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http://rosettacode.org/wiki/Factors_of_an_integer | Factors of an integer |
Basic Data Operation
This is a basic data operation. It represents a fundamental action on a basic data type.
You may see other such operations in the Basic Data Operations category, or:
Integer Operations
Arithmetic |
Comparison
Boolean Operations
Bitwise |
Logical
String Operations
Concatenation |
Interpolation |
Comparison |
Matching
Memory Operations
Pointers & references |
Addresses
Task
Compute the factors of a positive integer.
These factors are the positive integers by which the number being factored can be divided to yield a positive integer result.
(Though the concepts function correctly for zero and negative integers, the set of factors of zero has countably infinite members, and the factors of negative integers can be obtained from the factors of related positive numbers without difficulty; this task does not require handling of either of these cases).
Note that every prime number has two factors: 1 and itself.
Related tasks
count in factors
prime decomposition
Sieve of Eratosthenes
primality by trial division
factors of a Mersenne number
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
sequence: smallest number greater than previous term with exactly n divisors
| #Perl | Perl | sub factors
{
my($n) = @_;
return grep { $n % $_ == 0 }(1 .. $n);
}
print join ' ',factors(64), "\n"; |
http://rosettacode.org/wiki/Extreme_floating_point_values | Extreme floating point values | The IEEE floating point specification defines certain 'extreme' floating point values such as minus zero, -0.0, a value distinct from plus zero; not a number, NaN; and plus and minus infinity.
The task is to use expressions involving other 'normal' floating point values in your language to calculate these, (and maybe other), extreme floating point values in your language and assign them to variables.
Print the values of these variables if possible; and show some arithmetic with these values and variables.
If your language can directly enter these extreme floating point values then show it.
See also
What Every Computer Scientist Should Know About Floating-Point Arithmetic
Related tasks
Infinity
Detect division by zero
Literals/Floating point
| #Go | Go | package main
import (
"fmt"
"math"
)
func main() {
// compute "extreme values" from non-extreme values
var zero float64 // zero is handy.
var negZero, posInf, negInf, nan float64 // values to compute.
negZero = zero * -1
posInf = 1 / zero
negInf = -1 / zero
nan = zero / zero
// print extreme values stored in variables
fmt.Println(negZero, posInf, negInf, nan)
// directly obtain extreme values
fmt.Println(math.Float64frombits(1<<63),
math.Inf(1), math.Inf(-1), math.NaN())
// validate some arithmetic on extreme values
fmt.Println()
validateNaN(negInf+posInf, "-Inf + Inf")
validateNaN(0*posInf, "0 * Inf")
validateNaN(posInf/posInf, "Inf / Inf")
// mod is specifically named in "What every computer scientist..."
// Go math package doc lists many special cases for other package functions.
validateNaN(math.Mod(posInf, 1), "Inf % 1")
validateNaN(1+nan, "1 + NaN")
validateZero(1/posInf, "1 / Inf")
validateGT(posInf, math.MaxFloat64, "Inf > max value")
validateGT(-math.MaxFloat64, negInf, "-Inf < max neg value")
validateNE(nan, nan, "NaN != NaN")
validateEQ(negZero, 0, "-0 == 0")
}
func validateNaN(n float64, op string) {
if math.IsNaN(n) {
fmt.Println(op, "-> NaN")
} else {
fmt.Println("!!! Expected NaN from", op, " Found", n)
}
}
func validateZero(n float64, op string) {
if n == 0 {
fmt.Println(op, "-> 0")
} else {
fmt.Println("!!! Expected 0 from", op, " Found", n)
}
}
func validateGT(a, b float64, op string) {
if a > b {
fmt.Println(op)
} else {
fmt.Println("!!! Expected", op, " Found not true.")
}
}
func validateNE(a, b float64, op string) {
if a == b {
fmt.Println("!!! Expected", op, " Found not true.")
} else {
fmt.Println(op)
}
}
func validateEQ(a, b float64, op string) {
if a == b {
fmt.Println(op)
} else {
fmt.Println("!!! Expected", op, " Found not true.")
}
} |
http://rosettacode.org/wiki/Extend_your_language | Extend your language | Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Some programming languages allow you to extend the language. While this can be done to a certain degree in most languages (e.g. by using macros), other languages go much further. Most notably in the Forth and Lisp families, programming per se is done by extending the language without any formal distinction between built-in and user-defined elements.
If your language supports it, show how to introduce a new flow control mechanism. A practical and useful example is a four-way branch:
Occasionally, code must be written that depends on two conditions, resulting in up to four branches (depending on whether both, only the first, only the second, or none of the conditions are "true"). In a C-like language this could look like the following:
if (condition1isTrue) {
if (condition2isTrue)
bothConditionsAreTrue();
else
firstConditionIsTrue();
}
else if (condition2isTrue)
secondConditionIsTrue();
else
noConditionIsTrue();
Besides being rather cluttered, the statement(s) for 'condition2isTrue' must be written down twice. If 'condition2isTrue' were a lengthy and involved expression, it would be quite unreadable, and the code generated by the compiler might be unnecessarily large.
This can be improved by introducing a new keyword if2. It is similar to if, but takes two conditional statements instead of one, and up to three 'else' statements. One proposal (in pseudo-C syntax) might be:
if2 (condition1isTrue) (condition2isTrue)
bothConditionsAreTrue();
else1
firstConditionIsTrue();
else2
secondConditionIsTrue();
else
noConditionIsTrue();
Pick the syntax which suits your language. The keywords 'else1' and 'else2' are just examples. The new conditional expression should look, nest and behave analogously to the language's built-in 'if' statement.
| #Agda | Agda |
data Bool : Set where
true : Bool
false : Bool
if_then_else : ∀ {l} {A : Set l} -> Bool -> A -> A -> A
if true then t else e = t
if false then t else e = e
if2_,_then_else1_else2_else_ : ∀ {l} {A : Set l} -> (b1 b2 : Bool) -> (t e1 e2 e : A) -> A
if2 true , true then t else1 e1 else2 e2 else e = t
if2 true , false then t else1 e1 else2 e2 else e = e1
if2 false , true then t else1 e1 else2 e2 else e = e2
if2 false , false then t else1 e1 else2 e2 else e = e
example : Bool
example = if2 true , false then true else1 false else2 true else false
|
http://rosettacode.org/wiki/Extend_your_language | Extend your language | Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Some programming languages allow you to extend the language. While this can be done to a certain degree in most languages (e.g. by using macros), other languages go much further. Most notably in the Forth and Lisp families, programming per se is done by extending the language without any formal distinction between built-in and user-defined elements.
If your language supports it, show how to introduce a new flow control mechanism. A practical and useful example is a four-way branch:
Occasionally, code must be written that depends on two conditions, resulting in up to four branches (depending on whether both, only the first, only the second, or none of the conditions are "true"). In a C-like language this could look like the following:
if (condition1isTrue) {
if (condition2isTrue)
bothConditionsAreTrue();
else
firstConditionIsTrue();
}
else if (condition2isTrue)
secondConditionIsTrue();
else
noConditionIsTrue();
Besides being rather cluttered, the statement(s) for 'condition2isTrue' must be written down twice. If 'condition2isTrue' were a lengthy and involved expression, it would be quite unreadable, and the code generated by the compiler might be unnecessarily large.
This can be improved by introducing a new keyword if2. It is similar to if, but takes two conditional statements instead of one, and up to three 'else' statements. One proposal (in pseudo-C syntax) might be:
if2 (condition1isTrue) (condition2isTrue)
bothConditionsAreTrue();
else1
firstConditionIsTrue();
else2
secondConditionIsTrue();
else
noConditionIsTrue();
Pick the syntax which suits your language. The keywords 'else1' and 'else2' are just examples. The new conditional expression should look, nest and behave analogously to the language's built-in 'if' statement.
| #ALGOL_68 | ALGOL 68 | # operator to turn two boolean values into an integer - name inspired by the COBOL sample #
PRIO ALSO = 1;
OP ALSO = ( BOOL a, b )INT: IF a AND b THEN 1 ELIF a THEN 2 ELIF b THEN 3 ELSE 4 FI;
# using the above operator, we can use the standard CASE construct to provide the #
# required construct, e.g.: #
BOOL a := TRUE, b := FALSE;
CASE a ALSO b
IN print( ( "both: a and b are TRUE", newline ) )
, print( ( "first: only a is TRUE", newline ) )
, print( ( "second: only b is TRUE", newline ) )
, print( ( "neither: a and b are FALSE", newline ) )
ESAC |
http://rosettacode.org/wiki/Factorial_base_numbers_indexing_permutations_of_a_collection | Factorial base numbers indexing permutations of a collection | You need a random arrangement of a deck of cards, you are sick of lame ways of doing this. This task is a super-cool way of doing this using factorial base numbers.
The first 25 factorial base numbers in increasing order are: 0.0.0, 0.0.1, 0.1.0, 0.1.1, 0.2.0, 0.2.1, 1.0.0, 1.0.1, 1.1.0, 1.1.1,1.2.0, 1.2.1, 2.0.0, 2.0.1, 2.1.0, 2.1.1, 2.2.0, 2.2.1, 3.0.0, 3.0.1, 3.1.0, 3.1.1, 3.2.0, 3.2.1, 1.0.0.0
Observe that the least significant digit is base 2 the next base 3, in general an n-digit factorial base number has digits n..1 in base n+1..2.
I want to produce a 1 to 1 mapping between these numbers and permutations:-
0.0.0 -> 0123
0.0.1 -> 0132
0.1.0 -> 0213
0.1.1 -> 0231
0.2.0 -> 0312
0.2.1 -> 0321
1.0.0 -> 1023
1.0.1 -> 1032
1.1.0 -> 1203
1.1.1 -> 1230
1.2.0 -> 1302
1.2.1 -> 1320
2.0.0 -> 2013
2.0.1 -> 2031
2.1.0 -> 2103
2.1.1 -> 2130
2.2.0 -> 2301
2.2.1 -> 2310
3.0.0 -> 3012
3.0.1 -> 3021
3.1.0 -> 3102
3.1.1 -> 3120
3.2.0 -> 3201
3.2.1 -> 3210
The following psudo-code will do this:
Starting with m=0 and Ω, an array of elements to be permutated, for each digit g starting with the most significant digit in the factorial base number.
If g is greater than zero, rotate the elements from m to m+g in Ω (see example)
Increment m and repeat the first step using the next most significant digit until the factorial base number is exhausted.
For example: using the factorial base number 2.0.1 and Ω = 0 1 2 3 where place 0 in both is the most significant (left-most) digit/element.
Step 1: m=0 g=2; Rotate places 0 through 2. 0 1 2 3 becomes 2 0 1 3
Step 2: m=1 g=0; No action.
Step 3: m=2 g=1; Rotate places 2 through 3. 2 0 1 3 becomes 2 0 3 1
Let me work 2.0.1 and 0123
step 1 n=0 g=2 Ω=2013
step 2 n=1 g=0 so no action
step 3 n=2 g=1 Ω=2031
The task:
First use your function to recreate the above table.
Secondly use your function to generate all permutaions of 11 digits, perhaps count them don't display them, compare this method with
methods in rc's permutations task.
Thirdly here following are two ramdom 51 digit factorial base numbers I prepared earlier:
39.49.7.47.29.30.2.12.10.3.29.37.33.17.12.31.29.34.17.25.2.4.25.4.1.14.20.6.21.18.1.1.1.4.0.5.15.12.4.3.10.10.9.1.6.5.5.3.0.0.0
51.48.16.22.3.0.19.34.29.1.36.30.12.32.12.29.30.26.14.21.8.12.1.3.10.4.7.17.6.21.8.12.15.15.13.15.7.3.12.11.9.5.5.6.6.3.4.0.3.2.1
use your function to crate the corresponding permutation of the following shoe of cards:
A♠K♠Q♠J♠10♠9♠8♠7♠6♠5♠4♠3♠2♠A♥K♥Q♥J♥10♥9♥8♥7♥6♥5♥4♥3♥2♥A♦K♦Q♦J♦10♦9♦8♦7♦6♦5♦4♦3♦2♦A♣K♣Q♣J♣10♣9♣8♣7♣6♣5♣4♣3♣2♣
Finally create your own 51 digit factorial base number and produce the corresponding permutation of the above shoe
| #Wren | Wren | import "random" for Random
import "/math" for Int
import "/fmt" for Fmt
var genFactBaseNums = Fn.new { |size, countOnly|
var results = []
var count = 0
var n = 0
while (true) {
var radix = 2
var res = null
if (!countOnly) res = List.filled(size, 0)
var k = n
while (k > 0) {
var div = (k/radix).floor
var rem = k % radix
if (!countOnly) {
if (radix <= size + 1) res[size-radix+1] = rem
}
k = div
radix = radix + 1
}
if (radix > size+2) break
count = count + 1
if (!countOnly) results.add(res)
n = n + 1
}
return [results, count]
}
var mapToPerms = Fn.new { |factNums|
var perms = []
var psize = factNums[0].count + 1
var start = List.filled(psize, 0)
for (i in 0...psize) start[i] = i
for (fn in factNums) {
var perm = start.toList
for (m in 0...fn.count) {
var g = fn[m]
if (g != 0) {
var first = m
var last = m + g
for (i in 1..g) {
var temp = perm[first]
for (j in first+1..last) perm[j-1] = perm[j]
perm[last] = temp
}
}
}
perms.add(perm)
}
return perms
}
var join = Fn.new { |ints, sep| ints.map { |i| i.toString }.join(sep) }
var undot = Fn.new { |s| s.split(".").map { |ss| Num.fromString(ss) }.toList }
var rand = Random.new()
// Recreate the table.
var factNums = genFactBaseNums.call(3, false)[0]
var perms = mapToPerms.call(factNums)
var i = 0
for (fn in factNums) {
Fmt.print("$s -> $s", join.call(fn, "."), join.call(perms[i], ""))
i = i + 1
}
// Check that the number of perms generated is equal to 11! (this takes a while).
var count = genFactBaseNums.call(10, true)[1]
Fmt.print("\nPermutations generated = $,d", count)
Fmt.print("compared to 11! which = $,d", Int.factorial(11))
System.print()
// Generate shuffles for the 2 given 51 digit factorial base numbers.
var fbn51s = [
"39.49.7.47.29.30.2.12.10.3.29.37.33.17.12.31.29.34.17.25.2.4.25.4.1.14.20.6.21.18.1.1.1.4.0.5.15.12.4.3.10.10.9.1.6.5.5.3.0.0.0",
"51.48.16.22.3.0.19.34.29.1.36.30.12.32.12.29.30.26.14.21.8.12.1.3.10.4.7.17.6.21.8.12.15.15.13.15.7.3.12.11.9.5.5.6.6.3.4.0.3.2.1"
]
factNums = [undot.call(fbn51s[0]), undot.call(fbn51s[1])]
perms = mapToPerms.call(factNums)
var shoe = "A♠K♠Q♠J♠T♠9♠8♠7♠6♠5♠4♠3♠2♠A♥K♥Q♥J♥T♥9♥8♥7♥6♥5♥4♥3♥2♥A♦K♦Q♦J♦T♦9♦8♦7♦6♦5♦4♦3♦2♦A♣K♣Q♣J♣T♣9♣8♣7♣6♣5♣4♣3♣2♣".toList
var cards = List.filled(52, null)
for (i in 0..51) {
cards[i] = shoe[2*i..2*i+1].join()
if (cards[i][0] == "T") cards[i] = "10" + cards[i][1..-1]
}
i = 0
for (fbn51 in fbn51s) {
System.print(fbn51)
for (d in perms[i]) System.write(cards[d])
System.print("\n")
i = i + 1
}
// Create a random 51 digit factorial base number and produce a shuffle from that.
var fbn51 = List.filled(51, 0)
for (i in 0..50) fbn51[i] = rand.int(52-i)
System.print(join.call(fbn51, "."))
perms = mapToPerms.call([fbn51])
for (d in perms[0]) System.write(cards[d])
System.print() |
http://rosettacode.org/wiki/Factorions | Factorions |
Definition
A factorion is a natural number that equals the sum of the factorials of its digits.
Example
145 is a factorion in base 10 because:
1! + 4! + 5! = 1 + 24 + 120 = 145
It can be shown (see talk page) that no factorion in base 10 can exceed 1,499,999.
Task
Write a program in your language to demonstrate, by calculating and printing out the factorions, that:
There are 3 factorions in base 9
There are 4 factorions in base 10
There are 5 factorions in base 11
There are 2 factorions in base 12 (up to the same upper bound as for base 10)
See also
Wikipedia article
OEIS:A014080 - Factorions in base 10
OEIS:A193163 - Factorions in base n
| #Go | Go | package main
import (
"fmt"
"strconv"
)
func main() {
// cache factorials from 0 to 11
var fact [12]uint64
fact[0] = 1
for n := uint64(1); n < 12; n++ {
fact[n] = fact[n-1] * n
}
for b := 9; b <= 12; b++ {
fmt.Printf("The factorions for base %d are:\n", b)
for i := uint64(1); i < 1500000; i++ {
digits := strconv.FormatUint(i, b)
sum := uint64(0)
for _, digit := range digits {
if digit < 'a' {
sum += fact[digit-'0']
} else {
sum += fact[digit+10-'a']
}
}
if sum == i {
fmt.Printf("%d ", i)
}
}
fmt.Println("\n")
}
} |
http://rosettacode.org/wiki/Factorions | Factorions |
Definition
A factorion is a natural number that equals the sum of the factorials of its digits.
Example
145 is a factorion in base 10 because:
1! + 4! + 5! = 1 + 24 + 120 = 145
It can be shown (see talk page) that no factorion in base 10 can exceed 1,499,999.
Task
Write a program in your language to demonstrate, by calculating and printing out the factorions, that:
There are 3 factorions in base 9
There are 4 factorions in base 10
There are 5 factorions in base 11
There are 2 factorions in base 12 (up to the same upper bound as for base 10)
See also
Wikipedia article
OEIS:A014080 - Factorions in base 10
OEIS:A193163 - Factorions in base n
| #Haskell | Haskell | import Text.Printf (printf)
import Data.List (unfoldr)
import Control.Monad (guard)
factorion :: Int -> Int -> Bool
factorion b n = f b n == n
where
f b = sum . map (product . enumFromTo 1) . unfoldr (\x -> guard (x > 0) >> pure (x `mod` b, x `div` b))
main :: IO ()
main = mapM_ (uncurry (printf "Factorions for base %2d: %s\n") . (\(a, b) -> (b, result a b)))
[(3,9), (4,10), (5,11), (2,12)]
where
factorions b = filter (factorion b) [1..]
result n = show . take n . factorions |
http://rosettacode.org/wiki/Filter | Filter | Task
Select certain elements from an Array into a new Array in a generic way.
To demonstrate, select all even numbers from an Array.
As an option, give a second solution which filters destructively,
by modifying the original Array rather than creating a new Array.
| #Tcl | Tcl | set l {56 21 71 27 39 62 87 76 82 94 45 83 65 45 28 90 52 44 1 89}
puts [lmap x $l {if {$x % 2} continue; set x}] |
http://rosettacode.org/wiki/FizzBuzz | FizzBuzz | Task
Write a program that prints the integers from 1 to 100 (inclusive).
But:
for multiples of three, print Fizz (instead of the number)
for multiples of five, print Buzz (instead of the number)
for multiples of both three and five, print FizzBuzz (instead of the number)
The FizzBuzz problem was presented as the lowest level of comprehension required to illustrate adequacy.
Also see
(a blog) dont-overthink-fizzbuzz
(a blog) fizzbuzz-the-programmers-stairway-to-heaven
| #PL.2FI | PL/I | do i = 1 to 100;
select;
when (mod(i,15) = 0) put skip list ('FizzBuzz');
when (mod(i,3) = 0) put skip list ('Fizz');
when (mod(i,5) = 0) put skip list ('Buzz');
otherwise put skip list (i);
end;
end; |
http://rosettacode.org/wiki/Extensible_prime_generator | Extensible prime generator | Task
Write a generator of prime numbers, in order, that will automatically adjust to accommodate the generation of any reasonably high prime.
The routine should demonstrably rely on either:
Being based on an open-ended counter set to count without upper limit other than system or programming language limits. In this case, explain where this counter is in the code.
Being based on a limit that is extended automatically. In this case, choose a small limit that ensures the limit will be passed when generating some of the values to be asked for below.
If other methods of creating an extensible prime generator are used, the algorithm's means of extensibility/lack of limits should be stated.
The routine should be used to:
Show the first twenty primes.
Show the primes between 100 and 150.
Show the number of primes between 7,700 and 8,000.
Show the 10,000th prime.
Show output on this page.
Note: You may reference code already on this site if it is written to be imported/included, then only the code necessary for import and the performance of this task need be shown. (It is also important to leave a forward link on the referenced tasks entry so that later editors know that the code is used for multiple tasks).
Note 2: If a languages in-built prime generator is extensible or is guaranteed to generate primes up to a system limit, (231 or memory overflow for example), then this may be used as long as an explanation of the limits of the prime generator is also given. (Which may include a link to/excerpt from, language documentation).
Note 3:The task is written so it may be useful in solving the task Emirp primes as well as others (depending on its efficiency).
Reference
Prime Numbers. Website with large count of primes.
| #AutoHotkey | AutoHotkey | SetBatchLines, -1
p := 1 ;p functions as the counter
Loop, 10000 {
p := NextPrime(p)
if (A_Index < 21)
a .= p ", "
if (p < 151 && p > 99)
b .= p ", "
if (p < 8001 && p > 7699)
c++
}
MsgBox, % "First twenty primes: " RTrim(a, ", ")
. "`nPrimes between 100 and 150: " RTrim(b, ", ")
. "`nNumber of primes between 7,700 and 8,000: " RTrim(c, ", ")
. "`nThe 10,000th prime: " p
NextPrime(n) {
Loop
if (IsPrime(++n))
return n
}
IsPrime(n) {
if (n < 2)
return, 0
else if (n < 4)
return, 1
else if (!Mod(n, 2))
return, 0
else if (n < 9)
return 1
else if (!Mod(n, 3))
return, 0
else {
r := Floor(Sqrt(n))
f := 5
while (f <= r) {
if (!Mod(n, f))
return, 0
if (!Mod(n, (f + 2)))
return, 0
f += 6
}
return, 1
}
} |
http://rosettacode.org/wiki/Fibonacci_sequence | Fibonacci sequence | The Fibonacci sequence is a sequence Fn of natural numbers defined recursively:
F0 = 0
F1 = 1
Fn = Fn-1 + Fn-2, if n>1
Task
Write a function to generate the nth Fibonacci number.
Solutions can be iterative or recursive (though recursive solutions are generally considered too slow and are mostly used as an exercise in recursion).
The sequence is sometimes extended into negative numbers by using a straightforward inverse of the positive definition:
Fn = Fn+2 - Fn+1, if n<0
support for negative n in the solution is optional.
Related tasks
Fibonacci n-step number sequences
Leonardo numbers
References
Wikipedia, Fibonacci number
Wikipedia, Lucas number
MathWorld, Fibonacci Number
Some identities for r-Fibonacci numbers
OEIS Fibonacci numbers
OEIS Lucas numbers
| #Free_Pascal | Free Pascal | type
/// domain for Fibonacci function
/// where result is within nativeUInt
// You can not name it fibonacciDomain,
// since the Fibonacci function itself
// is defined for all whole numbers
// but the result beyond F(n) exceeds high(nativeUInt).
fibonacciLeftInverseRange =
{$ifdef CPU64} 0..93 {$else} 0..47 {$endif};
{**
implements Fibonacci sequence iteratively
\param n the index of the Fibonacci number to calculate
\returns the Fibonacci value at n
}
function fibonacci(const n: fibonacciLeftInverseRange): nativeUInt;
type
/// more meaningful identifiers than simple integers
relativePosition = (previous, current, next);
var
/// temporary iterator variable
i: longword;
/// holds preceding fibonacci values
f: array[relativePosition] of nativeUInt;
begin
f[previous] := 0;
f[current] := 1;
// note, in Pascal for-loop-limits are inclusive
for i := 1 to n do
begin
f[next] := f[previous] + f[current];
f[previous] := f[current];
f[current] := f[next];
end;
// assign to previous, bc f[current] = f[next] for next iteration
fibonacci := f[previous];
end; |
http://rosettacode.org/wiki/Factors_of_an_integer | Factors of an integer |
Basic Data Operation
This is a basic data operation. It represents a fundamental action on a basic data type.
You may see other such operations in the Basic Data Operations category, or:
Integer Operations
Arithmetic |
Comparison
Boolean Operations
Bitwise |
Logical
String Operations
Concatenation |
Interpolation |
Comparison |
Matching
Memory Operations
Pointers & references |
Addresses
Task
Compute the factors of a positive integer.
These factors are the positive integers by which the number being factored can be divided to yield a positive integer result.
(Though the concepts function correctly for zero and negative integers, the set of factors of zero has countably infinite members, and the factors of negative integers can be obtained from the factors of related positive numbers without difficulty; this task does not require handling of either of these cases).
Note that every prime number has two factors: 1 and itself.
Related tasks
count in factors
prime decomposition
Sieve of Eratosthenes
primality by trial division
factors of a Mersenne number
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
sequence: smallest number greater than previous term with exactly n divisors
| #Phix | Phix | ?factors(12345,1)
|
http://rosettacode.org/wiki/Extreme_floating_point_values | Extreme floating point values | The IEEE floating point specification defines certain 'extreme' floating point values such as minus zero, -0.0, a value distinct from plus zero; not a number, NaN; and plus and minus infinity.
The task is to use expressions involving other 'normal' floating point values in your language to calculate these, (and maybe other), extreme floating point values in your language and assign them to variables.
Print the values of these variables if possible; and show some arithmetic with these values and variables.
If your language can directly enter these extreme floating point values then show it.
See also
What Every Computer Scientist Should Know About Floating-Point Arithmetic
Related tasks
Infinity
Detect division by zero
Literals/Floating point
| #Groovy | Groovy | def negInf = -1.0d / 0.0d; //also Double.NEGATIVE_INFINITY
def inf = 1.0d / 0.0d; //also Double.POSITIVE_INFINITY
def nan = 0.0d / 0.0d; //also Double.NaN
def negZero = -2.0d / inf;
println(" Negative inf: " + negInf);
println(" Positive inf: " + inf);
println(" NaN: " + nan);
println(" Negative 0: " + negZero);
println(" inf + -inf: " + (inf + negInf));
println(" 0 * NaN: " + (0 * nan));
println(" NaN == NaN: " + (nan == nan));
println("NaN equals NaN: " + (nan.equals(nan))); |
http://rosettacode.org/wiki/Extreme_floating_point_values | Extreme floating point values | The IEEE floating point specification defines certain 'extreme' floating point values such as minus zero, -0.0, a value distinct from plus zero; not a number, NaN; and plus and minus infinity.
The task is to use expressions involving other 'normal' floating point values in your language to calculate these, (and maybe other), extreme floating point values in your language and assign them to variables.
Print the values of these variables if possible; and show some arithmetic with these values and variables.
If your language can directly enter these extreme floating point values then show it.
See also
What Every Computer Scientist Should Know About Floating-Point Arithmetic
Related tasks
Infinity
Detect division by zero
Literals/Floating point
| #haskell | haskell |
main = do
let inf = 1/0
let minus_inf = -1/0
let minus_zero = -1/inf
let nan = 0/0
putStrLn ("Positive infinity = "++(show inf))
putStrLn ("Negative infinity = "++(show minus_inf))
putStrLn ("Negative zero = "++(show minus_zero))
putStrLn ("Not a number = "++(show nan))
--Some Arithmetic
putStrLn ("inf + 2.0 = "++(show (inf+2.0)))
putStrLn ("inf - 10 = "++(show (inf-10)))
putStrLn ("inf - inf = "++(show (inf-inf)))
putStrLn ("inf * 0 = "++(show (inf * 0)))
putStrLn ("nan + 1.0= "++(show (nan+1.0)))
putStrLn ("nan + nan = "++(show (nan + nan)))
--Some Comparisons
putStrLn ("nan == nan = "++(show (nan == nan)))
putStrLn ("0.0 == - 0.0 = "++(show (0.0 == minus_zero)))
putStrLn ("inf == inf = "++(show (inf == inf)))
|
http://rosettacode.org/wiki/Extend_your_language | Extend your language | Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Some programming languages allow you to extend the language. While this can be done to a certain degree in most languages (e.g. by using macros), other languages go much further. Most notably in the Forth and Lisp families, programming per se is done by extending the language without any formal distinction between built-in and user-defined elements.
If your language supports it, show how to introduce a new flow control mechanism. A practical and useful example is a four-way branch:
Occasionally, code must be written that depends on two conditions, resulting in up to four branches (depending on whether both, only the first, only the second, or none of the conditions are "true"). In a C-like language this could look like the following:
if (condition1isTrue) {
if (condition2isTrue)
bothConditionsAreTrue();
else
firstConditionIsTrue();
}
else if (condition2isTrue)
secondConditionIsTrue();
else
noConditionIsTrue();
Besides being rather cluttered, the statement(s) for 'condition2isTrue' must be written down twice. If 'condition2isTrue' were a lengthy and involved expression, it would be quite unreadable, and the code generated by the compiler might be unnecessarily large.
This can be improved by introducing a new keyword if2. It is similar to if, but takes two conditional statements instead of one, and up to three 'else' statements. One proposal (in pseudo-C syntax) might be:
if2 (condition1isTrue) (condition2isTrue)
bothConditionsAreTrue();
else1
firstConditionIsTrue();
else2
secondConditionIsTrue();
else
noConditionIsTrue();
Pick the syntax which suits your language. The keywords 'else1' and 'else2' are just examples. The new conditional expression should look, nest and behave analogously to the language's built-in 'if' statement.
| #ALGOL_W | ALGOL W | begin
% executes pBoth, p1, p2 or pNeither %
% depending on whether c1 and c2 are true, c1 is true, c2 is true %
% neither c1 nor c2 are true %
procedure if2 ( logical value c1, c2
; procedure pBoth, p1, p2, pNeither
);
if c1 and c2 then pBoth
else if c1 then p1
else if c2 then p2
else pNeither
;
begin
logical a, b;
a := true;
b := false;
if2( a, b
, write( "both: a and b are TRUE" )
, write( "first: only a is TRUE" )
, write( "second: only b is TRUE" )
, write( "neither: a and b are FALSE" )
)
end
end. |
http://rosettacode.org/wiki/Extend_your_language | Extend your language | Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Some programming languages allow you to extend the language. While this can be done to a certain degree in most languages (e.g. by using macros), other languages go much further. Most notably in the Forth and Lisp families, programming per se is done by extending the language without any formal distinction between built-in and user-defined elements.
If your language supports it, show how to introduce a new flow control mechanism. A practical and useful example is a four-way branch:
Occasionally, code must be written that depends on two conditions, resulting in up to four branches (depending on whether both, only the first, only the second, or none of the conditions are "true"). In a C-like language this could look like the following:
if (condition1isTrue) {
if (condition2isTrue)
bothConditionsAreTrue();
else
firstConditionIsTrue();
}
else if (condition2isTrue)
secondConditionIsTrue();
else
noConditionIsTrue();
Besides being rather cluttered, the statement(s) for 'condition2isTrue' must be written down twice. If 'condition2isTrue' were a lengthy and involved expression, it would be quite unreadable, and the code generated by the compiler might be unnecessarily large.
This can be improved by introducing a new keyword if2. It is similar to if, but takes two conditional statements instead of one, and up to three 'else' statements. One proposal (in pseudo-C syntax) might be:
if2 (condition1isTrue) (condition2isTrue)
bothConditionsAreTrue();
else1
firstConditionIsTrue();
else2
secondConditionIsTrue();
else
noConditionIsTrue();
Pick the syntax which suits your language. The keywords 'else1' and 'else2' are just examples. The new conditional expression should look, nest and behave analogously to the language's built-in 'if' statement.
| #Arturo | Arturo | if2: function [cond1 cond2 both one two none][
case []
when? [and? cond1 cond2] -> do both
when? [cond1] -> do one
when? [cond2] -> do two
else -> do none
]
if2 false true [print "both"]
[print "only first"]
[print "only second"]
[print "none"] |
http://rosettacode.org/wiki/Exponentiation_order | Exponentiation order | This task will demonstrate the order of exponentiation (xy) when there are multiple exponents.
(Many programming languages, especially those with extended─precision integer arithmetic, usually support one of **, ^, ↑ or some such for exponentiation.)
Task requirements
Show the result of a language's evaluation of multiple exponentiation (either as an integer or floating point).
If your language's exponentiation operator is not one of the usual ones, please comment on how to recognize it.
Using whatever operator or syntax your language supports (if any), show the results in three lines (with identification):
5**3**2
(5**3)**2
5**(3**2)
If there are other methods (or formats) of multiple exponentiations, show them as well.
See also
MathWorld entry: exponentiation
Related tasks
exponentiation operator
arbitrary-precision integers (included)
Exponentiation with infix operators in (or operating on) the base
| #11l | 11l | print(5 ^ 3 ^ 2)
print((5 ^ 3) ^ 2)
print(5 ^ (3 ^ 2)) |
http://rosettacode.org/wiki/Factorial_base_numbers_indexing_permutations_of_a_collection | Factorial base numbers indexing permutations of a collection | You need a random arrangement of a deck of cards, you are sick of lame ways of doing this. This task is a super-cool way of doing this using factorial base numbers.
The first 25 factorial base numbers in increasing order are: 0.0.0, 0.0.1, 0.1.0, 0.1.1, 0.2.0, 0.2.1, 1.0.0, 1.0.1, 1.1.0, 1.1.1,1.2.0, 1.2.1, 2.0.0, 2.0.1, 2.1.0, 2.1.1, 2.2.0, 2.2.1, 3.0.0, 3.0.1, 3.1.0, 3.1.1, 3.2.0, 3.2.1, 1.0.0.0
Observe that the least significant digit is base 2 the next base 3, in general an n-digit factorial base number has digits n..1 in base n+1..2.
I want to produce a 1 to 1 mapping between these numbers and permutations:-
0.0.0 -> 0123
0.0.1 -> 0132
0.1.0 -> 0213
0.1.1 -> 0231
0.2.0 -> 0312
0.2.1 -> 0321
1.0.0 -> 1023
1.0.1 -> 1032
1.1.0 -> 1203
1.1.1 -> 1230
1.2.0 -> 1302
1.2.1 -> 1320
2.0.0 -> 2013
2.0.1 -> 2031
2.1.0 -> 2103
2.1.1 -> 2130
2.2.0 -> 2301
2.2.1 -> 2310
3.0.0 -> 3012
3.0.1 -> 3021
3.1.0 -> 3102
3.1.1 -> 3120
3.2.0 -> 3201
3.2.1 -> 3210
The following psudo-code will do this:
Starting with m=0 and Ω, an array of elements to be permutated, for each digit g starting with the most significant digit in the factorial base number.
If g is greater than zero, rotate the elements from m to m+g in Ω (see example)
Increment m and repeat the first step using the next most significant digit until the factorial base number is exhausted.
For example: using the factorial base number 2.0.1 and Ω = 0 1 2 3 where place 0 in both is the most significant (left-most) digit/element.
Step 1: m=0 g=2; Rotate places 0 through 2. 0 1 2 3 becomes 2 0 1 3
Step 2: m=1 g=0; No action.
Step 3: m=2 g=1; Rotate places 2 through 3. 2 0 1 3 becomes 2 0 3 1
Let me work 2.0.1 and 0123
step 1 n=0 g=2 Ω=2013
step 2 n=1 g=0 so no action
step 3 n=2 g=1 Ω=2031
The task:
First use your function to recreate the above table.
Secondly use your function to generate all permutaions of 11 digits, perhaps count them don't display them, compare this method with
methods in rc's permutations task.
Thirdly here following are two ramdom 51 digit factorial base numbers I prepared earlier:
39.49.7.47.29.30.2.12.10.3.29.37.33.17.12.31.29.34.17.25.2.4.25.4.1.14.20.6.21.18.1.1.1.4.0.5.15.12.4.3.10.10.9.1.6.5.5.3.0.0.0
51.48.16.22.3.0.19.34.29.1.36.30.12.32.12.29.30.26.14.21.8.12.1.3.10.4.7.17.6.21.8.12.15.15.13.15.7.3.12.11.9.5.5.6.6.3.4.0.3.2.1
use your function to crate the corresponding permutation of the following shoe of cards:
A♠K♠Q♠J♠10♠9♠8♠7♠6♠5♠4♠3♠2♠A♥K♥Q♥J♥10♥9♥8♥7♥6♥5♥4♥3♥2♥A♦K♦Q♦J♦10♦9♦8♦7♦6♦5♦4♦3♦2♦A♣K♣Q♣J♣10♣9♣8♣7♣6♣5♣4♣3♣2♣
Finally create your own 51 digit factorial base number and produce the corresponding permutation of the above shoe
| #zkl | zkl | fcn fpermute(omega,num){ // eg (0,1,2,3), (0,0,0)..(3,2,1)
omega=omega.copy(); // omega gonna be mutated
foreach m,g in ([0..].zip(num)){ if(g) omega.insert(m,omega.pop(m+g)) }
omega
} |
http://rosettacode.org/wiki/Factorions | Factorions |
Definition
A factorion is a natural number that equals the sum of the factorials of its digits.
Example
145 is a factorion in base 10 because:
1! + 4! + 5! = 1 + 24 + 120 = 145
It can be shown (see talk page) that no factorion in base 10 can exceed 1,499,999.
Task
Write a program in your language to demonstrate, by calculating and printing out the factorions, that:
There are 3 factorions in base 9
There are 4 factorions in base 10
There are 5 factorions in base 11
There are 2 factorions in base 12 (up to the same upper bound as for base 10)
See also
Wikipedia article
OEIS:A014080 - Factorions in base 10
OEIS:A193163 - Factorions in base n
| #J | J |
index=: $ #: I.@:,
factorion=: 10&$: :(] = [: +/ [: ! #.^:_1)&>
FACTORIONS=: 9 0 +"1 index Q=: 9 10 11 12 factorion/ i. 1500000
NB. base, factorion expressed in bases 10, and base
(,. ".@:((Num_j_,26}.Alpha_j_) {~ #.inv/)"1) FACTORIONS
9 1 1
9 2 2
9 41282 62558
10 1 1
10 2 2
10 145 145
10 40585 40585
11 1 1
11 2 2
11 26 24
11 48 44
11 40472 28453
12 1 1
12 2 2
NB. tallies of factorions in the bases
(9+i.4),.+/"1 Q
9 3
10 4
11 5
12 2
|
http://rosettacode.org/wiki/Factorions | Factorions |
Definition
A factorion is a natural number that equals the sum of the factorials of its digits.
Example
145 is a factorion in base 10 because:
1! + 4! + 5! = 1 + 24 + 120 = 145
It can be shown (see talk page) that no factorion in base 10 can exceed 1,499,999.
Task
Write a program in your language to demonstrate, by calculating and printing out the factorions, that:
There are 3 factorions in base 9
There are 4 factorions in base 10
There are 5 factorions in base 11
There are 2 factorions in base 12 (up to the same upper bound as for base 10)
See also
Wikipedia article
OEIS:A014080 - Factorions in base 10
OEIS:A193163 - Factorions in base n
| #Java | Java |
public class Factorion {
public static void main(String [] args){
System.out.println("Base 9:");
for(int i = 1; i <= 1499999; i++){
String iStri = String.valueOf(i);
int multiplied = operate(iStri,9);
if(multiplied == i){
System.out.print(i + "\t");
}
}
System.out.println("\nBase 10:");
for(int i = 1; i <= 1499999; i++){
String iStri = String.valueOf(i);
int multiplied = operate(iStri,10);
if(multiplied == i){
System.out.print(i + "\t");
}
}
System.out.println("\nBase 11:");
for(int i = 1; i <= 1499999; i++){
String iStri = String.valueOf(i);
int multiplied = operate(iStri,11);
if(multiplied == i){
System.out.print(i + "\t");
}
}
System.out.println("\nBase 12:");
for(int i = 1; i <= 1499999; i++){
String iStri = String.valueOf(i);
int multiplied = operate(iStri,12);
if(multiplied == i){
System.out.print(i + "\t");
}
}
}
public static int factorialRec(int n){
int result = 1;
return n == 0 ? result : result * n * factorialRec(n-1);
}
public static int operate(String s, int base){
int sum = 0;
String strx = fromDeci(base, Integer.parseInt(s));
for(int i = 0; i < strx.length(); i++){
if(strx.charAt(i) == 'A'){
sum += factorialRec(10);
}else if(strx.charAt(i) == 'B') {
sum += factorialRec(11);
}else if(strx.charAt(i) == 'C') {
sum += factorialRec(12);
}else {
sum += factorialRec(Integer.parseInt(String.valueOf(strx.charAt(i)), base));
}
}
return sum;
}
// Ln 57-71 from Geeks for Geeks @ https://www.geeksforgeeks.org/convert-base-decimal-vice-versa/
static char reVal(int num) {
if (num >= 0 && num <= 9)
return (char)(num + 48);
else
return (char)(num - 10 + 65);
}
static String fromDeci(int base, int num){
StringBuilder s = new StringBuilder();
while (num > 0) {
s.append(reVal(num % base));
num /= base;
}
return new String(new StringBuilder(s).reverse());
}
}
|
http://rosettacode.org/wiki/Filter | Filter | Task
Select certain elements from an Array into a new Array in a generic way.
To demonstrate, select all even numbers from an Array.
As an option, give a second solution which filters destructively,
by modifying the original Array rather than creating a new Array.
| #Toka | Toka | 10 cells is-array table
10 cells is-array even
{
variable source
[ swap source ! >r reset r> 0
[ i source @ array.get
dup 2 mod 0 <> [ drop ] ifTrue
] countedLoop
depth 0 swap [ i even array.put ] countedLoop
]
} is copy-even
10 0 [ i i table array.put ] countedLoop
table 10 copy-even |
http://rosettacode.org/wiki/FizzBuzz | FizzBuzz | Task
Write a program that prints the integers from 1 to 100 (inclusive).
But:
for multiples of three, print Fizz (instead of the number)
for multiples of five, print Buzz (instead of the number)
for multiples of both three and five, print FizzBuzz (instead of the number)
The FizzBuzz problem was presented as the lowest level of comprehension required to illustrate adequacy.
Also see
(a blog) dont-overthink-fizzbuzz
(a blog) fizzbuzz-the-programmers-stairway-to-heaven
| #PL.2FM | PL/M | 100H:
/* DECLARE OUTPUT IN TERMS OF CP/M -
PL/M DOES NOT COME WITH ANY STANDARD LIBRARY */
BDOS: PROCEDURE(FUNC, ARG);
DECLARE FUNC BYTE, ARG ADDRESS;
GO TO 5;
END BDOS;
PUT$STRING: PROCEDURE(STR);
DECLARE STR ADDRESS;
CALL BDOS(9, STR);
CALL BDOS(9, .(13,10,'$'));
END PUT$STRING;
/* PRINT A NUMBER */
PUT$NUMBER: PROCEDURE(N);
DECLARE S (5) BYTE INITIAL ('...$');
DECLARE P ADDRESS;
DECLARE (N, D, C BASED P) BYTE;
P = .S(3);
DIGIT:
P = P-1;
C = (N MOD 10) + '0';
N = N/10;
IF N > 0 THEN GO TO DIGIT;
CALL PUT$STRING(P);
END PUT$NUMBER;
/* FIZZBUZZ */
DECLARE N BYTE;
DO N = 1 TO 100;
IF N MOD 15 = 0 THEN
CALL PUT$STRING(.'FIZZBUZZ$');
ELSE IF N MOD 5 = 0 THEN
CALL PUT$STRING(.'BUZZ$');
ELSE IF N MOD 3 = 0 THEN
CALL PUT$STRING(.'FIZZ$');
ELSE
CALL PUT$NUMBER(N);
END;
/* EXIT TO CP/M */
CALL BDOS(0,0);
EOF |
http://rosettacode.org/wiki/Extensible_prime_generator | Extensible prime generator | Task
Write a generator of prime numbers, in order, that will automatically adjust to accommodate the generation of any reasonably high prime.
The routine should demonstrably rely on either:
Being based on an open-ended counter set to count without upper limit other than system or programming language limits. In this case, explain where this counter is in the code.
Being based on a limit that is extended automatically. In this case, choose a small limit that ensures the limit will be passed when generating some of the values to be asked for below.
If other methods of creating an extensible prime generator are used, the algorithm's means of extensibility/lack of limits should be stated.
The routine should be used to:
Show the first twenty primes.
Show the primes between 100 and 150.
Show the number of primes between 7,700 and 8,000.
Show the 10,000th prime.
Show output on this page.
Note: You may reference code already on this site if it is written to be imported/included, then only the code necessary for import and the performance of this task need be shown. (It is also important to leave a forward link on the referenced tasks entry so that later editors know that the code is used for multiple tasks).
Note 2: If a languages in-built prime generator is extensible or is guaranteed to generate primes up to a system limit, (231 or memory overflow for example), then this may be used as long as an explanation of the limits of the prime generator is also given. (Which may include a link to/excerpt from, language documentation).
Note 3:The task is written so it may be useful in solving the task Emirp primes as well as others (depending on its efficiency).
Reference
Prime Numbers. Website with large count of primes.
| #BQN | BQN | # Function that returns a new prime generator
PrimeGen ← {𝕤
i ← 0 # Counter: index of next prime to be output
primes ← ↕0
next ← 2
Sieve ← { p 𝕊 i‿n:
E ← {↕∘⌈⌾(((𝕩|-i)+𝕩×⊢)⁼)n-i} # Indices of multiples of 𝕩
i + / (1⥊˜n-i) E⊸{0¨⌾(𝕨⊸⊏)𝕩}´ p # Primes in segment [i,n)
}
{𝕤
{ i=≠primes ? # Extend if required
next ↩ ((2⋆24)⊸+ ⌊ ט) old←next # Sieve at most 16M new entries
primes ∾↩ (primes(⍋↑⊣)√next) Sieve old‿next
;@}
(i+↩1) ⊢ i⊑primes
}
}
_w_←{𝔽⍟𝔾∘𝔽_𝕣_𝔾∘𝔽⍟𝔾𝕩} # Looping utility for the session below |
http://rosettacode.org/wiki/Fibonacci_sequence | Fibonacci sequence | The Fibonacci sequence is a sequence Fn of natural numbers defined recursively:
F0 = 0
F1 = 1
Fn = Fn-1 + Fn-2, if n>1
Task
Write a function to generate the nth Fibonacci number.
Solutions can be iterative or recursive (though recursive solutions are generally considered too slow and are mostly used as an exercise in recursion).
The sequence is sometimes extended into negative numbers by using a straightforward inverse of the positive definition:
Fn = Fn+2 - Fn+1, if n<0
support for negative n in the solution is optional.
Related tasks
Fibonacci n-step number sequences
Leonardo numbers
References
Wikipedia, Fibonacci number
Wikipedia, Lucas number
MathWorld, Fibonacci Number
Some identities for r-Fibonacci numbers
OEIS Fibonacci numbers
OEIS Lucas numbers
| #FreeBASIC | FreeBASIC | 'Fibonacci extended
'Freebasic version 24 Windows
Dim Shared ADDQmod(0 To 19) As Ubyte
Dim Shared ADDbool(0 To 19) As Ubyte
For z As Integer=0 To 19
ADDQmod(z)=(z Mod 10+48)
ADDbool(z)=(-(10<=z))
Next z
Function plusINT(NUM1 As String,NUM2 As String) As String
Dim As Byte flag
#macro finish()
three=Ltrim(three,"0")
If three="" Then Return "0"
If flag=1 Then Swap NUM2,NUM1
Return three
Exit Function
#endmacro
var lenf=Len(NUM1)
var lens=Len(NUM2)
If lens>lenf Then
Swap NUM2,NUM1
Swap lens,lenf
flag=1
End If
var diff=lenf-lens-Sgn(lenf-lens)
var three="0"+NUM1
var two=String(lenf-lens,"0")+NUM2
Dim As Integer n2
Dim As Ubyte addup,addcarry
addcarry=0
For n2=lenf-1 To diff Step -1
addup=two[n2]+NUM1[n2]-96
three[n2+1]=addQmod(addup+addcarry)
addcarry=addbool(addup+addcarry)
Next n2
If addcarry=0 Then
finish()
End If
If n2=-1 Then
three[0]=addcarry+48
finish()
End If
For n2=n2 To 0 Step -1
addup=two[n2]+NUM1[n2]-96
three[n2+1]=addQmod(addup+addcarry)
addcarry=addbool(addup+addcarry)
Next n2
three[0]=addcarry+48
finish()
End Function
Function fibonacci(n As Integer) As String
Dim As String sl,l,term
sl="0": l="1"
If n=1 Then Return "0"
If n=2 Then Return "1"
n=n-2
For x As Integer= 1 To n
term=plusINT(l,sl)
sl=l
l=term
Next x
Function =term
End Function
'============== EXAMPLE ===============
print "THE SEQUENCE TO 10:"
print
For n As Integer=1 To 10
Print "term";n;": "; fibonacci(n)
Next n
print
print "Selected Fibonacci number"
print "Fibonacci 500"
print
print fibonacci(500)
Sleep |
http://rosettacode.org/wiki/Factors_of_an_integer | Factors of an integer |
Basic Data Operation
This is a basic data operation. It represents a fundamental action on a basic data type.
You may see other such operations in the Basic Data Operations category, or:
Integer Operations
Arithmetic |
Comparison
Boolean Operations
Bitwise |
Logical
String Operations
Concatenation |
Interpolation |
Comparison |
Matching
Memory Operations
Pointers & references |
Addresses
Task
Compute the factors of a positive integer.
These factors are the positive integers by which the number being factored can be divided to yield a positive integer result.
(Though the concepts function correctly for zero and negative integers, the set of factors of zero has countably infinite members, and the factors of negative integers can be obtained from the factors of related positive numbers without difficulty; this task does not require handling of either of these cases).
Note that every prime number has two factors: 1 and itself.
Related tasks
count in factors
prime decomposition
Sieve of Eratosthenes
primality by trial division
factors of a Mersenne number
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
sequence: smallest number greater than previous term with exactly n divisors
| #Phixmonti | Phixmonti | /# Rosetta Code problem: http://rosettacode.org/wiki/Factors_of_an_integer
by Galileo, 05/2022 #/
include ..\Utilitys.pmt
def Factors >ps
( ( 1 tps 2 / ) for tps over mod if drop endif endfor ps> )
enddef
11 Factors
21 Factors
32 factors
45 factors
67 factors
96 factors
pstack |
http://rosettacode.org/wiki/Extreme_floating_point_values | Extreme floating point values | The IEEE floating point specification defines certain 'extreme' floating point values such as minus zero, -0.0, a value distinct from plus zero; not a number, NaN; and plus and minus infinity.
The task is to use expressions involving other 'normal' floating point values in your language to calculate these, (and maybe other), extreme floating point values in your language and assign them to variables.
Print the values of these variables if possible; and show some arithmetic with these values and variables.
If your language can directly enter these extreme floating point values then show it.
See also
What Every Computer Scientist Should Know About Floating-Point Arithmetic
Related tasks
Infinity
Detect division by zero
Literals/Floating point
| #Icon_and_Unicon | Icon and Unicon | Inf=: _
NegInf=: __
NB. Negative zero cannot be represented in J to be distinct from 0.
NaN=. _. |
http://rosettacode.org/wiki/Extreme_floating_point_values | Extreme floating point values | The IEEE floating point specification defines certain 'extreme' floating point values such as minus zero, -0.0, a value distinct from plus zero; not a number, NaN; and plus and minus infinity.
The task is to use expressions involving other 'normal' floating point values in your language to calculate these, (and maybe other), extreme floating point values in your language and assign them to variables.
Print the values of these variables if possible; and show some arithmetic with these values and variables.
If your language can directly enter these extreme floating point values then show it.
See also
What Every Computer Scientist Should Know About Floating-Point Arithmetic
Related tasks
Infinity
Detect division by zero
Literals/Floating point
| #J | J | Inf=: _
NegInf=: __
NB. Negative zero cannot be represented in J to be distinct from 0.
NaN=. _. |
http://rosettacode.org/wiki/Extend_your_language | Extend your language | Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Some programming languages allow you to extend the language. While this can be done to a certain degree in most languages (e.g. by using macros), other languages go much further. Most notably in the Forth and Lisp families, programming per se is done by extending the language without any formal distinction between built-in and user-defined elements.
If your language supports it, show how to introduce a new flow control mechanism. A practical and useful example is a four-way branch:
Occasionally, code must be written that depends on two conditions, resulting in up to four branches (depending on whether both, only the first, only the second, or none of the conditions are "true"). In a C-like language this could look like the following:
if (condition1isTrue) {
if (condition2isTrue)
bothConditionsAreTrue();
else
firstConditionIsTrue();
}
else if (condition2isTrue)
secondConditionIsTrue();
else
noConditionIsTrue();
Besides being rather cluttered, the statement(s) for 'condition2isTrue' must be written down twice. If 'condition2isTrue' were a lengthy and involved expression, it would be quite unreadable, and the code generated by the compiler might be unnecessarily large.
This can be improved by introducing a new keyword if2. It is similar to if, but takes two conditional statements instead of one, and up to three 'else' statements. One proposal (in pseudo-C syntax) might be:
if2 (condition1isTrue) (condition2isTrue)
bothConditionsAreTrue();
else1
firstConditionIsTrue();
else2
secondConditionIsTrue();
else
noConditionIsTrue();
Pick the syntax which suits your language. The keywords 'else1' and 'else2' are just examples. The new conditional expression should look, nest and behave analogously to the language's built-in 'if' statement.
| #ATS | ATS | (* Languages with pattern matching ALREADY HAVE THIS! *)
fn
func (pred1 : bool, pred2 : bool) : void =
case+ (pred1, pred2) of
| (true, true) => println! ("(true, true)")
| (true, false) => println! ("(true, false)")
| (false, true) => println! ("(false, true)")
| (false, false) => println! ("(false, false)")
implement
main0 () =
begin
func (true, true);
func (true, false);
func (false, true);
func (false, false)
end |
http://rosettacode.org/wiki/Extend_your_language | Extend your language | Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Some programming languages allow you to extend the language. While this can be done to a certain degree in most languages (e.g. by using macros), other languages go much further. Most notably in the Forth and Lisp families, programming per se is done by extending the language without any formal distinction between built-in and user-defined elements.
If your language supports it, show how to introduce a new flow control mechanism. A practical and useful example is a four-way branch:
Occasionally, code must be written that depends on two conditions, resulting in up to four branches (depending on whether both, only the first, only the second, or none of the conditions are "true"). In a C-like language this could look like the following:
if (condition1isTrue) {
if (condition2isTrue)
bothConditionsAreTrue();
else
firstConditionIsTrue();
}
else if (condition2isTrue)
secondConditionIsTrue();
else
noConditionIsTrue();
Besides being rather cluttered, the statement(s) for 'condition2isTrue' must be written down twice. If 'condition2isTrue' were a lengthy and involved expression, it would be quite unreadable, and the code generated by the compiler might be unnecessarily large.
This can be improved by introducing a new keyword if2. It is similar to if, but takes two conditional statements instead of one, and up to three 'else' statements. One proposal (in pseudo-C syntax) might be:
if2 (condition1isTrue) (condition2isTrue)
bothConditionsAreTrue();
else1
firstConditionIsTrue();
else2
secondConditionIsTrue();
else
noConditionIsTrue();
Pick the syntax which suits your language. The keywords 'else1' and 'else2' are just examples. The new conditional expression should look, nest and behave analogously to the language's built-in 'if' statement.
| #C | C | /* Four-way branch.
*
* if2 (firsttest, secondtest
* , bothtrue
* , firstrue
* , secondtrue
* , bothfalse
* )
*/
#define if2(firsttest,secondtest,bothtrue,firsttrue,secondtrue,bothfalse)\
switch(((firsttest)?0:2)+((secondtest)?0:1)) {\
case 0: bothtrue; break;\
case 1: firsttrue; break;\
case 2: secondtrue; break;\
case 3: bothfalse; break;\
}
|
http://rosettacode.org/wiki/Exponentiation_order | Exponentiation order | This task will demonstrate the order of exponentiation (xy) when there are multiple exponents.
(Many programming languages, especially those with extended─precision integer arithmetic, usually support one of **, ^, ↑ or some such for exponentiation.)
Task requirements
Show the result of a language's evaluation of multiple exponentiation (either as an integer or floating point).
If your language's exponentiation operator is not one of the usual ones, please comment on how to recognize it.
Using whatever operator or syntax your language supports (if any), show the results in three lines (with identification):
5**3**2
(5**3)**2
5**(3**2)
If there are other methods (or formats) of multiple exponentiations, show them as well.
See also
MathWorld entry: exponentiation
Related tasks
exponentiation operator
arbitrary-precision integers (included)
Exponentiation with infix operators in (or operating on) the base
| #Action.21 | Action! | INCLUDE "D2:REAL.ACT" ;from the Action! Tool Kit
PROC Main()
REAL r2,r3,r5,tmp1,tmp2
Put(125) PutE() ;clear screen
IntToReal(2,r2)
IntToReal(3,r3)
IntToReal(5,r5)
PrintE("There is no power operator in Action!")
PrintE("Power function for REAL type is used.")
PrintE("But the precision is insufficient.")
Power(r5,r3,tmp1)
Power(tmp1,r2,tmp2)
Print("(5^3)^2=")
PrintRE(tmp2)
Power(r3,r2,tmp1)
Power(r5,tmp1,tmp2)
Print("5^(3^2)=")
PrintRE(tmp2)
RETURN |
http://rosettacode.org/wiki/Exponentiation_order | Exponentiation order | This task will demonstrate the order of exponentiation (xy) when there are multiple exponents.
(Many programming languages, especially those with extended─precision integer arithmetic, usually support one of **, ^, ↑ or some such for exponentiation.)
Task requirements
Show the result of a language's evaluation of multiple exponentiation (either as an integer or floating point).
If your language's exponentiation operator is not one of the usual ones, please comment on how to recognize it.
Using whatever operator or syntax your language supports (if any), show the results in three lines (with identification):
5**3**2
(5**3)**2
5**(3**2)
If there are other methods (or formats) of multiple exponentiations, show them as well.
See also
MathWorld entry: exponentiation
Related tasks
exponentiation operator
arbitrary-precision integers (included)
Exponentiation with infix operators in (or operating on) the base
| #Ada | Ada | with Ada.Text_IO;
procedure Exponentation_Order is
use Ada.Text_IO;
begin
-- Put_Line ("5**3**2 : " & Natural'(5**3**2)'Image);
Put_Line ("(5**3)**2 : " & Natural'((5**3)**2)'Image);
Put_Line ("5**(3**2) : " & Natural'(5**(3**2))'Image);
end Exponentation_Order; |
http://rosettacode.org/wiki/Exponentiation_order | Exponentiation order | This task will demonstrate the order of exponentiation (xy) when there are multiple exponents.
(Many programming languages, especially those with extended─precision integer arithmetic, usually support one of **, ^, ↑ or some such for exponentiation.)
Task requirements
Show the result of a language's evaluation of multiple exponentiation (either as an integer or floating point).
If your language's exponentiation operator is not one of the usual ones, please comment on how to recognize it.
Using whatever operator or syntax your language supports (if any), show the results in three lines (with identification):
5**3**2
(5**3)**2
5**(3**2)
If there are other methods (or formats) of multiple exponentiations, show them as well.
See also
MathWorld entry: exponentiation
Related tasks
exponentiation operator
arbitrary-precision integers (included)
Exponentiation with infix operators in (or operating on) the base
| #ALGOL_68 | ALGOL 68 | print( ( "5**3**2: ", 5**3**2, newline ) );
print( ( "(5**3)**2: ", (5**3)**2, newline ) );
print( ( "5**(3**2): ", 5**(3**2), newline ) ) |
http://rosettacode.org/wiki/Factorions | Factorions |
Definition
A factorion is a natural number that equals the sum of the factorials of its digits.
Example
145 is a factorion in base 10 because:
1! + 4! + 5! = 1 + 24 + 120 = 145
It can be shown (see talk page) that no factorion in base 10 can exceed 1,499,999.
Task
Write a program in your language to demonstrate, by calculating and printing out the factorions, that:
There are 3 factorions in base 9
There are 4 factorions in base 10
There are 5 factorions in base 11
There are 2 factorions in base 12 (up to the same upper bound as for base 10)
See also
Wikipedia article
OEIS:A014080 - Factorions in base 10
OEIS:A193163 - Factorions in base n
| #Julia | Julia | isfactorian(n, base) = mapreduce(factorial, +, map(c -> parse(Int, c, base=16), split(string(n, base=base), ""))) == n
printallfactorian(base) = println("Factorians for base $base: ", [n for n in 1:100000 if isfactorian(n, base)])
foreach(printallfactorian, 9:12)
|
http://rosettacode.org/wiki/Factorions | Factorions |
Definition
A factorion is a natural number that equals the sum of the factorials of its digits.
Example
145 is a factorion in base 10 because:
1! + 4! + 5! = 1 + 24 + 120 = 145
It can be shown (see talk page) that no factorion in base 10 can exceed 1,499,999.
Task
Write a program in your language to demonstrate, by calculating and printing out the factorions, that:
There are 3 factorions in base 9
There are 4 factorions in base 10
There are 5 factorions in base 11
There are 2 factorions in base 12 (up to the same upper bound as for base 10)
See also
Wikipedia article
OEIS:A014080 - Factorions in base 10
OEIS:A193163 - Factorions in base n
| #Mathematica_.2F_Wolfram_Language | Mathematica / Wolfram Language | ClearAll[FactorionQ]
FactorionQ[n_,b_:10]:=Total[IntegerDigits[n,b]!]==n
Select[Range[1500000],FactorionQ[#,9]&]
Select[Range[1500000],FactorionQ[#,10]&]
Select[Range[1500000],FactorionQ[#,11]&]
Select[Range[1500000],FactorionQ[#,12]&] |
http://rosettacode.org/wiki/Factorions | Factorions |
Definition
A factorion is a natural number that equals the sum of the factorials of its digits.
Example
145 is a factorion in base 10 because:
1! + 4! + 5! = 1 + 24 + 120 = 145
It can be shown (see talk page) that no factorion in base 10 can exceed 1,499,999.
Task
Write a program in your language to demonstrate, by calculating and printing out the factorions, that:
There are 3 factorions in base 9
There are 4 factorions in base 10
There are 5 factorions in base 11
There are 2 factorions in base 12 (up to the same upper bound as for base 10)
See also
Wikipedia article
OEIS:A014080 - Factorions in base 10
OEIS:A193163 - Factorions in base n
| #Nim | Nim | from math import fac
from strutils import join
iterator digits(n, base: Natural): Natural =
## Yield the digits of "n" in base "base".
var n = n
while true:
yield n mod base
n = n div base
if n == 0: break
func isFactorion(n, base: Natural): bool =
## Return true if "n" is a factorion for base "base".
var s = 0
for d in n.digits(base):
inc s, fac(d)
result = s == n
func factorions(base, limit: Natural): seq[Natural] =
## Return the list of factorions for base "base" up to "limit".
for n in 1..limit:
if n.isFactorion(base):
result.add(n)
for base in 9..12:
echo "Factorions for base ", base, ':'
echo factorions(base, 1_500_000 - 1).join(" ") |
http://rosettacode.org/wiki/Filter | Filter | Task
Select certain elements from an Array into a new Array in a generic way.
To demonstrate, select all even numbers from an Array.
As an option, give a second solution which filters destructively,
by modifying the original Array rather than creating a new Array.
| #TUSCRIPT | TUSCRIPT |
$$ MODE TUSCRIPT
arr="1'4'9'16'25'36'49'64'81'100",even=""
LOOP nr=arr
rest=MOD (nr,2)
IF (rest==0) even=APPEND (even,nr)
ENDLOOP
PRINT even
|
http://rosettacode.org/wiki/FizzBuzz | FizzBuzz | Task
Write a program that prints the integers from 1 to 100 (inclusive).
But:
for multiples of three, print Fizz (instead of the number)
for multiples of five, print Buzz (instead of the number)
for multiples of both three and five, print FizzBuzz (instead of the number)
The FizzBuzz problem was presented as the lowest level of comprehension required to illustrate adequacy.
Also see
(a blog) dont-overthink-fizzbuzz
(a blog) fizzbuzz-the-programmers-stairway-to-heaven
| #PL.2FSQL | PL/SQL | BEGIN
FOR i IN 1 .. 100
LOOP
CASE
WHEN MOD(i, 15) = 0 THEN
DBMS_OUTPUT.put_line('FizzBuzz');
WHEN MOD(i, 5) = 0 THEN
DBMS_OUTPUT.put_line('Buzz');
WHEN MOD(i, 3) = 0 THEN
DBMS_OUTPUT.put_line('Fizz');
ELSE
DBMS_OUTPUT.put_line(i);
END CASE;
END LOOP;
END; |
http://rosettacode.org/wiki/Extensible_prime_generator | Extensible prime generator | Task
Write a generator of prime numbers, in order, that will automatically adjust to accommodate the generation of any reasonably high prime.
The routine should demonstrably rely on either:
Being based on an open-ended counter set to count without upper limit other than system or programming language limits. In this case, explain where this counter is in the code.
Being based on a limit that is extended automatically. In this case, choose a small limit that ensures the limit will be passed when generating some of the values to be asked for below.
If other methods of creating an extensible prime generator are used, the algorithm's means of extensibility/lack of limits should be stated.
The routine should be used to:
Show the first twenty primes.
Show the primes between 100 and 150.
Show the number of primes between 7,700 and 8,000.
Show the 10,000th prime.
Show output on this page.
Note: You may reference code already on this site if it is written to be imported/included, then only the code necessary for import and the performance of this task need be shown. (It is also important to leave a forward link on the referenced tasks entry so that later editors know that the code is used for multiple tasks).
Note 2: If a languages in-built prime generator is extensible or is guaranteed to generate primes up to a system limit, (231 or memory overflow for example), then this may be used as long as an explanation of the limits of the prime generator is also given. (Which may include a link to/excerpt from, language documentation).
Note 3:The task is written so it may be useful in solving the task Emirp primes as well as others (depending on its efficiency).
Reference
Prime Numbers. Website with large count of primes.
| #C | C | #include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define CHUNK_BYTES (32 << 8)
#define CHUNK_SIZE (CHUNK_BYTES << 6)
int field[CHUNK_BYTES];
#define GET(x) (field[(x)>>6] & 1<<((x)>>1&31))
#define SET(x) (field[(x)>>6] |= 1<<((x)>>1&31))
typedef unsigned uint;
typedef struct {
uint *e;
uint cap, len;
} uarray;
uarray primes, offset;
void push(uarray *a, uint n)
{
if (a->len >= a->cap) {
if (!(a->cap *= 2)) a->cap = 16;
a->e = realloc(a->e, sizeof(uint) * a->cap);
}
a->e[a->len++] = n;
}
uint low;
void init(void)
{
uint p, q;
unsigned char f[1<<16];
memset(f, 0, sizeof(f));
push(&primes, 2);
push(&offset, 0);
for (p = 3; p < 1<<16; p += 2) {
if (f[p]) continue;
for (q = p*p; q < 1<<16; q += 2*p) f[q] = 1;
push(&primes, p);
push(&offset, q);
}
low = 1<<16;
}
void sieve(void)
{
uint i, p, q, hi, ptop;
if (!low) init();
memset(field, 0, sizeof(field));
hi = low + CHUNK_SIZE;
ptop = sqrt(hi) * 2 + 1;
for (i = 1; (p = primes.e[i]*2) < ptop; i++) {
for (q = offset.e[i] - low; q < CHUNK_SIZE; q += p)
SET(q);
offset.e[i] = q + low;
}
for (p = 1; p < CHUNK_SIZE; p += 2)
if (!GET(p)) push(&primes, low + p);
low = hi;
}
int main(void)
{
uint i, p, c;
while (primes.len < 20) sieve();
printf("First 20:");
for (i = 0; i < 20; i++)
printf(" %u", primes.e[i]);
putchar('\n');
while (primes.e[primes.len-1] < 150) sieve();
printf("Between 100 and 150:");
for (i = 0; i < primes.len; i++) {
if ((p = primes.e[i]) >= 100 && p < 150)
printf(" %u", primes.e[i]);
}
putchar('\n');
while (primes.e[primes.len-1] < 8000) sieve();
for (i = c = 0; i < primes.len; i++)
if ((p = primes.e[i]) >= 7700 && p < 8000) c++;
printf("%u primes between 7700 and 8000\n", c);
for (c = 10; c <= 100000000; c *= 10) {
while (primes.len < c) sieve();
printf("%uth prime: %u\n", c, primes.e[c-1]);
}
return 0;
} |
http://rosettacode.org/wiki/Fibonacci_sequence | Fibonacci sequence | The Fibonacci sequence is a sequence Fn of natural numbers defined recursively:
F0 = 0
F1 = 1
Fn = Fn-1 + Fn-2, if n>1
Task
Write a function to generate the nth Fibonacci number.
Solutions can be iterative or recursive (though recursive solutions are generally considered too slow and are mostly used as an exercise in recursion).
The sequence is sometimes extended into negative numbers by using a straightforward inverse of the positive definition:
Fn = Fn+2 - Fn+1, if n<0
support for negative n in the solution is optional.
Related tasks
Fibonacci n-step number sequences
Leonardo numbers
References
Wikipedia, Fibonacci number
Wikipedia, Lucas number
MathWorld, Fibonacci Number
Some identities for r-Fibonacci numbers
OEIS Fibonacci numbers
OEIS Lucas numbers
| #Frink | Frink |
fibonacciN[n] :=
{
a = 0
b = 1
count = 0
while count < n
{
[a,b] = [b, a + b]
count = count + 1
}
return a
}
|
http://rosettacode.org/wiki/Factors_of_an_integer | Factors of an integer |
Basic Data Operation
This is a basic data operation. It represents a fundamental action on a basic data type.
You may see other such operations in the Basic Data Operations category, or:
Integer Operations
Arithmetic |
Comparison
Boolean Operations
Bitwise |
Logical
String Operations
Concatenation |
Interpolation |
Comparison |
Matching
Memory Operations
Pointers & references |
Addresses
Task
Compute the factors of a positive integer.
These factors are the positive integers by which the number being factored can be divided to yield a positive integer result.
(Though the concepts function correctly for zero and negative integers, the set of factors of zero has countably infinite members, and the factors of negative integers can be obtained from the factors of related positive numbers without difficulty; this task does not require handling of either of these cases).
Note that every prime number has two factors: 1 and itself.
Related tasks
count in factors
prime decomposition
Sieve of Eratosthenes
primality by trial division
factors of a Mersenne number
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
sequence: smallest number greater than previous term with exactly n divisors
| #PHP | PHP | function GetFactors($n){
$factors = array(1, $n);
for($i = 2; $i * $i <= $n; $i++){
if($n % $i == 0){
$factors[] = $i;
if($i * $i != $n)
$factors[] = $n/$i;
}
}
sort($factors);
return $factors;
} |
http://rosettacode.org/wiki/Execute_Computer/Zero | Execute Computer/Zero | Execute Computer/Zero is an implementation of Computer/zero Assembly.
Other implementations of Computer/zero Assembly.
Task
Create a Computer/zero Assembly emulator. You may consider this webpage as a reference implementation. Output the results of the sample programs "2+2" and "7*8" found there.
The virtual machine "bytecode" needs to be able to modify itself (or at least act as though it can) while the virtual machine is running, to be consistent with the reference implementation.
For output, it is sufficient to have the implementation of the STP opcode return the accumulator to your actual language, and then use your standard printing routines to output it.
Bonus Points
Run all 5 sample programs at the aforementioned website and output their results.
| #ALGOL_68 | ALGOL 68 | BEGIN # execute some ComputerZero programs #
# instructions #
INT nop = 0, lda = 1, sta = 2, add = 3, sub = 4, brz = 5, jmp = 6, stp = 7;
PROC instr = ( INT op, v )INT: ( 32 * op ) + v;
OP NOP = ( INT v )INT: instr( nop, v );
OP LDA = ( INT v )INT: instr( lda, v );
OP STA = ( INT v )INT: instr( sta, v );
OP ADD = ( INT v )INT: instr( add, v );
OP SUB = ( INT v )INT: instr( sub, v );
OP BRZ = ( INT v )INT: instr( brz, v );
OP JMP = ( INT v )INT: instr( jmp, v );
OP STP = ( INT v )INT: instr( stp, v );
# executes the program named name #
PROC execute = ( STRING name, []INT program )VOID:
BEGIN
[ 0 : 31 ]INT m; # the computer 32 has bytes of memory #
FOR i FROM LWB m TO UPB m DO m[ i ] := 0 OD;
# "assemble" the program #
INT m pos := -1;
FOR i FROM LWB program TO UPB program DO
m[ m pos +:= 1 ] := program[ i ]
OD;
# execute the program #
BOOL running := TRUE;
INT pc := 0;
INT a := 0;
WHILE running DO
INT op := m[ pc ] OVER 32;
INT operand := m[ pc ] MOD 32;
pc +:= 1 MODAB 32;
IF op = nop THEN SKIP
ELIF op = lda THEN a := m[ operand ]
ELIF op = sta THEN m[ operand ] := a
ELIF op = add THEN a +:= m[ operand ] MODAB 256
ELIF op = sub THEN a -:= m[ operand ] MODAB 256
ELIF op = brz THEN IF a = 0 THEN pc := operand FI
ELIF op = jmp THEN pc := operand
ELSE # stp #
running := FALSE;
print( ( " " * ( 12 - ( ( UPB name - LWB name ) + 1 ) ) ) );
print( ( name, ": ", whole( a, -3 ), newline ) )
FI
OD
END # execute # ;
# task test programs (from the Computer Zero website) #
# the unary NOP, LDA, STA, etc. operators are used to construct the #
# instructions; as parameterless operators aren't allowed, NOP and STP #
# must have a dummy parameter #
execute( "2+2", ( LDA 3, ADD 4, STP 0, 2, 2 )
);
execute( "7*8"
, ( LDA 12, ADD 10, STA 12, LDA 11, SUB 13, STA 11, BRZ 8, JMP 0
, LDA 12, STP 0, 8, 7, 0, 1
)
);
execute( "fibonacci"
, ( LDA 14, STA 15, ADD 13, STA 14, LDA 15, STA 13, LDA 16, SUB 17
, BRZ 11, STA 16, JMP 0, LDA 14, STP 0, 1, 1, 0
, 8, 1
)
);
execute( "linkedList"
, ( LDA 13, ADD 15, STA 5, ADD 16, STA 7, NOP 0, STA 14, NOP 0
, BRZ 11, STA 15, JMP 0, LDA 14, STP 0, LDA 0, 0, 28
, 1, 0, 0, 0, 6, 0, 2, 26
, 5, 20, 3, 30, 1, 22, 4, 24
)
);
execute( "prisoner"
, ( NOP 0, NOP 0, STP 0, 0, LDA 3, SUB 29, BRZ 18, LDA 3
, STA 29, BRZ 14, LDA 1, ADD 31, STA 1, JMP 2, LDA 0, ADD 31
, STA 0, JMP 2, LDA 3, STA 29, LDA 1, ADD 30, ADD 3, STA 1
, LDA 0, ADD 30, ADD 3, STA 0, JMP 2, 0, 1, 3
)
);
# subtractions yielding negative results #
execute( "0-255", ( LDA 3, SUB 4, STP 0, 0, 255 ) );
execute( "0-1", ( LDA 3, SUB 4, STP 0, 0, 1 ) );
# overflow on addition #
execute( "1+255", ( LDA 3, ADD 4, STP 0, 1, 255 ) )
END
|
http://rosettacode.org/wiki/Extreme_floating_point_values | Extreme floating point values | The IEEE floating point specification defines certain 'extreme' floating point values such as minus zero, -0.0, a value distinct from plus zero; not a number, NaN; and plus and minus infinity.
The task is to use expressions involving other 'normal' floating point values in your language to calculate these, (and maybe other), extreme floating point values in your language and assign them to variables.
Print the values of these variables if possible; and show some arithmetic with these values and variables.
If your language can directly enter these extreme floating point values then show it.
See also
What Every Computer Scientist Should Know About Floating-Point Arithmetic
Related tasks
Infinity
Detect division by zero
Literals/Floating point
| #Java | Java | public class Extreme {
public static void main(String[] args) {
double negInf = -1.0 / 0.0; //also Double.NEGATIVE_INFINITY
double inf = 1.0 / 0.0; //also Double.POSITIVE_INFINITY
double nan = 0.0 / 0.0; //also Double.NaN
double negZero = -2.0 / inf;
System.out.println("Negative inf: " + negInf);
System.out.println("Positive inf: " + inf);
System.out.println("NaN: " + nan);
System.out.println("Negative 0: " + negZero);
System.out.println("inf + -inf: " + (inf + negInf));
System.out.println("0 * NaN: " + (0 * nan));
System.out.println("NaN == NaN: " + (nan == nan));
}
} |
http://rosettacode.org/wiki/Execute_SNUSP | Execute SNUSP | Execute SNUSP is an implementation of SNUSP.
Other implementations of SNUSP.
RCSNUSP
SNUSP
An implementation need only properly implement the Core SNUSP instructions ('$', '\', '/', '+', '-', '<', '>', ',', '.', '!', and '?'). Modular SNUSP ('#', '@') and Bloated SNUSP (':', ';', '%', and '&') are also allowed, but not required. Any extra characters that you implement should be noted in the description of your implementation. Any cell size is allowed, EOF support is optional, as is whether you have bounded or unbounded memory.
| #11l | 11l | V HW = ‘
/++++!/===========?\>++.>+.+++++++..+++\
\+++\ | /+>+++++++>/ /++++++++++<<.++>./
$+++/ | \+++++++++>\ \+++++.>.+++.-----\
\==-<<<<+>+++/ /=.>.+>.--------.-/’
F snusp(store, code)
V ds = [Byte(0)] * store
V dp = 0
V cs = code.split("\n")
V ipr = 0
V ipc = 0
L(row) cs
ipc = row.findi(‘$’)
I ipc != -1
ipr = L.index
L.break
V id = 0
F step()
I @id [&] 1
@ipr += 1 - (@id [&] 2)
E
@ipc += 1 - (@id [&] 2)
L ipr >= 0 & ipr < cs.len & ipc >= 0 & ipc < cs[ipr].len
S cs[ipr][ipc]
‘>’
dp++
‘<’
dp--
‘+’
ds[dp]++
‘-’
ds[dp]--
‘.’
:stdout.write(Char(code' ds[dp]))
‘,’
ds[dp] = Byte(:stdin.read(1).code)
‘/’
id = (-)id
‘\’
id (+)= 1
‘!’
step()
‘?’
I !(ds[dp])
step()
step()
snusp(5, HW) |
http://rosettacode.org/wiki/Extend_your_language | Extend your language | Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Some programming languages allow you to extend the language. While this can be done to a certain degree in most languages (e.g. by using macros), other languages go much further. Most notably in the Forth and Lisp families, programming per se is done by extending the language without any formal distinction between built-in and user-defined elements.
If your language supports it, show how to introduce a new flow control mechanism. A practical and useful example is a four-way branch:
Occasionally, code must be written that depends on two conditions, resulting in up to four branches (depending on whether both, only the first, only the second, or none of the conditions are "true"). In a C-like language this could look like the following:
if (condition1isTrue) {
if (condition2isTrue)
bothConditionsAreTrue();
else
firstConditionIsTrue();
}
else if (condition2isTrue)
secondConditionIsTrue();
else
noConditionIsTrue();
Besides being rather cluttered, the statement(s) for 'condition2isTrue' must be written down twice. If 'condition2isTrue' were a lengthy and involved expression, it would be quite unreadable, and the code generated by the compiler might be unnecessarily large.
This can be improved by introducing a new keyword if2. It is similar to if, but takes two conditional statements instead of one, and up to three 'else' statements. One proposal (in pseudo-C syntax) might be:
if2 (condition1isTrue) (condition2isTrue)
bothConditionsAreTrue();
else1
firstConditionIsTrue();
else2
secondConditionIsTrue();
else
noConditionIsTrue();
Pick the syntax which suits your language. The keywords 'else1' and 'else2' are just examples. The new conditional expression should look, nest and behave analogously to the language's built-in 'if' statement.
| #C.23 | C# |
using System;
using System.Reflection;
namespace Extend_your_language
{
class Program
{
public static void Main(string[] args)
{
Console.WriteLine();
Console.WriteLine("Hello World!");
Console.WriteLine();
int x = 0;
int y = 0;
for(x=0;x<2;x++)
{
for(y=0;y<2;y++)
{
CONDITIONS( (x==0) , (y==0) ).
IF2 ("METHOD1").
ELSE1("METHOD2").
ELSE2("METHOD3").
ELSE ("METHOD4");
}
}
Console.WriteLine();
Console.Write("Press any key to continue . . . ");
Console.ReadKey(true);
}
public static void METHOD1()
{
Console.WriteLine("METHOD 1 executed - both are true");
}
public static void METHOD2()
{
Console.WriteLine("METHOD 2 executed - first is true");
}
public static void METHOD3()
{
Console.WriteLine("METHOD 3 executed - second is true");
}
public static void METHOD4()
{
Console.WriteLine("METHOD 4 executed - both are false");
}
static int CONDITIONS(bool condition1, bool condition2)
{
int c = 0;
if(condition1 && condition2)
c = 0;
else if(condition1)
c = 1;
else if(condition2)
c = 2;
else
c = 3;
return c;
}
}
public static class ExtensionMethods
{
public static int IF2(this int value, string method)
{
if(value == 0)
{
MethodInfo m = typeof(Program).GetMethod(method);
m.Invoke(null,null);
}
return value;
}
public static int ELSE1(this int value, string method)
{
if(value == 1)
{
MethodInfo m = typeof(Program).GetMethod(method);
m.Invoke(null,null);
}
return value;
}
public static int ELSE2(this int value, string method)
{
if(value == 2)
{
MethodInfo m = typeof(Program).GetMethod(method);
m.Invoke(null,null);
}
return value;
}
public static void ELSE(this int value, string method)
{
if(value == 3)
{
MethodInfo m = typeof(Program).GetMethod(method);
m.Invoke(null,null);
}
}
}
}
|
http://rosettacode.org/wiki/Exponentiation_order | Exponentiation order | This task will demonstrate the order of exponentiation (xy) when there are multiple exponents.
(Many programming languages, especially those with extended─precision integer arithmetic, usually support one of **, ^, ↑ or some such for exponentiation.)
Task requirements
Show the result of a language's evaluation of multiple exponentiation (either as an integer or floating point).
If your language's exponentiation operator is not one of the usual ones, please comment on how to recognize it.
Using whatever operator or syntax your language supports (if any), show the results in three lines (with identification):
5**3**2
(5**3)**2
5**(3**2)
If there are other methods (or formats) of multiple exponentiations, show them as well.
See also
MathWorld entry: exponentiation
Related tasks
exponentiation operator
arbitrary-precision integers (included)
Exponentiation with infix operators in (or operating on) the base
| #ALGOL_W | ALGOL W | begin
write( "5**3**2: ", round( 5 ** 3 ** 2 ) );
write( "(5**3)**2: ", round( ( 5 ** 3 ) ** 2 ) );
write( "5**(3**2): ", round( 5 ** round( 3 ** 2 ) ) )
end. |
http://rosettacode.org/wiki/Exponentiation_order | Exponentiation order | This task will demonstrate the order of exponentiation (xy) when there are multiple exponents.
(Many programming languages, especially those with extended─precision integer arithmetic, usually support one of **, ^, ↑ or some such for exponentiation.)
Task requirements
Show the result of a language's evaluation of multiple exponentiation (either as an integer or floating point).
If your language's exponentiation operator is not one of the usual ones, please comment on how to recognize it.
Using whatever operator or syntax your language supports (if any), show the results in three lines (with identification):
5**3**2
(5**3)**2
5**(3**2)
If there are other methods (or formats) of multiple exponentiations, show them as well.
See also
MathWorld entry: exponentiation
Related tasks
exponentiation operator
arbitrary-precision integers (included)
Exponentiation with infix operators in (or operating on) the base
| #APL | APL | 5*3*2
1953125
(5*3)*2
15625
5*(3*2)
1953125
|
http://rosettacode.org/wiki/Exponentiation_order | Exponentiation order | This task will demonstrate the order of exponentiation (xy) when there are multiple exponents.
(Many programming languages, especially those with extended─precision integer arithmetic, usually support one of **, ^, ↑ or some such for exponentiation.)
Task requirements
Show the result of a language's evaluation of multiple exponentiation (either as an integer or floating point).
If your language's exponentiation operator is not one of the usual ones, please comment on how to recognize it.
Using whatever operator or syntax your language supports (if any), show the results in three lines (with identification):
5**3**2
(5**3)**2
5**(3**2)
If there are other methods (or formats) of multiple exponentiations, show them as well.
See also
MathWorld entry: exponentiation
Related tasks
exponentiation operator
arbitrary-precision integers (included)
Exponentiation with infix operators in (or operating on) the base
| #AppleScript | AppleScript | set r1 to 5 ^ 3 ^ 2 -- Changes to 5 ^ (3 ^ 2) when compiled.
set r2 to (5 ^ 3) ^ 2
set r3 to 5 ^ (3 ^ 2)
return "5 ^ 3 ^ 2 = " & r1 & "
(5 ^ 3) ^ 2 = " & r2 & "
5 ^ (3 ^ 2) = " & r3 |
http://rosettacode.org/wiki/Factorions | Factorions |
Definition
A factorion is a natural number that equals the sum of the factorials of its digits.
Example
145 is a factorion in base 10 because:
1! + 4! + 5! = 1 + 24 + 120 = 145
It can be shown (see talk page) that no factorion in base 10 can exceed 1,499,999.
Task
Write a program in your language to demonstrate, by calculating and printing out the factorions, that:
There are 3 factorions in base 9
There are 4 factorions in base 10
There are 5 factorions in base 11
There are 2 factorions in base 12 (up to the same upper bound as for base 10)
See also
Wikipedia article
OEIS:A014080 - Factorions in base 10
OEIS:A193163 - Factorions in base n
| #OCaml | OCaml | let () =
(* cache factorials from 0 to 11 *)
let fact = Array.make 12 0 in
fact.(0) <- 1;
for n = 1 to pred 12 do
fact.(n) <- fact.(n-1) * n;
done;
for b = 9 to 12 do
Printf.printf "The factorions for base %d are:\n" b;
for i = 1 to pred 1_500_000 do
let sum = ref 0 in
let j = ref i in
while !j > 0 do
let d = !j mod b in
sum := !sum + fact.(d);
j := !j / b;
done;
if !sum = i then (print_int i; print_string " ")
done;
print_string "\n\n";
done |
http://rosettacode.org/wiki/Factorions | Factorions |
Definition
A factorion is a natural number that equals the sum of the factorials of its digits.
Example
145 is a factorion in base 10 because:
1! + 4! + 5! = 1 + 24 + 120 = 145
It can be shown (see talk page) that no factorion in base 10 can exceed 1,499,999.
Task
Write a program in your language to demonstrate, by calculating and printing out the factorions, that:
There are 3 factorions in base 9
There are 4 factorions in base 10
There are 5 factorions in base 11
There are 2 factorions in base 12 (up to the same upper bound as for base 10)
See also
Wikipedia article
OEIS:A014080 - Factorions in base 10
OEIS:A193163 - Factorions in base n
| #Pascal | Pascal | program munchhausennumber;
{$IFDEF FPC}{$MODE objFPC}{$Optimization,On,all}{$ELSE}{$APPTYPE CONSOLE}{$ENDIF}
uses
sysutils;
type
tdigit = byte;
const
MAXBASE = 17;
var
DgtPotDgt : array[0..MAXBASE-1] of NativeUint;
dgtCnt : array[0..MAXBASE-1] of NativeInt;
cnt: NativeUint;
function convertToString(n:NativeUint;base:byte):AnsiString;
const
cBASEDIGITS = '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvxyz';
var
r,dgt: NativeUint;
begin
IF base > length(cBASEDIGITS) then
EXIT('Base to big');
result := '';
repeat
r := n div base;
dgt := n-r*base;
result := cBASEDIGITS[dgt+1]+result;
n := r;
until n =0;
end;
function CheckSameDigits(n1,n2,base:NativeUInt):boolean;
var
i : NativeUInt;
Begin
fillchar(dgtCnt,SizeOf(dgtCnt),#0);
repeat
//increment digit of n1
i := n1;n1 := n1 div base;i := i-n1*base;inc(dgtCnt[i]);
//decrement digit of n2
i := n2;n2 := n2 div base;i := i-n2*base;dec(dgtCnt[i]);
until (n1=0) AND (n2= 0);
result := true;
For i := 2 to Base-1 do
result := result AND (dgtCnt[i]=0);
result := result AND (dgtCnt[0]+dgtCnt[1]=0);
end;
procedure Munch(number,DgtPowSum,minDigit:NativeUInt;digits,base:NativeInt);
var
i: NativeUint;
s1,s2: AnsiString;
begin
inc(cnt);
number := number*base;
IF digits > 1 then
Begin
For i := minDigit to base-1 do
Munch(number+i,DgtPowSum+DgtPotDgt[i],i,digits-1,base);
end
else
For i := minDigit to base-1 do
//number is always the arrangement of the digits leading to smallest number
IF (number+i)<= (DgtPowSum+DgtPotDgt[i]) then
IF CheckSameDigits(number+i,DgtPowSum+DgtPotDgt[i],base) then
iF number+i>0 then
begin
s1 := convertToString(DgtPowSum+DgtPotDgt[i],base);
s2 := convertToString(number+i,base);
If length(s1)= length(s2) then
writeln(Format('%*d %*s %*s',[Base-1,DgtPowSum+DgtPotDgt[i],Base-1,s1,Base-1,s2]));
end;
end;
//factorions
procedure InitDgtPotDgt(base:byte);
var
i: NativeUint;
Begin
DgtPotDgt[0]:= 1;
For i := 1 to Base-1 do
DgtPotDgt[i] := DgtPotDgt[i-1]*i;
DgtPotDgt[0]:= 0;
end;
{
//Munchhausen numbers
procedure InitDgtPotDgt;
var
i,k,dgtpow: NativeUint;
Begin
// digit ^ digit ,special case 0^0 here 0
DgtPotDgt[0]:= 0;
For i := 1 to Base-1 do
Begin
dgtpow := i;
For k := 2 to i do
dgtpow := dgtpow*i;
DgtPotDgt[i] := dgtpow;
end;
end;
}
var
base : byte;
begin
cnt := 0;
For base := 2 to MAXBASE do
begin
writeln('Base = ',base);
InitDgtPotDgt(base);
Munch(0,0,0,base,base);
end;
writeln('Check Count ',cnt);
end. |
http://rosettacode.org/wiki/Filter | Filter | Task
Select certain elements from an Array into a new Array in a generic way.
To demonstrate, select all even numbers from an Array.
As an option, give a second solution which filters destructively,
by modifying the original Array rather than creating a new Array.
| #UNIX_Shell | UNIX Shell | a=(1 2 3 4 5)
unset e[@]
for ((i=0;i<${#a[@]};i++)); do
[ $((a[$i]%2)) -eq 0 ] && e[$i]="${a[$i]}"
done |
http://rosettacode.org/wiki/FizzBuzz | FizzBuzz | Task
Write a program that prints the integers from 1 to 100 (inclusive).
But:
for multiples of three, print Fizz (instead of the number)
for multiples of five, print Buzz (instead of the number)
for multiples of both three and five, print FizzBuzz (instead of the number)
The FizzBuzz problem was presented as the lowest level of comprehension required to illustrate adequacy.
Also see
(a blog) dont-overthink-fizzbuzz
(a blog) fizzbuzz-the-programmers-stairway-to-heaven
| #Pony | Pony | use "collections"
actor Main
new create(env: Env) =>
for i in Range[I32](1, 100) do
env.out.print(fizzbuzz(i))
end
fun fizzbuzz(n: I32): String =>
if (n % 15) == 0 then
"FizzBuzz"
elseif (n % 5) == 0 then
"Buzz"
elseif (n % 3) == 0 then
"Fizz"
else
n.string()
end |
http://rosettacode.org/wiki/Extensible_prime_generator | Extensible prime generator | Task
Write a generator of prime numbers, in order, that will automatically adjust to accommodate the generation of any reasonably high prime.
The routine should demonstrably rely on either:
Being based on an open-ended counter set to count without upper limit other than system or programming language limits. In this case, explain where this counter is in the code.
Being based on a limit that is extended automatically. In this case, choose a small limit that ensures the limit will be passed when generating some of the values to be asked for below.
If other methods of creating an extensible prime generator are used, the algorithm's means of extensibility/lack of limits should be stated.
The routine should be used to:
Show the first twenty primes.
Show the primes between 100 and 150.
Show the number of primes between 7,700 and 8,000.
Show the 10,000th prime.
Show output on this page.
Note: You may reference code already on this site if it is written to be imported/included, then only the code necessary for import and the performance of this task need be shown. (It is also important to leave a forward link on the referenced tasks entry so that later editors know that the code is used for multiple tasks).
Note 2: If a languages in-built prime generator is extensible or is guaranteed to generate primes up to a system limit, (231 or memory overflow for example), then this may be used as long as an explanation of the limits of the prime generator is also given. (Which may include a link to/excerpt from, language documentation).
Note 3:The task is written so it may be useful in solving the task Emirp primes as well as others (depending on its efficiency).
Reference
Prime Numbers. Website with large count of primes.
| #C.2B.2B | C++ | #include <iostream>
#include <cstdint>
#include <queue>
#include <utility>
#include <vector>
#include <limits>
template<typename integer>
class prime_generator {
public:
integer next_prime();
integer count() const {
return count_;
}
private:
struct queue_item {
queue_item(integer prime, integer multiple, unsigned int wheel_index) :
prime_(prime), multiple_(multiple), wheel_index_(wheel_index) {}
integer prime_;
integer multiple_;
unsigned int wheel_index_;
};
struct cmp {
bool operator()(const queue_item& a, const queue_item& b) const {
return a.multiple_ > b.multiple_;
}
};
static integer wheel_next(unsigned int& index) {
integer offset = wheel_[index];
++index;
if (index == std::size(wheel_))
index = 0;
return offset;
}
typedef std::priority_queue<queue_item, std::vector<queue_item>, cmp> queue;
integer next_ = 11;
integer count_ = 0;
queue queue_;
unsigned int wheel_index_ = 0;
static const unsigned int wheel_[];
static const integer primes_[];
};
template<typename integer>
const unsigned int prime_generator<integer>::wheel_[] = {
2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2, 6, 4, 2,
6, 4, 6, 8, 4, 2, 4, 2, 4, 8, 6, 4, 6, 2, 4, 6,
2, 6, 6, 4, 2, 4, 6, 2, 6, 4, 2, 4, 2, 10, 2, 10
};
template<typename integer>
const integer prime_generator<integer>::primes_[] = {
2, 3, 5, 7
};
template<typename integer>
integer prime_generator<integer>::next_prime() {
if (count_ < std::size(primes_))
return primes_[count_++];
integer n = next_;
integer prev = 0;
while (!queue_.empty()) {
queue_item item = queue_.top();
if (prev != 0 && prev != item.multiple_)
n += wheel_next(wheel_index_);
if (item.multiple_ > n)
break;
else if (item.multiple_ == n) {
queue_.pop();
queue_item new_item(item);
new_item.multiple_ += new_item.prime_ * wheel_next(new_item.wheel_index_);
queue_.push(new_item);
}
else
throw std::overflow_error("prime_generator: overflow!");
prev = item.multiple_;
}
if (std::numeric_limits<integer>::max()/n > n)
queue_.emplace(n, n * n, wheel_index_);
next_ = n + wheel_next(wheel_index_);
++count_;
return n;
}
int main() {
typedef uint32_t integer;
prime_generator<integer> pgen;
std::cout << "First 20 primes:\n";
for (int i = 0; i < 20; ++i) {
integer p = pgen.next_prime();
if (i != 0)
std::cout << ", ";
std::cout << p;
}
std::cout << "\nPrimes between 100 and 150:\n";
for (int n = 0; ; ) {
integer p = pgen.next_prime();
if (p > 150)
break;
if (p >= 100) {
if (n != 0)
std::cout << ", ";
std::cout << p;
++n;
}
}
int count = 0;
for (;;) {
integer p = pgen.next_prime();
if (p > 8000)
break;
if (p >= 7700)
++count;
}
std::cout << "\nNumber of primes between 7700 and 8000: " << count << '\n';
for (integer n = 10000; n <= 10000000; n *= 10) {
integer prime;
while (pgen.count() != n)
prime = pgen.next_prime();
std::cout << n << "th prime: " << prime << '\n';
}
return 0;
} |
http://rosettacode.org/wiki/Fibonacci_sequence | Fibonacci sequence | The Fibonacci sequence is a sequence Fn of natural numbers defined recursively:
F0 = 0
F1 = 1
Fn = Fn-1 + Fn-2, if n>1
Task
Write a function to generate the nth Fibonacci number.
Solutions can be iterative or recursive (though recursive solutions are generally considered too slow and are mostly used as an exercise in recursion).
The sequence is sometimes extended into negative numbers by using a straightforward inverse of the positive definition:
Fn = Fn+2 - Fn+1, if n<0
support for negative n in the solution is optional.
Related tasks
Fibonacci n-step number sequences
Leonardo numbers
References
Wikipedia, Fibonacci number
Wikipedia, Lucas number
MathWorld, Fibonacci Number
Some identities for r-Fibonacci numbers
OEIS Fibonacci numbers
OEIS Lucas numbers
| #FRISC_Assembly | FRISC Assembly | FIBONACCI PUSH R1
PUSH R2
PUSH R3
MOVE 0, R1
MOVE 1, R2
FIB_LOOP SUB R0, 1, R0
JP_Z FIB_DONE
MOVE R2, R3
ADD R1, R2, R2
MOVE R3, R1
JP FIB_LOOP
FIB_DONE MOVE R2, R0
POP R3
POP R2
POP R1
RET |
http://rosettacode.org/wiki/Factors_of_an_integer | Factors of an integer |
Basic Data Operation
This is a basic data operation. It represents a fundamental action on a basic data type.
You may see other such operations in the Basic Data Operations category, or:
Integer Operations
Arithmetic |
Comparison
Boolean Operations
Bitwise |
Logical
String Operations
Concatenation |
Interpolation |
Comparison |
Matching
Memory Operations
Pointers & references |
Addresses
Task
Compute the factors of a positive integer.
These factors are the positive integers by which the number being factored can be divided to yield a positive integer result.
(Though the concepts function correctly for zero and negative integers, the set of factors of zero has countably infinite members, and the factors of negative integers can be obtained from the factors of related positive numbers without difficulty; this task does not require handling of either of these cases).
Note that every prime number has two factors: 1 and itself.
Related tasks
count in factors
prime decomposition
Sieve of Eratosthenes
primality by trial division
factors of a Mersenne number
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
sequence: smallest number greater than previous term with exactly n divisors
| #Picat | Picat | factors(N) = [[D,N // D] : D in 1..N.sqrt.floor, N mod D == 0].flatten.sort_remove_dups. |
http://rosettacode.org/wiki/Execute_Computer/Zero | Execute Computer/Zero | Execute Computer/Zero is an implementation of Computer/zero Assembly.
Other implementations of Computer/zero Assembly.
Task
Create a Computer/zero Assembly emulator. You may consider this webpage as a reference implementation. Output the results of the sample programs "2+2" and "7*8" found there.
The virtual machine "bytecode" needs to be able to modify itself (or at least act as though it can) while the virtual machine is running, to be consistent with the reference implementation.
For output, it is sufficient to have the implementation of the STP opcode return the accumulator to your actual language, and then use your standard printing routines to output it.
Bonus Points
Run all 5 sample programs at the aforementioned website and output their results.
| #J | J | OPS=: 'nop lda sta add sub brz jmp stp'
assemble1=: {{
y=. tolower y
ins=. {.;:y-.":i.10
cod=. 8|(;:OPS) i. ins
val=. {.0".y-.OPS
if. *cod do.
assert. 32>val
val+32*cod
else.
assert. 256>val
val
end.
}}
assemble=: {{
if. 0=L. y do.
delim=. {.((tolower y)-.(":i.10),;OPS),LF
y=. delim cut y
end.
code=. assemble1@> y
mem=: code (i.#code)} 32#0
}}
exec1=: {{
'cod val'=. 0 32#:pc{mem
pc=: 32|pc+1
select. cod
case. 0 do.
case. 1 do. acc=: val{mem
case. 2 do. mem=: acc val} mem
case. 3 do. acc=: 256|acc+val{mem
case. 4 do. acc=: 256|acc-val{mem
case. 5 do. pc=: 32|pc[^:(*acc) val
case. 6 do. pc=: 32|val
case. 7 do. pc=: __
end.
}}
exec=: {{
pc=: acc=: 0
while. 0<:pc do. exec1'' end.
acc
}} |
http://rosettacode.org/wiki/Execute_Computer/Zero | Execute Computer/Zero | Execute Computer/Zero is an implementation of Computer/zero Assembly.
Other implementations of Computer/zero Assembly.
Task
Create a Computer/zero Assembly emulator. You may consider this webpage as a reference implementation. Output the results of the sample programs "2+2" and "7*8" found there.
The virtual machine "bytecode" needs to be able to modify itself (or at least act as though it can) while the virtual machine is running, to be consistent with the reference implementation.
For output, it is sufficient to have the implementation of the STP opcode return the accumulator to your actual language, and then use your standard printing routines to output it.
Bonus Points
Run all 5 sample programs at the aforementioned website and output their results.
| #Julia | Julia | mutable struct ComputerZero
ip::Int
ram::Vector{UInt8}
accum::UInt8
isready::Bool
end
function ComputerZero(program)
memory = zeros(UInt8, 32)
for i in 1:min(32, length(program))
memory[i] = program[i]
end
return ComputerZero(1, memory, 0, true)
end
NOP(c) = (c.ip = mod1(c.ip + 1, 32))
LDA(c) = (c.accum = c.ram[c.ram[c.ip] & 0b00011111 + 1]; c.ip = mod1(c.ip + 1, 32))
STA(c) = (c.ram[c.ram[c.ip] & 0b00011111 + 1] = c.accum; c.ip = mod1(c.ip + 1, 32))
ADD(c) = (c.accum += c.ram[c.ram[c.ip] & 0b00011111 + 1]; c.ip = mod1(c.ip + 1, 32))
SUB(c) = (c.accum -= c.ram[c.ram[c.ip] & 0b00011111 + 1]; c.ip = mod1(c.ip + 1, 32))
BRZ(c) = (c.ip = (c.accum == 0) ? c.ram[c.ip] & 0b00011111 + 1 : mod1(c.ip + 1, 32))
JMP(c) = (c.ip = c.ram[c.ip] & 0b00011111 + 1)
STP(c) = (println("Program completed with accumulator value $(c.accum).\n"); c.isready = false)
const step = [NOP, LDA, STA, ADD, SUB, BRZ, JMP, STP]
const instructions = ["NOP", "LDA", "STA", "ADD", "SUB", "BRZ", "JMP", "STP"]
const assemblywords = Dict(s => i - 1 for (i, s) in pairs(instructions))
function run(compzero::ComputerZero, debug = false)
while compzero.isready
instruction = compzero.ram[compzero.ip]
opcode, operand = instruction >> 5 + 1, instruction & 0b00011111
debug && println("op $(instructions[opcode]), operand $operand")
step[opcode](compzero)
end
return compzero.accum
end
run(program::Vector, debug = false) = run(ComputerZero(program), debug)
function compile(text::String)
bin, lines = UInt8[], 0
for line in strip.(split(text, "\n"))
lines += 1
if isempty(line)
push!(bin, 0)
elseif (m = match(r"(\w\w\w)\s+(\d\d?)", line)) != nothing
push!(bin, UInt8((assemblywords[m.captures[1]] << 5) | parse(UInt8, m.captures[2])))
elseif (m = match(r"(\w\w\w)", line)) != nothing
push!(bin, UInt8(assemblywords[m.match] << 5))
elseif (m = match(r"\d\d?", line)) != nothing
push!(bin, parse(UInt8, m.match))
else
error("Compilation error at line $lines: error parsing <$line>")
end
end
println("Compiled $lines lines.")
return bin
end
const testprograms = [
"""
LDA 3
ADD 4
STP
2
2
""",
"""
LDA 12
ADD 10
STA 12
LDA 11
SUB 13
STA 11
BRZ 8
JMP 0
LDA 12
STP
8
7
0
1
""",
"""
LDA 14
STA 15
ADD 13
STA 14
LDA 15
STA 13
LDA 16
SUB 17
BRZ 11
STA 16
JMP 0
LDA 14
STP
1
1
0
8
1
""",
"""
LDA 13
ADD 15
STA 5
ADD 16
STA 7
NOP
STA 14
NOP
BRZ 11
STA 15
JMP 0
LDA 14
STP
LDA 0
0
28
1
6
0
2
26
5
20
3
30
1
22
4
24
""",
"""
0
0
STP
NOP
LDA 3
SUB 29
BRZ 18
LDA 3
STA 29
BRZ 14
LDA 1
ADD 31
STA 1
JMP 2
LDA 0
ADD 31
STA 0
JMP 2
LDA 3
STA 29
LDA 0
ADD 30
ADD 3
STA 0
LDA 1
ADD 30
ADD 3
STA 1
JMP 2
0
1
3
"""
]
for t in testprograms
run(compile(t))
end
|
http://rosettacode.org/wiki/Extreme_floating_point_values | Extreme floating point values | The IEEE floating point specification defines certain 'extreme' floating point values such as minus zero, -0.0, a value distinct from plus zero; not a number, NaN; and plus and minus infinity.
The task is to use expressions involving other 'normal' floating point values in your language to calculate these, (and maybe other), extreme floating point values in your language and assign them to variables.
Print the values of these variables if possible; and show some arithmetic with these values and variables.
If your language can directly enter these extreme floating point values then show it.
See also
What Every Computer Scientist Should Know About Floating-Point Arithmetic
Related tasks
Infinity
Detect division by zero
Literals/Floating point
| #jq | jq | 0/0 #=> null
1e1000 #=> 1.7976931348623157e+308 |
http://rosettacode.org/wiki/Extreme_floating_point_values | Extreme floating point values | The IEEE floating point specification defines certain 'extreme' floating point values such as minus zero, -0.0, a value distinct from plus zero; not a number, NaN; and plus and minus infinity.
The task is to use expressions involving other 'normal' floating point values in your language to calculate these, (and maybe other), extreme floating point values in your language and assign them to variables.
Print the values of these variables if possible; and show some arithmetic with these values and variables.
If your language can directly enter these extreme floating point values then show it.
See also
What Every Computer Scientist Should Know About Floating-Point Arithmetic
Related tasks
Infinity
Detect division by zero
Literals/Floating point
| #Julia | Julia |
function showextremes()
values = [0.0, -0.0, Inf, -Inf, NaN]
println(1 ./ values)
end
showextremes()
@show Inf + 2.0
@show Inf + Inf
@show Inf - Inf
@show Inf * Inf
@show Inf / Inf
@show Inf * 0
@show 0 == -0
@show NaN == NaN
@show NaN === NaN
|
http://rosettacode.org/wiki/Execute_SNUSP | Execute SNUSP | Execute SNUSP is an implementation of SNUSP.
Other implementations of SNUSP.
RCSNUSP
SNUSP
An implementation need only properly implement the Core SNUSP instructions ('$', '\', '/', '+', '-', '<', '>', ',', '.', '!', and '?'). Modular SNUSP ('#', '@') and Bloated SNUSP (':', ';', '%', and '&') are also allowed, but not required. Any extra characters that you implement should be noted in the description of your implementation. Any cell size is allowed, EOF support is optional, as is whether you have bounded or unbounded memory.
| #Ada | Ada | #
# snusp.icn, A Modular SNUSP interpreter
#
$define VERSION 0.6
# allow a couple of cli options
link options
# directions
$define DRIGHT 1
$define DLEFT 2
$define DUP 3
$define DDOWN 4
record position(row, col)
global dir, ip, ram
procedure main(argv)
local ch, codespace, col, dp, fn, line
local row := 1
local wid := 0
local dirs := []
local ips := []
local opts, verbose, debug
opts := options(argv, "-h! -v! -d!", errorproc)
\opts["v"] & verbose := 1
\opts["h"] & show_help(verbose)
\opts["d"] & debug := 1
ip := position(1,1)
# initial direction
dir := DRIGHT
# prepare initial memory
ram := list(1, 0)
# prepare code field
codespace := []
fn := open(argv[1], "r") | &input
if (fn === &input) & \opts["h"] then return
while line := read(fn) do {
put(codespace, line)
wid := max(*line, wid)
}
if *codespace = 0 then return
every line := !codespace do {
codespace[row] := left(codespace[row], wid)
# track starting indicator
if /col := find("$", codespace[row]) then {
ip.row := row
ip.col := col
}
row +:= 1
}
if \verbose then {
write("Starting at ", ip.row, ", ", ip.col, " with codespace:")
every write(!codespace)
}
dp := 1
repeat {
if not (ch := codespace[ip.row][ip.col]) then break
if \debug then {
write(&errout, "dir: ", dir, " ch: ", ch, " [", ord(ch), "]",
" row: ", ip.row, " col: ", ip.col,
" dp: ", dp, " ram[dp]: ", ram[dp])
}
case ch of {
# six of the bf instructions
"+": ram[dp] +:= 1
"-": ram[dp] -:= 1
">": resize(dp +:= 1)
"<": dp -:= 1
".": writes(char(ram[dp]) | char(0))
",": ram[dp] := getche()
# direction change, LURD, RULD, SKIP, SKIPZ
"\\": { # LURD
case dir of {
DRIGHT: dir := DDOWN
DLEFT: dir := DUP
DUP: dir := DLEFT
DDOWN: dir := DRIGHT
}
}
"/": { # RULD
case dir of {
DRIGHT: dir := DUP
DLEFT: dir := DDOWN
DUP: dir := DRIGHT
DDOWN: dir := DLEFT
}
}
"!": step()
"?": { # skipz
if ram[dp] = 0 then {
step()
}
}
# modular SNUSP
"@": { # Enter
push(dirs, dir)
push(ips, copy(ip))
}
"#": { # Leave
if *dirs < 1 then break
dir := pop(dirs)
ip := pop(ips)
step()
}
}
step()
}
end
# advance the ip depending on direction
procedure step()
case dir of {
DRIGHT: ip.col +:= 1
DLEFT: ip.col -:= 1
DUP: ip.row -:= 1
DDOWN: ip.row +:= 1
}
end
# enlarge memory when needed
procedure resize(elements)
until *ram >= elements do put(ram, 0)
end
# quick help or verbose help
procedure show_help(verbose)
write("SNUSP interpeter in Unicon, version ", VERSION)
write("CORE and MODULAR, not yet BLOATED")
write()
write("Usage: unicon snusp.icn -x [filename] [-h|-v|-d]")
write(" -h, help")
write(" -v, verbose (and verbose help")
write(" -d, debug (step tracer)")
if \verbose then {
write()
write("Instructions:")
write(" + INCR, Increment current memory location")
write(" - DECR, Decrement current memory location")
write(" > RIGHT, Advance memory pointer")
write(" < LEFT, Retreat memory pointer")
write(" . WRITE, Output contents of current memory cell, in ASCII")
write(" , READ, Accept key and place byte value in current memory cell")
write(" \\ LURD, If going:")
write(" left, go up")
write(" up, go left")
write(" right, go down")
write(" down, go right")
write(" / RULD, If going:")
write(" right, go up")
write(" up, go right")
write(" left, go down")
write(" down, go left")
write(" !, SKIP, Move forward one step in current direction")
write(" ?, SKIPZ, If current memory cell is zero then SKIP")
write("Modular SNUSP adds:")
write(" @, ENTER, Push direction and instruction pointer")
write(" #, LEAVE, Pop direction and instruction pointer and SKIP")
write()
write("All other characters are NOOP, explicitly includes =,|,spc")
write(" $, can set the starting location; first one found")
write()
write("Hello world examples:")
write()
write("CORE SNUSP:")
write("/++++!/===========?\\>++.>+.+++++++..+++\\")
write("\\+++\\ | /+>+++++++>/ /++++++++++<<.++>./")
write("$+++/ | \\+++++++++>\\ \\+++++.>.+++.-----\\")
write(" \\==-<<<<+>+++/ /=.>.+>.--------.-/")
write()
write("Modular SNUSP:")
write(" /@@@@++++# #+++@@\ #-----@@@\\n")
write("$@\\H.@/e.+++++++l.l.+++o.>>++++.< .<@/w.@\\o.+++r.++@\\l.@\\d.>+.@/.#")
write(" \\@@@@=>++++>+++++<<@+++++# #---@@/!=========/!==/")
write()
}
end |
http://rosettacode.org/wiki/Extend_your_language | Extend your language | Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Some programming languages allow you to extend the language. While this can be done to a certain degree in most languages (e.g. by using macros), other languages go much further. Most notably in the Forth and Lisp families, programming per se is done by extending the language without any formal distinction between built-in and user-defined elements.
If your language supports it, show how to introduce a new flow control mechanism. A practical and useful example is a four-way branch:
Occasionally, code must be written that depends on two conditions, resulting in up to four branches (depending on whether both, only the first, only the second, or none of the conditions are "true"). In a C-like language this could look like the following:
if (condition1isTrue) {
if (condition2isTrue)
bothConditionsAreTrue();
else
firstConditionIsTrue();
}
else if (condition2isTrue)
secondConditionIsTrue();
else
noConditionIsTrue();
Besides being rather cluttered, the statement(s) for 'condition2isTrue' must be written down twice. If 'condition2isTrue' were a lengthy and involved expression, it would be quite unreadable, and the code generated by the compiler might be unnecessarily large.
This can be improved by introducing a new keyword if2. It is similar to if, but takes two conditional statements instead of one, and up to three 'else' statements. One proposal (in pseudo-C syntax) might be:
if2 (condition1isTrue) (condition2isTrue)
bothConditionsAreTrue();
else1
firstConditionIsTrue();
else2
secondConditionIsTrue();
else
noConditionIsTrue();
Pick the syntax which suits your language. The keywords 'else1' and 'else2' are just examples. The new conditional expression should look, nest and behave analogously to the language's built-in 'if' statement.
| #Clay | Clay | alias if2(cond1:Bool,
cond2:Bool,
both,
first,
second,
neither)
{
var res1 = cond1;
var res2 = cond2;
if (res1 and res2) return both;
if (res1) return first;
if (res2) return second;
return neither;
}
|
http://rosettacode.org/wiki/Extend_your_language | Extend your language | Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Some programming languages allow you to extend the language. While this can be done to a certain degree in most languages (e.g. by using macros), other languages go much further. Most notably in the Forth and Lisp families, programming per se is done by extending the language without any formal distinction between built-in and user-defined elements.
If your language supports it, show how to introduce a new flow control mechanism. A practical and useful example is a four-way branch:
Occasionally, code must be written that depends on two conditions, resulting in up to four branches (depending on whether both, only the first, only the second, or none of the conditions are "true"). In a C-like language this could look like the following:
if (condition1isTrue) {
if (condition2isTrue)
bothConditionsAreTrue();
else
firstConditionIsTrue();
}
else if (condition2isTrue)
secondConditionIsTrue();
else
noConditionIsTrue();
Besides being rather cluttered, the statement(s) for 'condition2isTrue' must be written down twice. If 'condition2isTrue' were a lengthy and involved expression, it would be quite unreadable, and the code generated by the compiler might be unnecessarily large.
This can be improved by introducing a new keyword if2. It is similar to if, but takes two conditional statements instead of one, and up to three 'else' statements. One proposal (in pseudo-C syntax) might be:
if2 (condition1isTrue) (condition2isTrue)
bothConditionsAreTrue();
else1
firstConditionIsTrue();
else2
secondConditionIsTrue();
else
noConditionIsTrue();
Pick the syntax which suits your language. The keywords 'else1' and 'else2' are just examples. The new conditional expression should look, nest and behave analogously to the language's built-in 'if' statement.
| #Clojure | Clojure |
(defmacro if2 [[cond1 cond2] bothTrue firstTrue secondTrue else]
`(let [cond1# ~cond1
cond2# ~cond2]
(if cond1# (if cond2# ~bothTrue ~firstTrue)
(if cond2# ~secondTrue ~else))))
|
http://rosettacode.org/wiki/Exponentiation_order | Exponentiation order | This task will demonstrate the order of exponentiation (xy) when there are multiple exponents.
(Many programming languages, especially those with extended─precision integer arithmetic, usually support one of **, ^, ↑ or some such for exponentiation.)
Task requirements
Show the result of a language's evaluation of multiple exponentiation (either as an integer or floating point).
If your language's exponentiation operator is not one of the usual ones, please comment on how to recognize it.
Using whatever operator or syntax your language supports (if any), show the results in three lines (with identification):
5**3**2
(5**3)**2
5**(3**2)
If there are other methods (or formats) of multiple exponentiations, show them as well.
See also
MathWorld entry: exponentiation
Related tasks
exponentiation operator
arbitrary-precision integers (included)
Exponentiation with infix operators in (or operating on) the base
| #Arturo | Arturo | print 5^3^2
print (5^3)^2
print 5^(3^2) |
http://rosettacode.org/wiki/Exponentiation_order | Exponentiation order | This task will demonstrate the order of exponentiation (xy) when there are multiple exponents.
(Many programming languages, especially those with extended─precision integer arithmetic, usually support one of **, ^, ↑ or some such for exponentiation.)
Task requirements
Show the result of a language's evaluation of multiple exponentiation (either as an integer or floating point).
If your language's exponentiation operator is not one of the usual ones, please comment on how to recognize it.
Using whatever operator or syntax your language supports (if any), show the results in three lines (with identification):
5**3**2
(5**3)**2
5**(3**2)
If there are other methods (or formats) of multiple exponentiations, show them as well.
See also
MathWorld entry: exponentiation
Related tasks
exponentiation operator
arbitrary-precision integers (included)
Exponentiation with infix operators in (or operating on) the base
| #AWK | AWK |
# syntax: GAWK -f EXPONENTIATION_ORDER.AWK
BEGIN {
printf("5^3^2 = %d\n",5^3^2)
printf("(5^3)^2 = %d\n",(5^3)^2)
printf("5^(3^2) = %d\n",5^(3^2))
exit(0)
}
|
http://rosettacode.org/wiki/Exponentiation_order | Exponentiation order | This task will demonstrate the order of exponentiation (xy) when there are multiple exponents.
(Many programming languages, especially those with extended─precision integer arithmetic, usually support one of **, ^, ↑ or some such for exponentiation.)
Task requirements
Show the result of a language's evaluation of multiple exponentiation (either as an integer or floating point).
If your language's exponentiation operator is not one of the usual ones, please comment on how to recognize it.
Using whatever operator or syntax your language supports (if any), show the results in three lines (with identification):
5**3**2
(5**3)**2
5**(3**2)
If there are other methods (or formats) of multiple exponentiations, show them as well.
See also
MathWorld entry: exponentiation
Related tasks
exponentiation operator
arbitrary-precision integers (included)
Exponentiation with infix operators in (or operating on) the base
| #BASIC | BASIC | 10 PRINT "5**3**2 = ";5**3**2
20 PRINT "(5**3)**2 = ";(5**3)**2
30 PRINT "5**(3**2) = ";5**(3**2) |
http://rosettacode.org/wiki/Factorions | Factorions |
Definition
A factorion is a natural number that equals the sum of the factorials of its digits.
Example
145 is a factorion in base 10 because:
1! + 4! + 5! = 1 + 24 + 120 = 145
It can be shown (see talk page) that no factorion in base 10 can exceed 1,499,999.
Task
Write a program in your language to demonstrate, by calculating and printing out the factorions, that:
There are 3 factorions in base 9
There are 4 factorions in base 10
There are 5 factorions in base 11
There are 2 factorions in base 12 (up to the same upper bound as for base 10)
See also
Wikipedia article
OEIS:A014080 - Factorions in base 10
OEIS:A193163 - Factorions in base n
| #Perl | Perl | use strict;
use warnings;
use ntheory qw/factorial todigits/;
my $limit = 1500000;
for my $b (9 .. 12) {
print "Factorions in base $b:\n";
$_ == factorial($_) and print "$_ " for 0..$b-1;
for my $i (1 .. int $limit/$b) {
my $sum;
my $prod = $i * $b;
for (reverse todigits($i, $b)) {
$sum += factorial($_);
$sum = 0 && last if $sum > $prod;
}
next if $sum == 0;
($sum + factorial($_) == $prod + $_) and print $prod+$_ . ' ' for 0..$b-1;
}
print "\n\n";
} |
http://rosettacode.org/wiki/Factorions | Factorions |
Definition
A factorion is a natural number that equals the sum of the factorials of its digits.
Example
145 is a factorion in base 10 because:
1! + 4! + 5! = 1 + 24 + 120 = 145
It can be shown (see talk page) that no factorion in base 10 can exceed 1,499,999.
Task
Write a program in your language to demonstrate, by calculating and printing out the factorions, that:
There are 3 factorions in base 9
There are 4 factorions in base 10
There are 5 factorions in base 11
There are 2 factorions in base 12 (up to the same upper bound as for base 10)
See also
Wikipedia article
OEIS:A014080 - Factorions in base 10
OEIS:A193163 - Factorions in base n
| #Phix | Phix | with javascript_semantics
for base=9 to 12 do
printf(1,"The factorions for base %d are: ", base)
for i=1 to 1499999 do
atom total = 0, j = i, d
while j>0 and total<=i do
d = remainder(j,base)
total += factorial(d)
j = floor(j/base)
end while
if total==i then printf(1,"%d ", i) end if
end for
printf(1,"\n")
end for
|
http://rosettacode.org/wiki/Filter | Filter | Task
Select certain elements from an Array into a new Array in a generic way.
To demonstrate, select all even numbers from an Array.
As an option, give a second solution which filters destructively,
by modifying the original Array rather than creating a new Array.
| #UnixPipes | UnixPipes | yes \ | cat -n | while read a; do ; expr $a % 2 >/dev/null && echo $a ; done |
http://rosettacode.org/wiki/FizzBuzz | FizzBuzz | Task
Write a program that prints the integers from 1 to 100 (inclusive).
But:
for multiples of three, print Fizz (instead of the number)
for multiples of five, print Buzz (instead of the number)
for multiples of both three and five, print FizzBuzz (instead of the number)
The FizzBuzz problem was presented as the lowest level of comprehension required to illustrate adequacy.
Also see
(a blog) dont-overthink-fizzbuzz
(a blog) fizzbuzz-the-programmers-stairway-to-heaven
| #Pop11 | Pop11 | lvars str;
for i from 1 to 100 do
if i rem 15 = 0 then
'FizzBuzz' -> str;
elseif i rem 3 = 0 then
'Fizz' -> str;
elseif i rem 5 = 0 then
'Buzz' -> str;
else
'' >< i -> str;
endif;
printf(str, '%s\n');
endfor; |
http://rosettacode.org/wiki/Extensible_prime_generator | Extensible prime generator | Task
Write a generator of prime numbers, in order, that will automatically adjust to accommodate the generation of any reasonably high prime.
The routine should demonstrably rely on either:
Being based on an open-ended counter set to count without upper limit other than system or programming language limits. In this case, explain where this counter is in the code.
Being based on a limit that is extended automatically. In this case, choose a small limit that ensures the limit will be passed when generating some of the values to be asked for below.
If other methods of creating an extensible prime generator are used, the algorithm's means of extensibility/lack of limits should be stated.
The routine should be used to:
Show the first twenty primes.
Show the primes between 100 and 150.
Show the number of primes between 7,700 and 8,000.
Show the 10,000th prime.
Show output on this page.
Note: You may reference code already on this site if it is written to be imported/included, then only the code necessary for import and the performance of this task need be shown. (It is also important to leave a forward link on the referenced tasks entry so that later editors know that the code is used for multiple tasks).
Note 2: If a languages in-built prime generator is extensible or is guaranteed to generate primes up to a system limit, (231 or memory overflow for example), then this may be used as long as an explanation of the limits of the prime generator is also given. (Which may include a link to/excerpt from, language documentation).
Note 3:The task is written so it may be useful in solving the task Emirp primes as well as others (depending on its efficiency).
Reference
Prime Numbers. Website with large count of primes.
| #Clojure | Clojure | ns test-project-intellij.core
(:gen-class)
(:require [clojure.string :as string]))
(def primes
" The following routine produces a infinite sequence of primes
(i.e. can be infinite since the evaluation is lazy in that it
only produces values as needed). The method is from clojure primes.clj library
which produces primes based upon O'Neill's paper:
'The Genuine Sieve of Eratosthenes'.
Produces primes based upon trial division on previously found primes up to
(sqrt number), and uses 'wheel' to avoid
testing numbers which are divisors of 2, 3, 5, or 7.
A full explanation of the method is available at:
[https://github.com/stuarthalloway/programming-clojure/pull/12] "
(concat
[2 3 5 7]
(lazy-seq
(let [primes-from ; generates primes by only checking if primes
; numbers which are not divisible by 2, 3, 5, or 7
(fn primes-from [n [f & r]]
(if (some #(zero? (rem n %))
(take-while #(<= (* % %) n) primes))
(recur (+ n f) r)
(lazy-seq (cons n (primes-from (+ n f) r)))))
; wheel provides offsets from previous number to insure we are not landing on a divisor of 2, 3, 5, 7
wheel (cycle [2 4 2 4 6 2 6 4 2 4 6 6 2 6 4 2
6 4 6 8 4 2 4 2 4 8 6 4 6 2 4 6
2 6 6 4 2 4 6 2 6 4 2 4 2 10 2 10])]
(primes-from 11 wheel)))))
(defn between [lo hi]
"Primes between lo and hi value "
(->> (take-while #(<= % hi) primes)
(filter #(>= % lo))
))
(println "First twenty:" (take 20 primes))
(println "Between 100 and 150:" (between 100 150))
(println "Number between 7,7700 and 8,000:" (count (between 7700 8000)))
(println "10,000th prime:" (nth primes (dec 10000))) ; decrement by one since nth starts counting from 0
} |
http://rosettacode.org/wiki/Fibonacci_sequence | Fibonacci sequence | The Fibonacci sequence is a sequence Fn of natural numbers defined recursively:
F0 = 0
F1 = 1
Fn = Fn-1 + Fn-2, if n>1
Task
Write a function to generate the nth Fibonacci number.
Solutions can be iterative or recursive (though recursive solutions are generally considered too slow and are mostly used as an exercise in recursion).
The sequence is sometimes extended into negative numbers by using a straightforward inverse of the positive definition:
Fn = Fn+2 - Fn+1, if n<0
support for negative n in the solution is optional.
Related tasks
Fibonacci n-step number sequences
Leonardo numbers
References
Wikipedia, Fibonacci number
Wikipedia, Lucas number
MathWorld, Fibonacci Number
Some identities for r-Fibonacci numbers
OEIS Fibonacci numbers
OEIS Lucas numbers
| #FunL | FunL | def
fib( 0 ) = 0
fib( 1 ) = 1
fib( n ) = fib( n - 1 ) + fib( n - 2 ) |
http://rosettacode.org/wiki/Factors_of_an_integer | Factors of an integer |
Basic Data Operation
This is a basic data operation. It represents a fundamental action on a basic data type.
You may see other such operations in the Basic Data Operations category, or:
Integer Operations
Arithmetic |
Comparison
Boolean Operations
Bitwise |
Logical
String Operations
Concatenation |
Interpolation |
Comparison |
Matching
Memory Operations
Pointers & references |
Addresses
Task
Compute the factors of a positive integer.
These factors are the positive integers by which the number being factored can be divided to yield a positive integer result.
(Though the concepts function correctly for zero and negative integers, the set of factors of zero has countably infinite members, and the factors of negative integers can be obtained from the factors of related positive numbers without difficulty; this task does not require handling of either of these cases).
Note that every prime number has two factors: 1 and itself.
Related tasks
count in factors
prime decomposition
Sieve of Eratosthenes
primality by trial division
factors of a Mersenne number
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
sequence: smallest number greater than previous term with exactly n divisors
| #PicoLisp | PicoLisp | (de factors (N)
(filter
'((D) (=0 (% N D)))
(range 1 N) ) ) |
http://rosettacode.org/wiki/Execute_Computer/Zero | Execute Computer/Zero | Execute Computer/Zero is an implementation of Computer/zero Assembly.
Other implementations of Computer/zero Assembly.
Task
Create a Computer/zero Assembly emulator. You may consider this webpage as a reference implementation. Output the results of the sample programs "2+2" and "7*8" found there.
The virtual machine "bytecode" needs to be able to modify itself (or at least act as though it can) while the virtual machine is running, to be consistent with the reference implementation.
For output, it is sufficient to have the implementation of the STP opcode return the accumulator to your actual language, and then use your standard printing routines to output it.
Bonus Points
Run all 5 sample programs at the aforementioned website and output their results.
| #Phix | Phix | with javascript_semantics
constant NOP = 0b000_00000, -- no operation
LDA = 0b001_00000, -- load accumulator, a := memory [xxxxx]
STA = 0b010_00000, -- store accumulator, memory [xxxxx] := a
ADD = 0b011_00000, -- add, a := a + memory [xxxxx]
SUB = 0b100_00000, -- subtract, a := a – memory [xxxxx]
BRZ = 0b101_00000, -- branch on zero, if a = 0 then goto xxxxx
JMP = 0b110_00000, -- jump, goto xxxxx
STP = 0b111_00000, -- stop
OP = 0b111_00000, -- operator mask
ARG = 0b000_11111 -- memory location (0 based)
procedure execute(string name, sequence program)
sequence memory = deep_copy(program)
integer pc = 1, a = 0
while true do
integer ppc = memory[pc],
op = and_bitsu(ppc,OP),
arg = and_bits(ppc,ARG)+1
pc += 1
switch op do
case NOP: break
case LDA: a = memory[arg]
case STA: memory[arg] = a
case ADD: a = and_bitsu(a+memory[arg],#FF)
case SUB: a = and_bitsu(a-memory[arg],#FF)
case BRZ: if a=0 then pc = arg end if
case JMP: pc = arg
case STP: printf(1,"%12s: %3d\n",{name,a})
return
end switch
end while
end procedure
execute("2+2",{LDA+3, ADD+4, STP, 2,2})
execute("7*8",{LDA+12, ADD+10, STA+12, LDA+11, SUB+13, STA+11, BRZ+8, JMP+0,
LDA+12, STP, 8,7,0,1})
execute("fibonacci",{LDA+14, STA+15, ADD+13, STA+14, LDA+15, STA+13, LDA+16,
SUB+17, BRZ+11, STA+16, JMP+0, LDA+14, STP, 1,1,0,8,1})
execute("linkedList",{LDA+13, ADD+15, STA+5, ADD+16, STA+7, NOP, STA+14, NOP,
BRZ+11, STA+15, JMP+0, LDA+14, STP, LDA+0, 0,28,1,0,0,
0,6,0,2,26,5,20,3,30,1,22,4,24})
execute("prisoner",{NOP, NOP, STP, 0, LDA+3, SUB+29, BRZ+18, LDA+3, STA+29,
BRZ+14, LDA+1, ADD+31, STA+1, JMP+2, LDA+0, ADD+31, STA+0,
JMP+2, LDA+3, STA+29, LDA+1, ADD+30, ADD+3, STA+1, LDA+0,
ADD+30, ADD+3, STA+0, JMP+2, 0, 1, 3})
-- subtractions yielding negative results
execute("0-255",{LDA+3,SUB+4,STP, 0,255})
execute("0-1",{LDA+3, SUB+4, STP, 0,1})
-- overflow on addition
execute("1+255",{LDA+3, ADD+4, STP, 1, 255})
|
http://rosettacode.org/wiki/Extreme_floating_point_values | Extreme floating point values | The IEEE floating point specification defines certain 'extreme' floating point values such as minus zero, -0.0, a value distinct from plus zero; not a number, NaN; and plus and minus infinity.
The task is to use expressions involving other 'normal' floating point values in your language to calculate these, (and maybe other), extreme floating point values in your language and assign them to variables.
Print the values of these variables if possible; and show some arithmetic with these values and variables.
If your language can directly enter these extreme floating point values then show it.
See also
What Every Computer Scientist Should Know About Floating-Point Arithmetic
Related tasks
Infinity
Detect division by zero
Literals/Floating point
| #Kotlin | Kotlin | // version 1.0.5-2
@Suppress("DIVISION_BY_ZERO", "FLOAT_LITERAL_CONFORMS_ZERO")
fun main(args: Array<String>) {
val inf = 1.0 / 0.0
val negInf = -1.0 / 0.0
val nan = 0.0 / 0.0
val negZero = -1.0e-325
println("*** Indirect ***\n")
println("Infinity : $inf")
println("Negative infinity : $negInf")
println("Not a number : $nan")
println("Negative zero : $negZero")
println("\n*** Direct ***\n")
println("Infinity : ${Double.POSITIVE_INFINITY}")
println("Negative infinity : ${Double.NEGATIVE_INFINITY}")
println("Not a number : ${Double.NaN}")
println("Negative zero : ${-0.0}")
println("\n*** Calculations ***\n")
println("inf * inf : ${inf * inf}")
println("inf + negInf : ${inf + negInf}")
println("nan / nan : ${nan / nan}")
println("negZero + 0.0 : ${negZero + 0.0}")
} |
http://rosettacode.org/wiki/Extreme_floating_point_values | Extreme floating point values | The IEEE floating point specification defines certain 'extreme' floating point values such as minus zero, -0.0, a value distinct from plus zero; not a number, NaN; and plus and minus infinity.
The task is to use expressions involving other 'normal' floating point values in your language to calculate these, (and maybe other), extreme floating point values in your language and assign them to variables.
Print the values of these variables if possible; and show some arithmetic with these values and variables.
If your language can directly enter these extreme floating point values then show it.
See also
What Every Computer Scientist Should Know About Floating-Point Arithmetic
Related tasks
Infinity
Detect division by zero
Literals/Floating point
| #Lua | Lua |
local inf=math.huge
local minusInf=-math.huge
local NaN=0/0
local negativeZeroSorta=-1E-240
|
http://rosettacode.org/wiki/Execute_SNUSP | Execute SNUSP | Execute SNUSP is an implementation of SNUSP.
Other implementations of SNUSP.
RCSNUSP
SNUSP
An implementation need only properly implement the Core SNUSP instructions ('$', '\', '/', '+', '-', '<', '>', ',', '.', '!', and '?'). Modular SNUSP ('#', '@') and Bloated SNUSP (':', ';', '%', and '&') are also allowed, but not required. Any extra characters that you implement should be noted in the description of your implementation. Any cell size is allowed, EOF support is optional, as is whether you have bounded or unbounded memory.
| #ALGOL_68 | ALGOL 68 | #
# snusp.icn, A Modular SNUSP interpreter
#
$define VERSION 0.6
# allow a couple of cli options
link options
# directions
$define DRIGHT 1
$define DLEFT 2
$define DUP 3
$define DDOWN 4
record position(row, col)
global dir, ip, ram
procedure main(argv)
local ch, codespace, col, dp, fn, line
local row := 1
local wid := 0
local dirs := []
local ips := []
local opts, verbose, debug
opts := options(argv, "-h! -v! -d!", errorproc)
\opts["v"] & verbose := 1
\opts["h"] & show_help(verbose)
\opts["d"] & debug := 1
ip := position(1,1)
# initial direction
dir := DRIGHT
# prepare initial memory
ram := list(1, 0)
# prepare code field
codespace := []
fn := open(argv[1], "r") | &input
if (fn === &input) & \opts["h"] then return
while line := read(fn) do {
put(codespace, line)
wid := max(*line, wid)
}
if *codespace = 0 then return
every line := !codespace do {
codespace[row] := left(codespace[row], wid)
# track starting indicator
if /col := find("$", codespace[row]) then {
ip.row := row
ip.col := col
}
row +:= 1
}
if \verbose then {
write("Starting at ", ip.row, ", ", ip.col, " with codespace:")
every write(!codespace)
}
dp := 1
repeat {
if not (ch := codespace[ip.row][ip.col]) then break
if \debug then {
write(&errout, "dir: ", dir, " ch: ", ch, " [", ord(ch), "]",
" row: ", ip.row, " col: ", ip.col,
" dp: ", dp, " ram[dp]: ", ram[dp])
}
case ch of {
# six of the bf instructions
"+": ram[dp] +:= 1
"-": ram[dp] -:= 1
">": resize(dp +:= 1)
"<": dp -:= 1
".": writes(char(ram[dp]) | char(0))
",": ram[dp] := getche()
# direction change, LURD, RULD, SKIP, SKIPZ
"\\": { # LURD
case dir of {
DRIGHT: dir := DDOWN
DLEFT: dir := DUP
DUP: dir := DLEFT
DDOWN: dir := DRIGHT
}
}
"/": { # RULD
case dir of {
DRIGHT: dir := DUP
DLEFT: dir := DDOWN
DUP: dir := DRIGHT
DDOWN: dir := DLEFT
}
}
"!": step()
"?": { # skipz
if ram[dp] = 0 then {
step()
}
}
# modular SNUSP
"@": { # Enter
push(dirs, dir)
push(ips, copy(ip))
}
"#": { # Leave
if *dirs < 1 then break
dir := pop(dirs)
ip := pop(ips)
step()
}
}
step()
}
end
# advance the ip depending on direction
procedure step()
case dir of {
DRIGHT: ip.col +:= 1
DLEFT: ip.col -:= 1
DUP: ip.row -:= 1
DDOWN: ip.row +:= 1
}
end
# enlarge memory when needed
procedure resize(elements)
until *ram >= elements do put(ram, 0)
end
# quick help or verbose help
procedure show_help(verbose)
write("SNUSP interpeter in Unicon, version ", VERSION)
write("CORE and MODULAR, not yet BLOATED")
write()
write("Usage: unicon snusp.icn -x [filename] [-h|-v|-d]")
write(" -h, help")
write(" -v, verbose (and verbose help")
write(" -d, debug (step tracer)")
if \verbose then {
write()
write("Instructions:")
write(" + INCR, Increment current memory location")
write(" - DECR, Decrement current memory location")
write(" > RIGHT, Advance memory pointer")
write(" < LEFT, Retreat memory pointer")
write(" . WRITE, Output contents of current memory cell, in ASCII")
write(" , READ, Accept key and place byte value in current memory cell")
write(" \\ LURD, If going:")
write(" left, go up")
write(" up, go left")
write(" right, go down")
write(" down, go right")
write(" / RULD, If going:")
write(" right, go up")
write(" up, go right")
write(" left, go down")
write(" down, go left")
write(" !, SKIP, Move forward one step in current direction")
write(" ?, SKIPZ, If current memory cell is zero then SKIP")
write("Modular SNUSP adds:")
write(" @, ENTER, Push direction and instruction pointer")
write(" #, LEAVE, Pop direction and instruction pointer and SKIP")
write()
write("All other characters are NOOP, explicitly includes =,|,spc")
write(" $, can set the starting location; first one found")
write()
write("Hello world examples:")
write()
write("CORE SNUSP:")
write("/++++!/===========?\\>++.>+.+++++++..+++\\")
write("\\+++\\ | /+>+++++++>/ /++++++++++<<.++>./")
write("$+++/ | \\+++++++++>\\ \\+++++.>.+++.-----\\")
write(" \\==-<<<<+>+++/ /=.>.+>.--------.-/")
write()
write("Modular SNUSP:")
write(" /@@@@++++# #+++@@\ #-----@@@\\n")
write("$@\\H.@/e.+++++++l.l.+++o.>>++++.< .<@/w.@\\o.+++r.++@\\l.@\\d.>+.@/.#")
write(" \\@@@@=>++++>+++++<<@+++++# #---@@/!=========/!==/")
write()
}
end |
http://rosettacode.org/wiki/Extend_your_language | Extend your language | Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Some programming languages allow you to extend the language. While this can be done to a certain degree in most languages (e.g. by using macros), other languages go much further. Most notably in the Forth and Lisp families, programming per se is done by extending the language without any formal distinction between built-in and user-defined elements.
If your language supports it, show how to introduce a new flow control mechanism. A practical and useful example is a four-way branch:
Occasionally, code must be written that depends on two conditions, resulting in up to four branches (depending on whether both, only the first, only the second, or none of the conditions are "true"). In a C-like language this could look like the following:
if (condition1isTrue) {
if (condition2isTrue)
bothConditionsAreTrue();
else
firstConditionIsTrue();
}
else if (condition2isTrue)
secondConditionIsTrue();
else
noConditionIsTrue();
Besides being rather cluttered, the statement(s) for 'condition2isTrue' must be written down twice. If 'condition2isTrue' were a lengthy and involved expression, it would be quite unreadable, and the code generated by the compiler might be unnecessarily large.
This can be improved by introducing a new keyword if2. It is similar to if, but takes two conditional statements instead of one, and up to three 'else' statements. One proposal (in pseudo-C syntax) might be:
if2 (condition1isTrue) (condition2isTrue)
bothConditionsAreTrue();
else1
firstConditionIsTrue();
else2
secondConditionIsTrue();
else
noConditionIsTrue();
Pick the syntax which suits your language. The keywords 'else1' and 'else2' are just examples. The new conditional expression should look, nest and behave analogously to the language's built-in 'if' statement.
| #COBOL | COBOL |
EVALUATE EXPRESSION-1 ALSO EXPRESSION-2
WHEN TRUE ALSO TRUE
DISPLAY 'Both are true.'
WHEN TRUE ALSO FALSE
DISPLAY 'Expression 1 is true.'
WHEN FALSE ALSO TRUE
DISPLAY 'Expression 2 is true.'
WHEN OTHER
DISPLAY 'Neither is true.'
END-EVALUATE
|
http://rosettacode.org/wiki/Exponentiation_order | Exponentiation order | This task will demonstrate the order of exponentiation (xy) when there are multiple exponents.
(Many programming languages, especially those with extended─precision integer arithmetic, usually support one of **, ^, ↑ or some such for exponentiation.)
Task requirements
Show the result of a language's evaluation of multiple exponentiation (either as an integer or floating point).
If your language's exponentiation operator is not one of the usual ones, please comment on how to recognize it.
Using whatever operator or syntax your language supports (if any), show the results in three lines (with identification):
5**3**2
(5**3)**2
5**(3**2)
If there are other methods (or formats) of multiple exponentiations, show them as well.
See also
MathWorld entry: exponentiation
Related tasks
exponentiation operator
arbitrary-precision integers (included)
Exponentiation with infix operators in (or operating on) the base
| #Bracmat | Bracmat | put$str$("5^3^2: " 5^3^2 "\n(5^3)^2: " (5^3)^2 "\n5^(3^2): " 5^(3^2) \n) |
http://rosettacode.org/wiki/Exponentiation_order | Exponentiation order | This task will demonstrate the order of exponentiation (xy) when there are multiple exponents.
(Many programming languages, especially those with extended─precision integer arithmetic, usually support one of **, ^, ↑ or some such for exponentiation.)
Task requirements
Show the result of a language's evaluation of multiple exponentiation (either as an integer or floating point).
If your language's exponentiation operator is not one of the usual ones, please comment on how to recognize it.
Using whatever operator or syntax your language supports (if any), show the results in three lines (with identification):
5**3**2
(5**3)**2
5**(3**2)
If there are other methods (or formats) of multiple exponentiations, show them as well.
See also
MathWorld entry: exponentiation
Related tasks
exponentiation operator
arbitrary-precision integers (included)
Exponentiation with infix operators in (or operating on) the base
| #C | C | #include<stdio.h>
#include<math.h>
int main()
{
printf("(5 ^ 3) ^ 2 = %.0f",pow(pow(5,3),2));
printf("\n5 ^ (3 ^ 2) = %.0f",pow(5,pow(3,2)));
return 0;
} |
http://rosettacode.org/wiki/Exponentiation_order | Exponentiation order | This task will demonstrate the order of exponentiation (xy) when there are multiple exponents.
(Many programming languages, especially those with extended─precision integer arithmetic, usually support one of **, ^, ↑ or some such for exponentiation.)
Task requirements
Show the result of a language's evaluation of multiple exponentiation (either as an integer or floating point).
If your language's exponentiation operator is not one of the usual ones, please comment on how to recognize it.
Using whatever operator or syntax your language supports (if any), show the results in three lines (with identification):
5**3**2
(5**3)**2
5**(3**2)
If there are other methods (or formats) of multiple exponentiations, show them as well.
See also
MathWorld entry: exponentiation
Related tasks
exponentiation operator
arbitrary-precision integers (included)
Exponentiation with infix operators in (or operating on) the base
| #C.2B.2B | C++ | #include <iostream>
#include <cmath>
int main() {
std::cout << "(5 ^ 3) ^ 2 = " << (uint) pow(pow(5,3), 2) << std::endl;
std::cout << "5 ^ (3 ^ 2) = "<< (uint) pow(5, (pow(3, 2)));
return EXIT_SUCCESS;
} |
http://rosettacode.org/wiki/Factorions | Factorions |
Definition
A factorion is a natural number that equals the sum of the factorials of its digits.
Example
145 is a factorion in base 10 because:
1! + 4! + 5! = 1 + 24 + 120 = 145
It can be shown (see talk page) that no factorion in base 10 can exceed 1,499,999.
Task
Write a program in your language to demonstrate, by calculating and printing out the factorions, that:
There are 3 factorions in base 9
There are 4 factorions in base 10
There are 5 factorions in base 11
There are 2 factorions in base 12 (up to the same upper bound as for base 10)
See also
Wikipedia article
OEIS:A014080 - Factorions in base 10
OEIS:A193163 - Factorions in base n
| #PureBasic | PureBasic | Declare main()
If OpenConsole() : main() : Else : End 1 : EndIf
Input() : End
Procedure main()
Define.i n,b,d,i,j,sum
Dim fact.i(12)
fact(0)=1
For n=1 To 11 : fact(n)=fact(n-1)*n : Next
For b=9 To 12
PrintN("The factorions for base "+Str(b)+" are: ")
For i=1 To 1500000-1
sum=0 : j=i
While j>0
d=j%b : sum+fact(d) : j/b
Wend
If sum=i : Print(Str(i)+" ") : EndIf
Next
Print(~"\n\n")
Next
EndProcedure |
http://rosettacode.org/wiki/Factorions | Factorions |
Definition
A factorion is a natural number that equals the sum of the factorials of its digits.
Example
145 is a factorion in base 10 because:
1! + 4! + 5! = 1 + 24 + 120 = 145
It can be shown (see talk page) that no factorion in base 10 can exceed 1,499,999.
Task
Write a program in your language to demonstrate, by calculating and printing out the factorions, that:
There are 3 factorions in base 9
There are 4 factorions in base 10
There are 5 factorions in base 11
There are 2 factorions in base 12 (up to the same upper bound as for base 10)
See also
Wikipedia article
OEIS:A014080 - Factorions in base 10
OEIS:A193163 - Factorions in base n
| #Python | Python | fact = [1] # cache factorials from 0 to 11
for n in range(1, 12):
fact.append(fact[n-1] * n)
for b in range(9, 12+1):
print(f"The factorions for base {b} are:")
for i in range(1, 1500000):
fact_sum = 0
j = i
while j > 0:
d = j % b
fact_sum += fact[d]
j = j//b
if fact_sum == i:
print(i, end=" ")
print("\n")
|
http://rosettacode.org/wiki/Filter | Filter | Task
Select certain elements from an Array into a new Array in a generic way.
To demonstrate, select all even numbers from an Array.
As an option, give a second solution which filters destructively,
by modifying the original Array rather than creating a new Array.
| #Ursala | Ursala | #import std
#import nat
x = <89,36,13,15,41,39,21,3,15,92,16,59,52,88,33,65,54,88,93,43>
#cast %nL
y = (not remainder\2)*~ x |
http://rosettacode.org/wiki/FizzBuzz | FizzBuzz | Task
Write a program that prints the integers from 1 to 100 (inclusive).
But:
for multiples of three, print Fizz (instead of the number)
for multiples of five, print Buzz (instead of the number)
for multiples of both three and five, print FizzBuzz (instead of the number)
The FizzBuzz problem was presented as the lowest level of comprehension required to illustrate adequacy.
Also see
(a blog) dont-overthink-fizzbuzz
(a blog) fizzbuzz-the-programmers-stairway-to-heaven
| #PostScript | PostScript | 1 1 100 {
/c false def
dup 3 mod 0 eq { (Fizz) print /c true def } if
dup 5 mod 0 eq { (Buzz) print /c true def } if
c {pop}{( ) cvs print} ifelse
(\n) print
} for |
http://rosettacode.org/wiki/Extensible_prime_generator | Extensible prime generator | Task
Write a generator of prime numbers, in order, that will automatically adjust to accommodate the generation of any reasonably high prime.
The routine should demonstrably rely on either:
Being based on an open-ended counter set to count without upper limit other than system or programming language limits. In this case, explain where this counter is in the code.
Being based on a limit that is extended automatically. In this case, choose a small limit that ensures the limit will be passed when generating some of the values to be asked for below.
If other methods of creating an extensible prime generator are used, the algorithm's means of extensibility/lack of limits should be stated.
The routine should be used to:
Show the first twenty primes.
Show the primes between 100 and 150.
Show the number of primes between 7,700 and 8,000.
Show the 10,000th prime.
Show output on this page.
Note: You may reference code already on this site if it is written to be imported/included, then only the code necessary for import and the performance of this task need be shown. (It is also important to leave a forward link on the referenced tasks entry so that later editors know that the code is used for multiple tasks).
Note 2: If a languages in-built prime generator is extensible or is guaranteed to generate primes up to a system limit, (231 or memory overflow for example), then this may be used as long as an explanation of the limits of the prime generator is also given. (Which may include a link to/excerpt from, language documentation).
Note 3:The task is written so it may be useful in solving the task Emirp primes as well as others (depending on its efficiency).
Reference
Prime Numbers. Website with large count of primes.
| #CoffeeScript | CoffeeScript |
primes = () ->
yield 2
yield 3
sieve = ([] for i in [1..3])
sieve[0].push 3
[r, s] = [3, 9]
pos = 1
n = 5
loop
isPrime = true
if sieve[pos].length > 0 # this entry has a list of factors
isPrime = false
sieve[(pos + m) % sieve.length].push m for m in sieve[pos]
sieve[pos] = []
if n is s # n is the next square
if isPrime
isPrime = false # r divides n, so not actually prime
sieve[(pos + r) % sieve.length].push r # however, r is prime
r += 2
s = r*r
yield n if isPrime
n += 2
pos += 1
if pos is sieve.length
sieve.push [] # array size must exceed largest prime found
sieve.push [] # adding two entries keeps size = O(sqrt n)
pos = 0
undefined # prevent CoffeeScript from aggregating values
module.exports = {
primes
}
|
http://rosettacode.org/wiki/Fibonacci_sequence | Fibonacci sequence | The Fibonacci sequence is a sequence Fn of natural numbers defined recursively:
F0 = 0
F1 = 1
Fn = Fn-1 + Fn-2, if n>1
Task
Write a function to generate the nth Fibonacci number.
Solutions can be iterative or recursive (though recursive solutions are generally considered too slow and are mostly used as an exercise in recursion).
The sequence is sometimes extended into negative numbers by using a straightforward inverse of the positive definition:
Fn = Fn+2 - Fn+1, if n<0
support for negative n in the solution is optional.
Related tasks
Fibonacci n-step number sequences
Leonardo numbers
References
Wikipedia, Fibonacci number
Wikipedia, Lucas number
MathWorld, Fibonacci Number
Some identities for r-Fibonacci numbers
OEIS Fibonacci numbers
OEIS Lucas numbers
| #Futhark | Futhark |
fun main(n: int): int =
loop((a,b) = (0,1)) = for _i < n do
(b, a + b)
in a
|
http://rosettacode.org/wiki/Factors_of_an_integer | Factors of an integer |
Basic Data Operation
This is a basic data operation. It represents a fundamental action on a basic data type.
You may see other such operations in the Basic Data Operations category, or:
Integer Operations
Arithmetic |
Comparison
Boolean Operations
Bitwise |
Logical
String Operations
Concatenation |
Interpolation |
Comparison |
Matching
Memory Operations
Pointers & references |
Addresses
Task
Compute the factors of a positive integer.
These factors are the positive integers by which the number being factored can be divided to yield a positive integer result.
(Though the concepts function correctly for zero and negative integers, the set of factors of zero has countably infinite members, and the factors of negative integers can be obtained from the factors of related positive numbers without difficulty; this task does not require handling of either of these cases).
Note that every prime number has two factors: 1 and itself.
Related tasks
count in factors
prime decomposition
Sieve of Eratosthenes
primality by trial division
factors of a Mersenne number
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
sequence: smallest number greater than previous term with exactly n divisors
| #PILOT | PILOT | T :Enter a number.
A :#n
C :factor = 1
T :The factors of #n are:
*Loop
C :remainder = n % factor
T ( remainder = 0 ) :#factor
J ( factor = n ) :*Finished
C :factor = factor + 1
J :*Loop
*Finished
END: |
http://rosettacode.org/wiki/Execute_Computer/Zero | Execute Computer/Zero | Execute Computer/Zero is an implementation of Computer/zero Assembly.
Other implementations of Computer/zero Assembly.
Task
Create a Computer/zero Assembly emulator. You may consider this webpage as a reference implementation. Output the results of the sample programs "2+2" and "7*8" found there.
The virtual machine "bytecode" needs to be able to modify itself (or at least act as though it can) while the virtual machine is running, to be consistent with the reference implementation.
For output, it is sufficient to have the implementation of the STP opcode return the accumulator to your actual language, and then use your standard printing routines to output it.
Bonus Points
Run all 5 sample programs at the aforementioned website and output their results.
| #Python | Python | """Computer/zero Assembly emulator. Requires Python >= 3.7"""
import re
from typing import Dict
from typing import Iterable
from typing import List
from typing import NamedTuple
from typing import Optional
from typing import Tuple
NOP = 0b000
LDA = 0b001
STA = 0b010
ADD = 0b011
SUB = 0b100
BRZ = 0b101
JMP = 0b110
STP = 0b111
OPCODES = {
"NOP": NOP,
"LDA": LDA,
"STA": STA,
"ADD": ADD,
"SUB": SUB,
"BRZ": BRZ,
"JMP": JMP,
"STP": STP,
}
RE_INSTRUCTION = re.compile(
r"\s*"
r"(?:(?P<label>\w+):)?"
r"\s*"
rf"(?P<opcode>{'|'.join(OPCODES)})?"
r"\s*"
r"(?P<argument>\w+)?"
r"\s*"
r"(?:;(?P<comment>[\w\s]+))?"
)
class AssemblySyntaxError(Exception):
pass
class Instruction(NamedTuple):
label: Optional[str]
opcode: Optional[str]
argument: Optional[str]
comment: Optional[str]
def parse(assembly: str) -> Tuple[List[Instruction], Dict[str, int]]:
instructions: List[Instruction] = []
labels: Dict[str, int] = {}
linenum: int = 0
for line in assembly.split("\n"):
match = RE_INSTRUCTION.match(line)
if not match:
raise AssemblySyntaxError(f"{line}: {linenum}")
instructions.append(Instruction(**match.groupdict()))
label = match.group(1)
if label:
labels[label] = linenum
linenum += 1
return instructions, labels
def compile(instructions: List[Instruction], labels: Dict[str, int]) -> Iterable[int]:
for instruction in instructions:
if instruction.argument is None:
argument = 0
elif instruction.argument.isnumeric():
argument = int(instruction.argument)
else:
argument = labels[instruction.argument]
if instruction.opcode:
yield OPCODES[instruction.opcode] << 5 | argument
else:
yield argument
def run(bytecode: bytes) -> int:
accumulator = 0
program_counter = 0
memory = list(bytecode)[:32] + [0 for _ in range(32 - len(bytecode))]
while program_counter < 32:
operation = memory[program_counter] >> 5
argument = memory[program_counter] & 0b11111
program_counter += 1
if operation == NOP:
continue
elif operation == LDA:
accumulator = memory[argument]
elif operation == STA:
memory[argument] = accumulator
elif operation == ADD:
accumulator = (accumulator + memory[argument]) % 256
elif operation == SUB:
accumulator = (accumulator - memory[argument]) % 256
elif operation == BRZ:
if accumulator == 0:
program_counter = argument
elif operation == JMP:
program_counter = argument
elif operation == STP:
break
else:
raise Exception(f"error: {operation} {argument}")
return accumulator
SAMPLES = [
"""\
LDA x
ADD y ; accumulator = x + y
STP
x: 2
y: 2
""",
"""\
loop: LDA prodt
ADD x
STA prodt
LDA y
SUB one
STA y
BRZ done
JMP loop
done: LDA prodt ; to display it
STP
x: 8
y: 7
prodt: 0
one: 1
""",
"""\
loop: LDA n
STA temp
ADD m
STA n
LDA temp
STA m
LDA count
SUB one
BRZ done
STA count
JMP loop
done: LDA n ; to display it
STP
m: 1
n: 1
temp: 0
count: 8 ; valid range: 1-11
one: 1
""",
"""\
start: LDA load
ADD car ; head of list
STA ldcar
ADD one
STA ldcdr ; next CONS cell
ldcar: NOP
STA value
ldcdr: NOP
BRZ done ; 0 stands for NIL
STA car
JMP start
done: LDA value ; CAR of last CONS
STP
load: LDA 0
value: 0
car: 28
one: 1
; order of CONS cells
; in memory
; does not matter
6
0 ; 0 stands for NIL
2 ; (CADR ls)
26 ; (CDDR ls) -- etc.
5
20
3
30
1 ; value of (CAR ls)
22 ; points to (CDR ls)
4
24
""",
"""\
p: 0 ; NOP in first round
c: 0
start: STP ; wait for p's move
pmove: NOP
LDA pmove
SUB cmove
BRZ same
LDA pmove
STA cmove ; tit for tat
BRZ cdeft
LDA c ; p defected, c did not
ADD three
STA c
JMP start
cdeft: LDA p
ADD three
STA p
JMP start
same: LDA pmove
STA cmove ; tit for tat
LDA p
ADD one
ADD pmove
STA p
LDA c
ADD one
ADD pmove
STA c
JMP start
cmove: 0 ; co-operate initially
one: 1
three: 3
""",
"""\
LDA 3
SUB 4
STP 0
0
255
""",
"""\
LDA 3
SUB 4
STP 0
0
1
""",
"""\
LDA 3
ADD 4
STP 0
1
255
""",
]
def main() -> None:
for sample in SAMPLES:
instructions, labels = parse(sample)
bytecode = bytes(compile(instructions, labels))
result = run(bytecode)
print(result)
if __name__ == "__main__":
main()
|
http://rosettacode.org/wiki/Extreme_floating_point_values | Extreme floating point values | The IEEE floating point specification defines certain 'extreme' floating point values such as minus zero, -0.0, a value distinct from plus zero; not a number, NaN; and plus and minus infinity.
The task is to use expressions involving other 'normal' floating point values in your language to calculate these, (and maybe other), extreme floating point values in your language and assign them to variables.
Print the values of these variables if possible; and show some arithmetic with these values and variables.
If your language can directly enter these extreme floating point values then show it.
See also
What Every Computer Scientist Should Know About Floating-Point Arithmetic
Related tasks
Infinity
Detect division by zero
Literals/Floating point
| #Mathematica_.2F_Wolfram_Language | Mathematica / Wolfram Language | Column@{ReleaseHold[
Function[expression,
Row@{HoldForm@InputForm@expression, " = ", Quiet@expression},
HoldAll] /@
Hold[1./0., 0./0., Limit[-Log[x], x -> 0], Limit[Log[x], x -> 0],
Infinity + 1, Infinity + Infinity, 2 Infinity,
Infinity - Infinity, 0 Infinity, ComplexInfinity + 1,
ComplexInfinity + ComplexInfinity, 2 ComplexInfinity,
0 ComplexInfinity, Indeterminate + 1, 0 Indeterminate]]} |
http://rosettacode.org/wiki/Extreme_floating_point_values | Extreme floating point values | The IEEE floating point specification defines certain 'extreme' floating point values such as minus zero, -0.0, a value distinct from plus zero; not a number, NaN; and plus and minus infinity.
The task is to use expressions involving other 'normal' floating point values in your language to calculate these, (and maybe other), extreme floating point values in your language and assign them to variables.
Print the values of these variables if possible; and show some arithmetic with these values and variables.
If your language can directly enter these extreme floating point values then show it.
See also
What Every Computer Scientist Should Know About Floating-Point Arithmetic
Related tasks
Infinity
Detect division by zero
Literals/Floating point
| #Maxima | Maxima | USER>write 3e145
30000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
USER>write 3e146
<MAXNUMBER>
|
http://rosettacode.org/wiki/Execute_SNUSP | Execute SNUSP | Execute SNUSP is an implementation of SNUSP.
Other implementations of SNUSP.
RCSNUSP
SNUSP
An implementation need only properly implement the Core SNUSP instructions ('$', '\', '/', '+', '-', '<', '>', ',', '.', '!', and '?'). Modular SNUSP ('#', '@') and Bloated SNUSP (':', ';', '%', and '&') are also allowed, but not required. Any extra characters that you implement should be noted in the description of your implementation. Any cell size is allowed, EOF support is optional, as is whether you have bounded or unbounded memory.
| #AutoHotkey | AutoHotkey | #
# snusp.icn, A Modular SNUSP interpreter
#
$define VERSION 0.6
# allow a couple of cli options
link options
# directions
$define DRIGHT 1
$define DLEFT 2
$define DUP 3
$define DDOWN 4
record position(row, col)
global dir, ip, ram
procedure main(argv)
local ch, codespace, col, dp, fn, line
local row := 1
local wid := 0
local dirs := []
local ips := []
local opts, verbose, debug
opts := options(argv, "-h! -v! -d!", errorproc)
\opts["v"] & verbose := 1
\opts["h"] & show_help(verbose)
\opts["d"] & debug := 1
ip := position(1,1)
# initial direction
dir := DRIGHT
# prepare initial memory
ram := list(1, 0)
# prepare code field
codespace := []
fn := open(argv[1], "r") | &input
if (fn === &input) & \opts["h"] then return
while line := read(fn) do {
put(codespace, line)
wid := max(*line, wid)
}
if *codespace = 0 then return
every line := !codespace do {
codespace[row] := left(codespace[row], wid)
# track starting indicator
if /col := find("$", codespace[row]) then {
ip.row := row
ip.col := col
}
row +:= 1
}
if \verbose then {
write("Starting at ", ip.row, ", ", ip.col, " with codespace:")
every write(!codespace)
}
dp := 1
repeat {
if not (ch := codespace[ip.row][ip.col]) then break
if \debug then {
write(&errout, "dir: ", dir, " ch: ", ch, " [", ord(ch), "]",
" row: ", ip.row, " col: ", ip.col,
" dp: ", dp, " ram[dp]: ", ram[dp])
}
case ch of {
# six of the bf instructions
"+": ram[dp] +:= 1
"-": ram[dp] -:= 1
">": resize(dp +:= 1)
"<": dp -:= 1
".": writes(char(ram[dp]) | char(0))
",": ram[dp] := getche()
# direction change, LURD, RULD, SKIP, SKIPZ
"\\": { # LURD
case dir of {
DRIGHT: dir := DDOWN
DLEFT: dir := DUP
DUP: dir := DLEFT
DDOWN: dir := DRIGHT
}
}
"/": { # RULD
case dir of {
DRIGHT: dir := DUP
DLEFT: dir := DDOWN
DUP: dir := DRIGHT
DDOWN: dir := DLEFT
}
}
"!": step()
"?": { # skipz
if ram[dp] = 0 then {
step()
}
}
# modular SNUSP
"@": { # Enter
push(dirs, dir)
push(ips, copy(ip))
}
"#": { # Leave
if *dirs < 1 then break
dir := pop(dirs)
ip := pop(ips)
step()
}
}
step()
}
end
# advance the ip depending on direction
procedure step()
case dir of {
DRIGHT: ip.col +:= 1
DLEFT: ip.col -:= 1
DUP: ip.row -:= 1
DDOWN: ip.row +:= 1
}
end
# enlarge memory when needed
procedure resize(elements)
until *ram >= elements do put(ram, 0)
end
# quick help or verbose help
procedure show_help(verbose)
write("SNUSP interpeter in Unicon, version ", VERSION)
write("CORE and MODULAR, not yet BLOATED")
write()
write("Usage: unicon snusp.icn -x [filename] [-h|-v|-d]")
write(" -h, help")
write(" -v, verbose (and verbose help")
write(" -d, debug (step tracer)")
if \verbose then {
write()
write("Instructions:")
write(" + INCR, Increment current memory location")
write(" - DECR, Decrement current memory location")
write(" > RIGHT, Advance memory pointer")
write(" < LEFT, Retreat memory pointer")
write(" . WRITE, Output contents of current memory cell, in ASCII")
write(" , READ, Accept key and place byte value in current memory cell")
write(" \\ LURD, If going:")
write(" left, go up")
write(" up, go left")
write(" right, go down")
write(" down, go right")
write(" / RULD, If going:")
write(" right, go up")
write(" up, go right")
write(" left, go down")
write(" down, go left")
write(" !, SKIP, Move forward one step in current direction")
write(" ?, SKIPZ, If current memory cell is zero then SKIP")
write("Modular SNUSP adds:")
write(" @, ENTER, Push direction and instruction pointer")
write(" #, LEAVE, Pop direction and instruction pointer and SKIP")
write()
write("All other characters are NOOP, explicitly includes =,|,spc")
write(" $, can set the starting location; first one found")
write()
write("Hello world examples:")
write()
write("CORE SNUSP:")
write("/++++!/===========?\\>++.>+.+++++++..+++\\")
write("\\+++\\ | /+>+++++++>/ /++++++++++<<.++>./")
write("$+++/ | \\+++++++++>\\ \\+++++.>.+++.-----\\")
write(" \\==-<<<<+>+++/ /=.>.+>.--------.-/")
write()
write("Modular SNUSP:")
write(" /@@@@++++# #+++@@\ #-----@@@\\n")
write("$@\\H.@/e.+++++++l.l.+++o.>>++++.< .<@/w.@\\o.+++r.++@\\l.@\\d.>+.@/.#")
write(" \\@@@@=>++++>+++++<<@+++++# #---@@/!=========/!==/")
write()
}
end |
http://rosettacode.org/wiki/Extend_your_language | Extend your language | Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Some programming languages allow you to extend the language. While this can be done to a certain degree in most languages (e.g. by using macros), other languages go much further. Most notably in the Forth and Lisp families, programming per se is done by extending the language without any formal distinction between built-in and user-defined elements.
If your language supports it, show how to introduce a new flow control mechanism. A practical and useful example is a four-way branch:
Occasionally, code must be written that depends on two conditions, resulting in up to four branches (depending on whether both, only the first, only the second, or none of the conditions are "true"). In a C-like language this could look like the following:
if (condition1isTrue) {
if (condition2isTrue)
bothConditionsAreTrue();
else
firstConditionIsTrue();
}
else if (condition2isTrue)
secondConditionIsTrue();
else
noConditionIsTrue();
Besides being rather cluttered, the statement(s) for 'condition2isTrue' must be written down twice. If 'condition2isTrue' were a lengthy and involved expression, it would be quite unreadable, and the code generated by the compiler might be unnecessarily large.
This can be improved by introducing a new keyword if2. It is similar to if, but takes two conditional statements instead of one, and up to three 'else' statements. One proposal (in pseudo-C syntax) might be:
if2 (condition1isTrue) (condition2isTrue)
bothConditionsAreTrue();
else1
firstConditionIsTrue();
else2
secondConditionIsTrue();
else
noConditionIsTrue();
Pick the syntax which suits your language. The keywords 'else1' and 'else2' are just examples. The new conditional expression should look, nest and behave analogously to the language's built-in 'if' statement.
| #Common_Lisp | Common Lisp | (defmacro if2 (cond1 cond2 both first second &rest neither)
(let ((res1 (gensym))
(res2 (gensym)))
`(let ((,res1 ,cond1)
(,res2 ,cond2))
(cond ((and ,res1 ,res2) ,both)
(,res1 ,first)
(,res2 ,second)
(t ,@neither))))) |
http://rosettacode.org/wiki/Exponentiation_order | Exponentiation order | This task will demonstrate the order of exponentiation (xy) when there are multiple exponents.
(Many programming languages, especially those with extended─precision integer arithmetic, usually support one of **, ^, ↑ or some such for exponentiation.)
Task requirements
Show the result of a language's evaluation of multiple exponentiation (either as an integer or floating point).
If your language's exponentiation operator is not one of the usual ones, please comment on how to recognize it.
Using whatever operator or syntax your language supports (if any), show the results in three lines (with identification):
5**3**2
(5**3)**2
5**(3**2)
If there are other methods (or formats) of multiple exponentiations, show them as well.
See also
MathWorld entry: exponentiation
Related tasks
exponentiation operator
arbitrary-precision integers (included)
Exponentiation with infix operators in (or operating on) the base
| #C.23 | C# | using System;
namespace exponents
{
class Program
{
static void Main(string[] args)
{
/*
* Like C, C# does not have an exponent operator.
* Exponentiation is done via Math.Pow, which
* only takes two arguments
*/
Console.WriteLine(Math.Pow(Math.Pow(5, 3), 2));
Console.WriteLine(Math.Pow(5, Math.Pow(3, 2)));
Console.Read();
}
}
}
|
http://rosettacode.org/wiki/Exponentiation_order | Exponentiation order | This task will demonstrate the order of exponentiation (xy) when there are multiple exponents.
(Many programming languages, especially those with extended─precision integer arithmetic, usually support one of **, ^, ↑ or some such for exponentiation.)
Task requirements
Show the result of a language's evaluation of multiple exponentiation (either as an integer or floating point).
If your language's exponentiation operator is not one of the usual ones, please comment on how to recognize it.
Using whatever operator or syntax your language supports (if any), show the results in three lines (with identification):
5**3**2
(5**3)**2
5**(3**2)
If there are other methods (or formats) of multiple exponentiations, show them as well.
See also
MathWorld entry: exponentiation
Related tasks
exponentiation operator
arbitrary-precision integers (included)
Exponentiation with infix operators in (or operating on) the base
| #Clojure | Clojure | (use 'clojure.math.numeric-tower)
;; (5**3)**2
(expt (expt 5 3) 2) ; => 15625
;; 5**(3**2)
(expt 5 (expt 3 2)) ; => 1953125
;; (5**3)**2 alternative: use reduce
(reduce expt [5 3 2]) ; => 15625
;; 5**(3**2) alternative: evaluating right-to-left with reduce requires a small modification
(defn rreduce [f coll] (reduce #(f %2 %) (reverse coll)))
(rreduce expt [5 3 2]) ; => 1953125 |
http://rosettacode.org/wiki/Factorions | Factorions |
Definition
A factorion is a natural number that equals the sum of the factorials of its digits.
Example
145 is a factorion in base 10 because:
1! + 4! + 5! = 1 + 24 + 120 = 145
It can be shown (see talk page) that no factorion in base 10 can exceed 1,499,999.
Task
Write a program in your language to demonstrate, by calculating and printing out the factorions, that:
There are 3 factorions in base 9
There are 4 factorions in base 10
There are 5 factorions in base 11
There are 2 factorions in base 12 (up to the same upper bound as for base 10)
See also
Wikipedia article
OEIS:A014080 - Factorions in base 10
OEIS:A193163 - Factorions in base n
| #Quackery | Quackery | [ table ] is results ( n --> s )
4 times
[ ' [ stack [ ] ]
copy
' results put ]
[ results dup take
rot join swap put ] is addresult ( n n --> )
[ table 9 10 11 12 ] is radix ( n --> n )
[ table 1 ] is ! ( n --> n )
1 11 times
[ i^ 1+ * dup
' ! put ]
drop
[ dip dup
0 temp put
[ tuck /mod !
temp tally
swap over 0 =
until ]
2drop
temp take = ] is factorion ( n n --> b )
1500000 times
[ i^ 4 times
[ dup
i^ radix
factorion if
[ dup i^
addresult ] ]
drop ]
4 times
[ say "Factorions for base "
i^ radix echo say ": "
i^ results take echo cr ] |
http://rosettacode.org/wiki/Factorions | Factorions |
Definition
A factorion is a natural number that equals the sum of the factorials of its digits.
Example
145 is a factorion in base 10 because:
1! + 4! + 5! = 1 + 24 + 120 = 145
It can be shown (see talk page) that no factorion in base 10 can exceed 1,499,999.
Task
Write a program in your language to demonstrate, by calculating and printing out the factorions, that:
There are 3 factorions in base 9
There are 4 factorions in base 10
There are 5 factorions in base 11
There are 2 factorions in base 12 (up to the same upper bound as for base 10)
See also
Wikipedia article
OEIS:A014080 - Factorions in base 10
OEIS:A193163 - Factorions in base n
| #Racket | Racket | #lang racket
(define fact
(curry list-ref (for/fold ([result (list 1)] #:result (reverse result))
([x (in-range 1 20)])
(cons (* x (first result)) result))))
(for ([b (in-range 9 13)])
(printf "The factorions for base ~a are:\n" b)
(for ([i (in-range 1 1500000)])
(let loop ([sum 0] [n i])
(cond
[(positive? n) (loop (+ sum (fact (modulo n b))) (quotient n b))]
[(= sum i) (printf "~a " i)])))
(newline)) |
http://rosettacode.org/wiki/Factorions | Factorions |
Definition
A factorion is a natural number that equals the sum of the factorials of its digits.
Example
145 is a factorion in base 10 because:
1! + 4! + 5! = 1 + 24 + 120 = 145
It can be shown (see talk page) that no factorion in base 10 can exceed 1,499,999.
Task
Write a program in your language to demonstrate, by calculating and printing out the factorions, that:
There are 3 factorions in base 9
There are 4 factorions in base 10
There are 5 factorions in base 11
There are 2 factorions in base 12 (up to the same upper bound as for base 10)
See also
Wikipedia article
OEIS:A014080 - Factorions in base 10
OEIS:A193163 - Factorions in base n
| #Raku | Raku | constant @factorial = 1, |[\*] 1..*;
constant $limit = 1500000;
constant $bases = 9 .. 12;
my @result;
$bases.map: -> $base {
@result[$base] = "\nFactorions in base $base:\n1 2";
sink (1 .. $limit div $base).map: -> $i {
my $product = $i * $base;
my $partial;
for $i.polymod($base xx *) {
$partial += @factorial[$_];
last if $partial > $product
}
next if $partial > $product;
my $sum;
for ^$base {
last if ($sum = $partial + @factorial[$_]) > $product + $_;
@result[$base] ~= " $sum" and last if $sum == $product + $_
}
}
}
.say for @result[$bases]; |
http://rosettacode.org/wiki/Filter | Filter | Task
Select certain elements from an Array into a new Array in a generic way.
To demonstrate, select all even numbers from an Array.
As an option, give a second solution which filters destructively,
by modifying the original Array rather than creating a new Array.
| #V | V | [even? dup 2 / >int 2 * - zero?].
[1 2 3 4 5 6 7 8 9] [even?] filter
=[2 4 6 8] |
http://rosettacode.org/wiki/FizzBuzz | FizzBuzz | Task
Write a program that prints the integers from 1 to 100 (inclusive).
But:
for multiples of three, print Fizz (instead of the number)
for multiples of five, print Buzz (instead of the number)
for multiples of both three and five, print FizzBuzz (instead of the number)
The FizzBuzz problem was presented as the lowest level of comprehension required to illustrate adequacy.
Also see
(a blog) dont-overthink-fizzbuzz
(a blog) fizzbuzz-the-programmers-stairway-to-heaven
| #Potion | Potion |
1 to 100 (a):
if (a % 15 == 0):
'FizzBuzz'.
elsif (a % 3 == 0):
'Fizz'.
elsif (a % 5 == 0):
'Buzz'.
else: a. string print
"\n" print. |
http://rosettacode.org/wiki/Extensible_prime_generator | Extensible prime generator | Task
Write a generator of prime numbers, in order, that will automatically adjust to accommodate the generation of any reasonably high prime.
The routine should demonstrably rely on either:
Being based on an open-ended counter set to count without upper limit other than system or programming language limits. In this case, explain where this counter is in the code.
Being based on a limit that is extended automatically. In this case, choose a small limit that ensures the limit will be passed when generating some of the values to be asked for below.
If other methods of creating an extensible prime generator are used, the algorithm's means of extensibility/lack of limits should be stated.
The routine should be used to:
Show the first twenty primes.
Show the primes between 100 and 150.
Show the number of primes between 7,700 and 8,000.
Show the 10,000th prime.
Show output on this page.
Note: You may reference code already on this site if it is written to be imported/included, then only the code necessary for import and the performance of this task need be shown. (It is also important to leave a forward link on the referenced tasks entry so that later editors know that the code is used for multiple tasks).
Note 2: If a languages in-built prime generator is extensible or is guaranteed to generate primes up to a system limit, (231 or memory overflow for example), then this may be used as long as an explanation of the limits of the prime generator is also given. (Which may include a link to/excerpt from, language documentation).
Note 3:The task is written so it may be useful in solving the task Emirp primes as well as others (depending on its efficiency).
Reference
Prime Numbers. Website with large count of primes.
| #D | D | void main() {
import std.stdio, std.range, std.algorithm, sieve_of_eratosthenes3;
Prime prime;
writeln("First twenty primes:\n", 20.iota.map!prime);
writeln("Primes primes between 100 and 150:\n",
uint.max.iota.map!prime.until!q{a > 150}.filter!q{a > 99});
writeln("Number of primes between 7,700 and 8,000: ",
uint.max.iota.map!prime.until!q{a > 8_000}
.filter!q{a > 7_699}.walkLength);
writeln("10,000th prime: ", prime(9_999));
} |
http://rosettacode.org/wiki/Fibonacci_sequence | Fibonacci sequence | The Fibonacci sequence is a sequence Fn of natural numbers defined recursively:
F0 = 0
F1 = 1
Fn = Fn-1 + Fn-2, if n>1
Task
Write a function to generate the nth Fibonacci number.
Solutions can be iterative or recursive (though recursive solutions are generally considered too slow and are mostly used as an exercise in recursion).
The sequence is sometimes extended into negative numbers by using a straightforward inverse of the positive definition:
Fn = Fn+2 - Fn+1, if n<0
support for negative n in the solution is optional.
Related tasks
Fibonacci n-step number sequences
Leonardo numbers
References
Wikipedia, Fibonacci number
Wikipedia, Lucas number
MathWorld, Fibonacci Number
Some identities for r-Fibonacci numbers
OEIS Fibonacci numbers
OEIS Lucas numbers
| #FutureBasic | FutureBasic | window 1, @"Fibonacci Sequence", (0,0,480,620)
local fn Fibonacci( n as long ) as long
static long s1
static long s2
long temp
if ( n < 2 )
s1 = n
exit fn
else
temp = s1 + s2
s2 = s1
s1 = temp
exit fn
end if
end fn = s1
long i
CFTimeInterval t
t = fn CACurrentMediaTime
for i = 0 to 40
print i;@".\t";fn Fibonacci(i)
next i
print : printf @"Compute time: %.3f ms",(fn CACurrentMediaTime-t)*1000
HandleEvents |
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