, minbase, bases},
digs = IntegerDigits[n];
minbase = Max[digs] + 1;
bases = Range[minbase, 36];
Pick[bases, PrimeQ[FromDigits[digs, #] & /@ bases], True]
]
OtherBasePrimesPower[p_] := Module[{min, max, out, maxlen},
min = 10^p;
max = 10^(p + 1) - 1;
out = {#, OtherBasePrimes[#]} & /@ Range[min, max];
maxlen = Max[Length /@ out[[All, 2]]];
Select[out, Last /* Length /* EqualTo[maxlen]]
]
OtherBasePrimesPower[0] // Column
OtherBasePrimesPower[1] // Column
OtherBasePrimesPower[2] // Column
OtherBasePrimesPower[3] // Column
OtherBasePrimesPower[4] // Column
OtherBasePrimesPower[5] // Column |
http://rosettacode.org/wiki/N-queens_problem | N-queens problem |
Solve the eight queens puzzle.
You can extend the problem to solve the puzzle with a board of size NxN.
For the number of solutions for small values of N, see OEIS: A000170.
Related tasks
A* search algorithm
Solve a Hidato puzzle
Solve a Holy Knight's tour
Knight's tour
Peaceful chess queen armies
Solve a Hopido puzzle
Solve a Numbrix puzzle
Solve the no connection puzzle
| #Delphi | Delphi |
program N_queens_problem;
{$APPTYPE CONSOLE}
uses
System.SysUtils;
var
i: Integer;
q: boolean;
a: array[0..8] of boolean;
b: array[0..16] of boolean;
c: array[0..14] of boolean;
x: array[0..8] of Integer;
procedure TryMove(i: Integer);
begin
var j := 1;
while True do
begin
q := false;
if a[j] and b[i + j] and c[i - j + 7] then
begin
x[i] := j;
a[j] := false;
b[i + j] := false;
c[i - j + 7] := false;
if i < 8 then
begin
TryMove(i + 1);
if not q then
begin
a[j] := true;
b[i + j] := true;
c[i - j + 7] := true;
end;
end
else
q := true;
end;
if q or (j = 8) then
Break;
inc(j);
end;
end;
begin
for i := 1 to 8 do
a[i] := true;
for i := 2 to 16 do
b[i] := true;
for i := 0 to 14 do
c[i] := true;
TryMove(1);
if q then
for i := 1 to 8 do
writeln(i, ' ', x[i]);
readln;
end. |
http://rosettacode.org/wiki/Multiplicative_order | Multiplicative order | The multiplicative order of a relative to m is the least positive integer n such that a^n is 1 (modulo m).
Example
The multiplicative order of 37 relative to 1000 is 100 because 37^100 is 1 (modulo 1000), and no number smaller than 100 would do.
One possible algorithm that is efficient also for large numbers is the following: By the Chinese Remainder Theorem, it's enough to calculate the multiplicative order for each prime exponent p^k of m, and
combine the results with the least common multiple operation.
Now the order of a with regard to p^k must divide Φ(p^k). Call this number t, and determine it's factors q^e. Since each multiple of the order will also yield 1 when used as exponent for a, it's enough to find the least d such that (q^d)*(t/(q^e)) yields 1 when used as exponent.
Task
Implement a routine to calculate the multiplicative order along these lines. You may assume that routines to determine the factorization into prime powers are available in some library.
An algorithm for the multiplicative order can be found in Bach & Shallit, Algorithmic Number Theory, Volume I: Efficient Algorithms, The MIT Press, 1996:
Exercise 5.8, page 115:
Suppose you are given a prime p and a complete factorization
of p-1. Show how to compute the order of an
element a in (Z/(p))* using O((lg p)4/(lg lg p)) bit
operations.
Solution, page 337:
Let the prime factorization of p-1 be q1e1q2e2...qkek . We use the following observation:
if x^((p-1)/qifi) = 1 (mod p) ,
and fi=ei or x^((p-1)/qifi+1) != 1 (mod p) , then qiei-fi||ordp x. (This follows by combining Exercises 5.1 and 2.10.)
Hence it suffices to find, for each i , the exponent fi such that the condition above holds.
This can be done as follows: first compute q1e1, q2e2, ... ,
qkek . This can be done using O((lg p)2) bit operations. Next, compute y1=(p-1)/q1e1, ... , yk=(p-1)/qkek .
This can be done using O((lg p)2) bit operations. Now, using the binary method,
compute x1=ay1(mod p), ... , xk=ayk(mod p) .
This can be done using O(k(lg p)3) bit operations, and k=O((lg p)/(lg lg p)) by Theorem 8.8.10.
Finally, for each i , repeatedly raise xi to the qi-th power (mod p) (as many as ei-1 times), checking to see when 1 is obtained.
This can be done using O((lg p)3) steps.
The total cost is dominated by O(k(lg p)3) , which is O((lg p)4/(lg lg p)).
| #Julia | Julia | using Primes
function factors(n)
f = [one(n)]
for (p,e) in factor(n)
f = reduce(vcat, [f*p^j for j in 1:e], init=f)
end
return length(f) == 1 ? [one(n), n] : sort!(f)
end
function multorder(a, m)
gcd(a,m) == 1 || error("$a and $m are not coprime")
res = one(m)
for (p,e) in factor(m)
m = p^e
t = div(m, p) * (p-1)
for f in factors(t)
if powermod(a, f, m) == 1
res = lcm(res, f)
break
end
end
end
res
end |
http://rosettacode.org/wiki/Multiplicative_order | Multiplicative order | The multiplicative order of a relative to m is the least positive integer n such that a^n is 1 (modulo m).
Example
The multiplicative order of 37 relative to 1000 is 100 because 37^100 is 1 (modulo 1000), and no number smaller than 100 would do.
One possible algorithm that is efficient also for large numbers is the following: By the Chinese Remainder Theorem, it's enough to calculate the multiplicative order for each prime exponent p^k of m, and
combine the results with the least common multiple operation.
Now the order of a with regard to p^k must divide Φ(p^k). Call this number t, and determine it's factors q^e. Since each multiple of the order will also yield 1 when used as exponent for a, it's enough to find the least d such that (q^d)*(t/(q^e)) yields 1 when used as exponent.
Task
Implement a routine to calculate the multiplicative order along these lines. You may assume that routines to determine the factorization into prime powers are available in some library.
An algorithm for the multiplicative order can be found in Bach & Shallit, Algorithmic Number Theory, Volume I: Efficient Algorithms, The MIT Press, 1996:
Exercise 5.8, page 115:
Suppose you are given a prime p and a complete factorization
of p-1. Show how to compute the order of an
element a in (Z/(p))* using O((lg p)4/(lg lg p)) bit
operations.
Solution, page 337:
Let the prime factorization of p-1 be q1e1q2e2...qkek . We use the following observation:
if x^((p-1)/qifi) = 1 (mod p) ,
and fi=ei or x^((p-1)/qifi+1) != 1 (mod p) , then qiei-fi||ordp x. (This follows by combining Exercises 5.1 and 2.10.)
Hence it suffices to find, for each i , the exponent fi such that the condition above holds.
This can be done as follows: first compute q1e1, q2e2, ... ,
qkek . This can be done using O((lg p)2) bit operations. Next, compute y1=(p-1)/q1e1, ... , yk=(p-1)/qkek .
This can be done using O((lg p)2) bit operations. Now, using the binary method,
compute x1=ay1(mod p), ... , xk=ayk(mod p) .
This can be done using O(k(lg p)3) bit operations, and k=O((lg p)/(lg lg p)) by Theorem 8.8.10.
Finally, for each i , repeatedly raise xi to the qi-th power (mod p) (as many as ei-1 times), checking to see when 1 is obtained.
This can be done using O((lg p)3) steps.
The total cost is dominated by O(k(lg p)3) , which is O((lg p)4/(lg lg p)).
| #Kotlin | Kotlin | // version 1.2.10
import java.math.BigInteger
val bigOne = BigInteger.ONE
val bigTwo = 2.toBigInteger()
val bigThree = 3.toBigInteger()
val bigTen = BigInteger.TEN
class PExp(val prime: BigInteger, val exp: Long)
fun moTest(a: BigInteger, n: BigInteger) {
if (!n.isProbablePrime(20)) {
println("Not computed. Modulus must be prime for this algorithm.")
return
}
if (a.bitLength() < 100) print("ord($a)") else print("ord([big])")
if (n.bitLength() < 100) print(" mod $n ") else print(" mod [big] ")
val mob = moBachShallit58(a, n, factor(n - bigOne))
println("= $mob")
}
fun moBachShallit58(a: BigInteger, n: BigInteger, pf: List<PExp>): BigInteger {
val n1 = n - bigOne
var mo = bigOne
for (pe in pf) {
val y = n1 / pe.prime.pow(pe.exp.toInt())
var o = 0L
var x = a.modPow(y, n.abs())
while (x > bigOne) {
x = x.modPow(pe.prime, n.abs())
o++
}
var o1 = o.toBigInteger()
o1 = pe.prime.pow(o1.toInt())
o1 /= mo.gcd(o1)
mo *= o1
}
return mo
}
fun factor(n: BigInteger): List<PExp> {
val pf = mutableListOf<PExp>()
var nn = n
var e = 0L
while (!nn.testBit(e.toInt())) e++
if (e > 0L) {
nn = nn shr e.toInt()
pf.add(PExp(bigTwo, e))
}
var s = bigSqrt(nn)
var d = bigThree
while (nn > bigOne) {
if (d > s) d = nn
e = 0L
while (true) {
val (q, r) = nn.divideAndRemainder(d)
if (r.bitLength() > 0) break
nn = q
e++
}
if (e > 0L) {
pf.add(PExp(d, e))
s = bigSqrt(nn)
}
d += bigTwo
}
return pf
}
fun bigSqrt(n: BigInteger): BigInteger {
var b = n
while (true) {
val a = b
b = (n / a + a) shr 1
if (b >= a) return a
}
}
fun main(args: Array<String>) {
moTest(37.toBigInteger(), 3343.toBigInteger())
var b = bigTen.pow(100) + bigOne
moTest(b, 7919.toBigInteger())
b = bigTen.pow(1000) + bigOne
moTest(b, BigInteger("15485863"))
b = bigTen.pow(10000) - bigOne
moTest(b, BigInteger("22801763489"))
moTest(BigInteger("1511678068"), BigInteger("7379191741"))
moTest(BigInteger("3047753288"), BigInteger("2257683301"))
} |
http://rosettacode.org/wiki/Nth_root | Nth root | Task
Implement the algorithm to compute the principal nth root
A
n
{\displaystyle {\sqrt[{n}]{A}}}
of a positive real number A, as explained at the Wikipedia page.
| #PowerShell | PowerShell | #NoTeS: This sample code does not validate inputs
# Thus, if there are errors the 'scary' red-text
# error messages will appear.
#
# This code will not work properly in floating point values of n,
# and negative values of A.
#
# Supports negative values of n by reciprocating the root.
$epsilon=1E-10 #Sample Epsilon (Precision)
function power($x,$e){ #As I said in the comment
$ret=1
for($i=1;$i -le $e;$i++){
$ret*=$x
}
return $ret
}
function root($y,$n){ #The main Function
if (0+$n -lt 0){$tmp=-$n} else {$tmp=$n} #This checks if n is negative.
$ans=1
do{
$d = ($y/(power $ans ($tmp-1)) - $ans)/$tmp
$ans+=$d
} while ($d -lt -$epsilon -or $d -gt $epsilon)
if (0+$n -lt 0){return 1/$ans} else {return $ans}
}
#Sample Inputs
root 625 2
root 2401 4
root 2 -2
root 1.23456789E-20 34
root 9.87654321E20 10 #Quite slow here, I admit...
((root 5 2)+1)/2 #Extra: Computes the golden ratio
((root 5 2)-1)/2 |
http://rosettacode.org/wiki/N%27th | N'th | Write a function/method/subroutine/... that when given an integer greater than or equal to zero returns a string of the number followed by an apostrophe then the ordinal suffix.
Example
Returns would include 1'st 2'nd 3'rd 11'th 111'th 1001'st 1012'th
Task
Use your routine to show here the output for at least the following (inclusive) ranges of integer inputs:
0..25, 250..265, 1000..1025
Note: apostrophes are now optional to allow correct apostrophe-less English.
| #JavaScript | JavaScript | console.log(function () {
var lstSuffix = 'th st nd rd th th th th th th'.split(' '),
fnOrdinalForm = function (n) {
return n.toString() + (
11 <= n % 100 && 13 >= n % 100 ?
"th" : lstSuffix[n % 10]
);
},
range = function (m, n) {
return Array.apply(
null, Array(n - m + 1)
).map(function (x, i) {
return m + i;
});
};
return [[0, 25], [250, 265], [1000, 1025]].map(function (tpl) {
return range.apply(null, tpl).map(fnOrdinalForm).join(' ');
}).join('\n\n');
}()); |
http://rosettacode.org/wiki/Non-decimal_radices/Convert | Non-decimal radices/Convert | Number base conversion is when you express a stored integer in an integer base, such as in octal (base 8) or binary (base 2). It also is involved when you take a string representing a number in a given base and convert it to the stored integer form. Normally, a stored integer is in binary, but that's typically invisible to the user, who normally enters or sees stored integers as decimal.
Task
Write a function (or identify the built-in function) which is passed a non-negative integer to convert, and another integer representing the base.
It should return a string containing the digits of the resulting number, without leading zeros except for the number 0 itself.
For the digits beyond 9, one should use the lowercase English alphabet, where the digit a = 9+1, b = a+1, etc.
For example: the decimal number 26 expressed in base 16 would be 1a.
Write a second function which is passed a string and an integer base, and it returns an integer representing that string interpreted in that base.
The programs may be limited by the word size or other such constraint of a given language. There is no need to do error checking for negatives, bases less than 2, or inappropriate digits.
| #Wren | Wren | import "/fmt" for Conv
System.print(Conv.itoa(26, 16))
System.print(Conv.atoi("1a", 16)) |
http://rosettacode.org/wiki/Non-decimal_radices/Convert | Non-decimal radices/Convert | Number base conversion is when you express a stored integer in an integer base, such as in octal (base 8) or binary (base 2). It also is involved when you take a string representing a number in a given base and convert it to the stored integer form. Normally, a stored integer is in binary, but that's typically invisible to the user, who normally enters or sees stored integers as decimal.
Task
Write a function (or identify the built-in function) which is passed a non-negative integer to convert, and another integer representing the base.
It should return a string containing the digits of the resulting number, without leading zeros except for the number 0 itself.
For the digits beyond 9, one should use the lowercase English alphabet, where the digit a = 9+1, b = a+1, etc.
For example: the decimal number 26 expressed in base 16 would be 1a.
Write a second function which is passed a string and an integer base, and it returns an integer representing that string interpreted in that base.
The programs may be limited by the word size or other such constraint of a given language. There is no need to do error checking for negatives, bases less than 2, or inappropriate digits.
| #XPL0 | XPL0 | include c:\cxpl\codes; \intrinsic 'code' declarations
string 0; \use zero-terminated string convention
func Num2Str(N, B); \Convert integer N to a numeric string in base B
int N, B;
char S(32); int I;
[I:= 31;
S(31):= 0; \terminate string
repeat I:= I-1;
N:= N/B;
S(I):= rem(0) + (if rem(0)<=9 then ^0 else ^a-10);
until N=0;
return @S(I); \BEWARE! very temporary string space
];
func Str2Num(S, B); \Convert numeric string S in base B to an integer
char S; int B;
int I, N;
[I:= 0; N:= 0;
while S(I) do
[N:= N*B + S(I) - (if S(I)<=^9 then ^0 else ^a-10); I:= I+1];
return N;
];
[Text(0, Num2Str(0, 10)); CrLf(0);
Text(0, Num2Str(26, 16)); CrLf(0);
Text(0, Num2Str($7FFF_FFFF, 2)); CrLf(0);
IntOut(0, Str2Num("0100", 2)); CrLf(0);
IntOut(0, Str2Num("1a", 16)); CrLf(0);
IntOut(0, Str2Num("deadbeef", 16)); CrLf(0);
] |
http://rosettacode.org/wiki/Narcissistic_decimal_number | Narcissistic decimal number | A Narcissistic decimal number is a non-negative integer,
n
{\displaystyle n}
, that is equal to the sum of the
m
{\displaystyle m}
-th powers of each of the digits in the decimal representation of
n
{\displaystyle n}
, where
m
{\displaystyle m}
is the number of digits in the decimal representation of
n
{\displaystyle n}
.
Narcissistic (decimal) numbers are sometimes called Armstrong numbers, named after Michael F. Armstrong.
They are also known as Plus Perfect numbers.
An example
if
n
{\displaystyle n}
is 153
then
m
{\displaystyle m}
, (the number of decimal digits) is 3
we have 13 + 53 + 33 = 1 + 125 + 27 = 153
and so 153 is a narcissistic decimal number
Task
Generate and show here the first 25 narcissistic decimal numbers.
Note:
0
1
=
0
{\displaystyle 0^{1}=0}
, the first in the series.
See also
the OEIS entry: Armstrong (or Plus Perfect, or narcissistic) numbers.
MathWorld entry: Narcissistic Number.
Wikipedia entry: Narcissistic number.
| #Rust | Rust |
fn is_narcissistic(x: u32) -> bool {
let digits: Vec<u32> = x
.to_string()
.chars()
.map(|c| c.to_digit(10).unwrap())
.collect();
digits
.iter()
.map(|d| d.pow(digits.len() as u32))
.sum::<u32>()
== x
}
fn main() {
let mut counter = 0;
let mut i = 0;
while counter < 25 {
if is_narcissistic(i) {
println!("{}", i);
counter += 1;
}
i += 1;
}
}
|
http://rosettacode.org/wiki/Narcissistic_decimal_number | Narcissistic decimal number | A Narcissistic decimal number is a non-negative integer,
n
{\displaystyle n}
, that is equal to the sum of the
m
{\displaystyle m}
-th powers of each of the digits in the decimal representation of
n
{\displaystyle n}
, where
m
{\displaystyle m}
is the number of digits in the decimal representation of
n
{\displaystyle n}
.
Narcissistic (decimal) numbers are sometimes called Armstrong numbers, named after Michael F. Armstrong.
They are also known as Plus Perfect numbers.
An example
if
n
{\displaystyle n}
is 153
then
m
{\displaystyle m}
, (the number of decimal digits) is 3
we have 13 + 53 + 33 = 1 + 125 + 27 = 153
and so 153 is a narcissistic decimal number
Task
Generate and show here the first 25 narcissistic decimal numbers.
Note:
0
1
=
0
{\displaystyle 0^{1}=0}
, the first in the series.
See also
the OEIS entry: Armstrong (or Plus Perfect, or narcissistic) numbers.
MathWorld entry: Narcissistic Number.
Wikipedia entry: Narcissistic number.
| #Scala | Scala | object NDN extends App {
val narc: Int => Int = n => (n.toString map (_.asDigit) map (math.pow(_, n.toString.size)) sum) toInt
val isNarc: Int => Boolean = i => i == narc(i)
println((Iterator from 0 filter isNarc take 25 toList) mkString(" "))
} |
http://rosettacode.org/wiki/Munchausen_numbers | Munchausen numbers | A Munchausen number is a natural number n the sum of whose digits (in base 10), each raised to the power of itself, equals n.
(Munchausen is also spelled: Münchhausen.)
For instance: 3435 = 33 + 44 + 33 + 55
Task
Find all Munchausen numbers between 1 and 5000.
Also see
The OEIS entry: A046253
The Wikipedia entry: Perfect digit-to-digit invariant, redirected from Munchausen Number
| #Java | Java |
public class Main {
public static void main(String[] args) {
for(int i = 0 ; i <= 5000 ; i++ ){
int val = String.valueOf(i).chars().map(x -> (int) Math.pow( x-48 ,x-48)).sum();
if( i == val){
System.out.println( i + " (munchausen)");
}
}
}
}
|
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