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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.
| #PowerShell | PowerShell | $array = -15..37
$array | Where-Object { $_ % 2 -eq 0 } |
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
| #Never | Never | func fizz_buzz() -> int
{
var i = 1;
for (i = 1; i <= 100; i = i + 1)
{
/* if (i % 15 == 0) */
if (i % 3 == 0 && i % 5 == 0)
{
prints("Fizz Buzz\n")
}
else if (i % 3 == 0)
{
prints("Fizz\n")
}
else if (i % 5 == 0)
{
prints("Buzz\n")
}
else
{
prints(i + "\n")
}
};
0
}
func main() -> int {
fizz_buzz();
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
| #DWScript | DWScript | function fib(N : Integer) : Integer;
begin
if N < 2 then Result := 1
else Result := fib(N-2) + fib(N-1);
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
| #LFE | LFE |
(defun factors (n)
(list-comp
((<- i (when (== 0 (rem n i))) (lists:seq 1 (trunc (/ n 2)))))
i))
|
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.
| #Prolog | Prolog | evens(D, Es) :- findall(E, (member(E, D), E mod 2 =:= 0), Es). |
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
| #NewLISP | NewLISP | (dotimes (i 100)
(println
(cond
((= 0 (% i 15)) "FizzBuzz")
((= 0 (% i 3)) "Fizz")
((= 0 (% i 5)) "Buzz")
('t i)))) |
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
| #Dyalect | Dyalect | func fib(n) {
if n < 2 {
return n
} else {
return fib(n - 1) + fib(n - 2)
}
}
print(fib(30)) |
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
| #Liberty_BASIC | Liberty BASIC | num = 10677106534462215678539721403561279
maxnFactors = 1000
dim primeFactors(maxnFactors), nPrimeFactors(maxnFactors)
global nDifferentPrimeNumbersFound, nFactors, iFactor
print "Start finding all factors of ";num; ":"
nDifferentPrimeNumbersFound=0
dummy = factorize(num,2)
nFactors = showPrimeFactors(num)
dim factors(nFactors)
dummy = generateFactors(1,1)
sort factors(), 0, nFactors-1
for i=1 to nFactors
print i;" ";factors(i-1)
next i
print "done"
wait
function factorize(iNum,offset)
factorFound=0
i = offset
do
if (iNum MOD i)=0 _
then
if primeFactors(nDifferentPrimeNumbersFound) = i _
then
nPrimeFactors(nDifferentPrimeNumbersFound) = nPrimeFactors(nDifferentPrimeNumbersFound) + 1
else
nDifferentPrimeNumbersFound = nDifferentPrimeNumbersFound + 1
primeFactors(nDifferentPrimeNumbersFound) = i
nPrimeFactors(nDifferentPrimeNumbersFound) = 1
end if
if iNum/i<>1 then dummy = factorize(iNum/i,i)
factorFound=1
end if
i=i+1
loop while factorFound=0 and i<=sqr(iNum)
if factorFound=0 _
then
nDifferentPrimeNumbersFound = nDifferentPrimeNumbersFound + 1
primeFactors(nDifferentPrimeNumbersFound) = iNum
nPrimeFactors(nDifferentPrimeNumbersFound) = 1
end if
end function
function showPrimeFactors(iNum)
showPrimeFactors=1
print iNum;" = ";
for i=1 to nDifferentPrimeNumbersFound
print primeFactors(i);"^";nPrimeFactors(i);
if i<nDifferentPrimeNumbersFound then print " * "; else print ""
showPrimeFactors = showPrimeFactors*(nPrimeFactors(i)+1)
next i
end function
function generateFactors(product,pIndex)
if pIndex>nDifferentPrimeNumbersFound _
then
factors(iFactor) = product
iFactor=iFactor+1
else
for i=0 to nPrimeFactors(pIndex)
dummy = generateFactors(product*primeFactors(pIndex)^i,pIndex+1)
next i
end if
end function |
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.
| #PureBasic | PureBasic | Dim Tal.i(9)
Dim Evens.i(0)
;- Set up an array with random numbers
For i=0 To ArraySize(Tal())
Tal(i)=Random(100)
Next
;- Pick out all Even and save them
j=0
For i=0 To ArraySize(Tal())
If Tal(i)%2=0
ReDim Evens(j) ; extend the Array as we find new Even's
Evens(j)=tal(i)
j+1
EndIf
Next
;- Display the result
PrintN("List of Randoms")
For i=0 To ArraySize(Tal())
Print(Str(Tal(i))+" ")
Next
PrintN(#CRLF$+#CRLF$+"List of Even(s)")
For i=0 To ArraySize(Evens())
Print(Str(Evens(i))+" ")
Next |
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
| #NewtonScript | NewtonScript | for i := 1 to 100 do
begin
if i mod 15 = 0 then
print("FizzBuzz")
else if i mod 3 = 0 then
print("Fizz")
else if i mod 5 = 0 then
print("Buzz")
else
print(i);
print("\n")
end |
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
| #E | E | def fib(n) {
var s := [0, 1]
for _ in 0..!n {
def [a, b] := s
s := [b, a+b]
}
return s[0]
} |
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
| #Lingo | Lingo | on factors(n)
res = [1]
repeat with i = 2 to n/2
if n mod i = 0 then res.add(i)
end repeat
res.add(n)
return res
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.
| #Python | Python | values = range(10)
evens = [x for x in values if not x & 1]
ievens = (x for x in values if not x & 1) # lazy
# alternately but less idiomatic:
evens = filter(lambda x: not x & 1, values) |
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
| #Nickle | Nickle | /* Fizzbuzz in nickle */
void function fizzbuzz(size) {
for (int i = 1; i < size; i++) {
if (i % 15 == 0) { printf("Fizzbuzz\n"); }
else if (i % 5 == 0) { printf("Buzz\n"); }
else if (i % 3 == 0) { printf("Fizz\n"); }
else { printf("%i\n", i); }
}
}
fizzbuzz(1000); |
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
| #EasyLang | EasyLang | func fib n . res .
if n < 2
res = n
.
prev = 0
val = 1
for _ range n - 1
res = prev + val
prev = val
val = res
.
.
call fib 36 r
print r |
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
| #Logo | Logo | to factors :n
output filter [equal? 0 modulo :n ?] iseq 1 :n
end
show factors 28 ; [1 2 4 7 14 28] |
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.
| #Quackery | Quackery | [ [] ]'[ rot
witheach
[ tuck over do iff
[ dip [ nested join ] ]
else nip ]
drop ] is only ( [ --> [ )
[ 1 & not ] is even ( n --> b )
[] 10 times [ 10 random join ]
say "Ten arbitrary digits: " dup echo cr
say "Only the even digits: " only even echo cr |
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
| #Nim | Nim | for i in 1..100:
if i mod 15 == 0:
echo("FizzBuzz")
elif i mod 3 == 0:
echo("Fizz")
elif i mod 5 == 0:
echo("Buzz")
else:
echo(i) |
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
| #EchoLisp | EchoLisp |
(define (fib n)
(if (< n 2) n
(+ (fib (- n 2)) (fib (1- n)))))
(remember 'fib #(0 1))
(for ((i 12)) (write (fib i)))
0 1 1 2 3 5 8 13 21 34 55 89
|
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
| #Lua | Lua | function Factors( n )
local f = {}
for i = 1, n/2 do
if n % i == 0 then
f[#f+1] = i
end
end
f[#f+1] = n
return f
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.
| #Q | Q | x where 0=x mod 2 |
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
| #Nix | Nix | with (import <nixpkgs> { }).lib;
with builtins;
let
fizzbuzz = { x ? 1 }:
''
${if (mod x 15 == 0) then
"FizzBuzz"
else if (mod x 3 == 0) then
"Fizz"
else if (mod x 5 == 0) then
"Buzz"
else
(toString x)}
'' + (if (x < 100) then
fizzbuzz { x = x + 1; } else "");
in
fizzbuzz { } |
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
| #ECL | ECL | //Calculates Fibonacci sequence up to n steps using Binet's closed form solution
FibFunction(UNSIGNED2 n) := FUNCTION
REAL Sqrt5 := Sqrt(5);
REAL Phi := (1+Sqrt(5))/2;
REAL Phi_Inv := 1/Phi;
UNSIGNED FibValue := ROUND( ( POWER(Phi,n)-POWER(Phi_Inv,n) ) /Sqrt5);
RETURN FibValue;
END;
FibSeries(UNSIGNED2 n) := FUNCTION
Fib_Layout := RECORD
UNSIGNED5 FibNum;
UNSIGNED5 FibValue;
END;
FibSeq := DATASET(n+1,
TRANSFORM
( Fib_Layout
, SELF.FibNum := COUNTER-1
, SELF.FibValue := IF(SELF.FibNum<2,SELF.FibNum, FibFunction(SELF.FibNum) )
)
);
RETURN FibSeq;
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
| #M2000_Interpreter | M2000 Interpreter |
\\ Factors of an integer
\\ For act as BASIC's FOR (if N<1 no loop start)
FORM 60,40
SET SWITCHES "+FOR"
MODULE LikeBasic {
10 INPUT N%
20 FOR I%=1 TO N%
30 IF N%/I%=INT(N%/I%) THEN PRINT I%,
40 NEXT I%
50 PRINT
}
CALL LikeBasic
SET SWITCHES "-FOR"
MODULE LikeM2000 {
DEF DECIMAL N%, I%
INPUT N%
IF N%<1 THEN EXIT
FOR I%=1 TO N% {
IF N% MOD I%=0 THEN PRINT I%,
}
PRINT
}
CALL LikeM2000
|
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.
| #R | R | a <- 1:100
evennums <- a[ a%%2 == 0 ]
print(evennums) |
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
| #Oberon-2 | Oberon-2 | MODULE FizzBuzz;
IMPORT Out;
VAR i: INTEGER;
BEGIN
FOR i := 1 TO 100 DO
IF i MOD 15 = 0 THEN
Out.String("FizzBuzz")
ELSIF i MOD 5 = 0 THEN
Out.String("Buzz")
ELSIF i MOD 3 = 0 THEN
Out.String("Fizz")
ELSE
Out.Int(i,0)
END;
Out.Ln
END
END FizzBuzz. |
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
| #EDSAC_order_code | EDSAC order code | [ Fibonacci sequence
==================
A program for the EDSAC
Calculates the nth Fibonacci
number and displays it at the
top of storage tank 3
The default value of n is 10
To calculate other Fibonacci
numbers, set the starting value
of the count to n-2
Works with Initial Orders 2 ]
T56K [ set load point ]
GK [ set theta ]
[ Orders ]
[ 0 ] T20@ [ a = 0 ]
A17@ [ a += y ]
U18@ [ temp = a ]
A16@ [ a += x ]
T17@ [ y = a; a = 0 ]
A18@ [ a += temp ]
T16@ [ x = a; a = 0 ]
A19@ [ a = count ]
S15@ [ a -= 1 ]
U19@ [ count = a ]
E@ [ if a>=0 go to θ ]
T20@ [ a = 0 ]
A17@ [ a += y ]
T96F [ C(96) = a; a = 0]
ZF [ halt ]
[ Data ]
[ 15 ] P0D [ const: 1 ]
[ 16 ] P0F [ var: x = 0 ]
[ 17 ] P0D [ var: y = 1 ]
[ 18 ] P0F [ var: temp = 0 ]
[ 19 ] P4F [ var: count = 8 ]
[ 20 ] P0F [ used to clear a ]
EZPF [ begin execution ] |
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
| #Maple | Maple |
numtheory:-divisors(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.
| #Racket | Racket |
-> (filter even? '(0 1 2 3 4 5 6 7 8 9))
'(0 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
| #Objeck | Objeck | bundle Default {
class Fizz {
function : Main(args : String[]) ~ Nil {
for(i := 0; i <= 100; i += 1;) {
if(i % 15 = 0) {
"FizzBuzz"->PrintLine();
}
else if(i % 3 = 0) {
"Fizz"->PrintLine();
}
else if(i % 5 = 0) {
"Buzz"->PrintLine();
}
else {
i->PrintLine();
};
};
}
}
} |
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
| #Eiffel | Eiffel |
class
APPLICATION
create
make
feature
fibonacci (n: INTEGER): INTEGER
require
non_negative: n >= 0
local
i, n2, n1, tmp: INTEGER
do
n2 := 0
n1 := 1
from
i := 1
until
i >= n
loop
tmp := n1
n1 := n2 + n1
n2 := tmp
i := i + 1
end
Result := n1
if n = 0 then
Result := 0
end
end
feature {NONE} -- Initialization
make
-- Run application.
do
print (fibonacci (0))
print (" ")
print (fibonacci (1))
print (" ")
print (fibonacci (2))
print (" ")
print (fibonacci (3))
print (" ")
print (fibonacci (4))
print ("%N")
end
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
| #Mathematica_.2F_Wolfram_Language | Mathematica / Wolfram Language | Factorize[n_Integer] := Divisors[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.
| #Raku | Raku | my @a = 1..6;
my @even = grep * %% 2, @a; |
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
| #Objective-C | Objective-C | // FizzBuzz in Objective-C
#import <Foundation/Foundation.h>
int main(int argc, char* argv[]) {
for (NSInteger i=1; I <= 101; i++) {
if (i % 15 == 0) {
NSLog(@"FizzBuzz\n");
} else if (i % 3 == 0) {
NSLog(@"Fizz\n");
} else if (i % 5 == 0) {
NSLog(@"Buzz\n");
} else {
NSLog(@"%li\n", i);
}
}
} |
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
| #Ela | Ela | fib = fib' 0 1
where fib' a b 0 = a
fib' a b n = fib' b (a + b) (n - 1) |
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
| #MATLAB_.2F_Octave | MATLAB / Octave | function fact(n);
f = factor(n); % prime decomposition
K = dec2bin(0:2^length(f)-1)-'0'; % generate all possible permutations
F = ones(1,2^length(f));
for k = 1:size(K)
F(k) = prod(f(~K(k,:))); % and compute products
end;
F = unique(F); % eliminate duplicates
printf('There are %i factors for %i.\n',length(F),n);
disp(F);
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.
| #Raven | Raven | [ 0 1 2 3 4 5 6 7 8 9 ] as nums
group nums each
dup 1 & if drop
list as evens |
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
| #OCaml | OCaml | let fizzbuzz i =
match i mod 3, i mod 5 with
0, 0 -> "FizzBuzz"
| 0, _ -> "Fizz"
| _, 0 -> "Buzz"
| _ -> string_of_int i
let _ =
for i = 1 to 100 do print_endline (fizzbuzz i) done |
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
| #Elena | Elena | import extensions;
fibu(n)
{
int[] ac := new int[]{ 0,1 };
if (n < 2)
{
^ ac[n]
}
else
{
for(int i := 2, i <= n, i+=1)
{
int t := ac[1];
ac[1] := ac[0] + ac[1];
ac[0] := t
};
^ ac[1]
}
}
public program()
{
for(int i := 0, i <= 10, i+=1)
{
console.printLine(fibu(i))
}
} |
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
| #Maxima | Maxima | (%i96) divisors(100);
(%o96) {1,2,4,5,10,20,25,50,100} |
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.
| #REBOL | REBOL | a: [] repeat i 100 [append a i] ; Build and load array.
evens: [] repeat element a [if even? element [append evens element]]
print mold evens |
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
| #Octave | Octave | for i = 1:100
if ( mod(i,15) == 0 )
disp("FizzBuzz");
elseif ( mod(i, 3) == 0 )
disp("Fizz")
elseif ( mod(i, 5) == 0 )
disp("Buzz")
else
disp(i)
endif
endfor |
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
| #Elixir | Elixir | defmodule Fibonacci do
def fib(0), do: 0
def fib(1), do: 1
def fib(n), do: fib(0, 1, n-2)
def fib(_, prv, -1), do: prv
def fib(prvprv, prv, n) do
next = prv + prvprv
fib(prv, next, n-1)
end
end
IO.inspect Enum.map(0..10, fn i-> Fibonacci.fib(i) 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
| #MAXScript | MAXScript |
fn factors n =
(
return (for i = 1 to n+1 where mod n i == 0 collect i)
)
|
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.
| #Red | Red | Red []
orig: [] repeat i 10 [append orig i]
?? orig
cpy: [] forall orig [if even? orig/1 [append cpy orig/1]]
;; or - because we know each second element is even :- )
;; cpy: extract next orig 2
?? cpy
remove-each ele orig [odd? ele] ;; destructive
?? orig
|
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
| #Oforth | Oforth | : fizzbuzz
| i |
100 loop: i [
null
i 3 mod ifZero: [ "Fizz" + ]
i 5 mod ifZero: [ "Buzz" + ]
dup ifNull: [ drop i ] .
] ; |
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
| #Elm | Elm | fibonacci : Int -> Int
fibonacci n = if n < 2 then
n
else
fibonacci(n - 2) + fibonacci(n - 1) |
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
| #Mercury | Mercury | :- module fac.
:- interface.
:- import_module io.
:- pred main(io::di, io::uo) is det.
:- implementation.
:- import_module float, int, list, math, string.
main(!IO) :-
io.command_line_arguments(Args, !IO),
list.filter_map(string.to_int, Args, CleanArgs),
list.foldl((pred(Arg::in, !.IO::di, !:IO::uo) is det :-
factor(Arg, X),
io.format("factor(%d, [", [i(Arg)], !IO),
io.write_list(X, ",", io.write_int, !IO),
io.write_string("])\n", !IO)
), CleanArgs, !IO).
:- pred factor(int::in, list(int)::out) is det.
factor(N, Factors) :-
Limit = float.truncate_to_int(math.sqrt(float(N))),
factor(N, 2, Limit, [], Unsorted),
list.sort_and_remove_dups([1, N | Unsorted], Factors).
:- pred factor(int, int, int, list(int), list(int)).
:- mode factor(in, in, in, in, out) is det.
factor(N, X, Limit, !Accumulator) :-
( if X > Limit
then true
else ( if 0 = N mod X
then !:Accumulator = [X, N / X | !.Accumulator]
else true ),
factor(N, X + 1, Limit, !Accumulator) ).
:- func factor(int) = list(int).
%:- mode factor(in) = out is det.
factor(N) = Factors :- factor(N, Factors).
:- end_module fac. |
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.
| #REXX | REXX | /*REXX program selects all even numbers from an array and puts them ──► a new array.*/
parse arg N seed . /*obtain optional arguments from the CL*/
if N=='' | N=="," then N= 50 /*Not specified? Then use the default.*/
if datatype(seed,'W') then call random ,,seed /*use the RANDOM seed for repeatability*/
old.= /*the OLD array, all are null so far. */
new.= /* " NEW " " " " " " */
do i=1 for N /*generate N random numbers ──► OLD */
old.i= random(1, 99999) /*generate random number 1 ──► 99999*/
end /*i*/
#= 0 /*number of elements in NEW (so far).*/
do j=1 for N /*process the elements of the OLD array*/
if old.j//2 \== 0 then iterate /*if element isn't even, then skip it.*/
#= # + 1 /*bump the number of NEW elements. */
new.#= old.j /*assign the number to the NEW array.*/
end /*j*/
do k=1 for # /*display all the NEW numbers. */
say right('new.'k, 20) "=" right(new.k, 9) /*display a line (an array element). */
end /*k*/ /*stick a fork in it, we're all 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
| #Ol | Ol |
(for-each (lambda (x)
(print (cond
((and (zero? (mod x 3)) (zero? (mod x 5)))
"FizzBuzz")
((zero? (mod x 3))
"Fizz")
((zero? (mod x 5))
"Buzz")
(else
x))))
(iota 100))
|
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
| #Emacs_Lisp | Emacs Lisp | (defun fib (n a b c)
(cond
((< c n) (fib n b (+ a b) (+ 1 c)))
((= c n) b)
(t a)))
(defun fibonacci (n)
(if (< n 2)
n
(fib n 0 1 1))) |
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
| #min | min | (mod 0 ==) :divisor?
(() 0 shorten) :new
(new (over swons 'pred dip) pick times nip) :iota
(
:n
n sqrt int iota ; Only consider numbers up to sqrt(n).
(n swap divisor?) filter =f1
f1 (n swap div) map reverse =f2 ; "Mirror" the list of divisors at sqrt(n).
(f1 last f2 first ==) (f2 rest #f2) when ; Handle perfect squares.
f1 f2 concat
) :factors
24 factors puts!
9 factors puts!
11 factors puts! |
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.
| #Ring | Ring |
aList = [1, 2, 3, 4, 5, 6]
bArray = list(3)
see evenSelect(aList)
func evenSelect aArray
i = 0
for n = 1 to len(aArray)
if (aArray[n] % 2) = 0
i = i + 1
bArray[i] = aArray[n] ok
next
return bArray
|
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
| #OOC | OOC | fizz: func (n: Int) -> Bool {
if(n % 3 == 0) {
printf("Fizz")
return true
}
return false
}
buzz: func (n: Int) -> Bool {
if(n % 5 == 0) {
printf("Buzz")
return true
}
return false
}
main: func {
for(n in 1..100) {
fizz:= fizz(n)
buzz:= buzz(n)
fizz || buzz || printf("%d", n)
println()
}
} |
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
| #Erlang | Erlang |
-module(fib).
-export([fib/1]).
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
| #MiniScript | MiniScript | factors = function(n)
result = [1]
for i in range(2, n)
if n % i == 0 then result.push i
end for
return result
end function
while true
n = val(input("Number to factor (0 to quit)? "))
if n <= 0 then break
print factors(n)
end while |
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.
| #Ruby | Ruby | # Enumerable#select returns a new array.
ary = [1, 2, 3, 4, 5, 6]
even_ary = ary.select {|elem| elem.even?}
p even_ary # => [2, 4, 6]
# Enumerable#select also works with Range.
range = 1..6
even_ary = range.select {|elem| elem.even?}
p even_ary # => [2, 4, 6] |
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
| #Order | Order | #include <order/interpreter.h>
// Get FB for one number
#define ORDER_PP_DEF_8fizzbuzz ORDER_PP_FN( \
8fn(8N, \
8let((8F, 8fn(8N, 8G, \
8is_0(8remainder(8N, 8G)))), \
8cond((8ap(8F, 8N, 15), 8quote(fizzbuzz)) \
(8ap(8F, 8N, 3), 8quote(fizz)) \
(8ap(8F, 8N, 5), 8quote(buzz)) \
(8else, 8N)))) )
// Print E followed by a comma (composable, 8print is not a function)
#define ORDER_PP_DEF_8print_el ORDER_PP_FN( \
8fn(8E, 8print(8E 8comma)) )
ORDER_PP( // foreach instead of map, to print but return nothing
8seq_for_each(8compose(8print_el, 8fizzbuzz), 8seq_iota(1, 100))
) |
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
| #ERRE | ERRE | !-------------------------------------------
! derived from my book "PROGRAMMARE IN ERRE"
! iterative solution
!-------------------------------------------
PROGRAM FIBONACCI
!$DOUBLE
!VAR F1#,F2#,TEMP#,COUNT%,N%
BEGIN !main
INPUT("Number",N%)
F1=0
F2=1
REPEAT
TEMP=F2
F2=F1+F2
F1=TEMP
COUNT%=COUNT%+1
UNTIL COUNT%=N%
PRINT("FIB(";N%;")=";F2)
! Obviously a FOR loop or a WHILE loop can
! be used to solve this problem
END PROGRAM
|
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
| #.D0.9C.D0.9A-61.2F52 | МК-61/52 | П9 1 П6 КИП6 ИП9 ИП6 / П8 ^ [x]
x#0 21 - x=0 03 ИП6 С/П ИП8 П9 БП
04 1 С/П БП 21
|
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.
| #Run_BASIC | Run BASIC | dim a1(100)
count = 100
for i = 1 to 100
a1(i) = int(rnd(0)*100)+1
count = count - (a1(i) mod 2)
next
'dim the extract and fill it
dim a2(count)
for i = 1 to 100
if not(a1(i) mod 2) then
n = n+1
a2(n) = a1(i)
end if
next
for i = 1 to count
print a2(i)
next |
http://rosettacode.org/wiki/Faces_from_a_mesh | Faces from a mesh | A mesh defining a surface has uniquely numbered vertices, and named,
simple-polygonal faces described usually by an ordered list of edge numbers
going around the face,
For example:
External image of two faces
Rough textual version without edges:
1
17
7 A
B
11
23
A is the triangle (1, 11, 7), or equally (7, 11, 1), going anti-clockwise, or
any of all the rotations of those ordered vertices.
1
7 A
11
B is the four-sided face (1, 17, 23, 11), or equally (23, 17, 1, 11) or any
of their rotations.
1
17
B
11
23
Let's call the above the perimeter format as it traces around the perimeter.
A second format
A separate algorithm returns polygonal faces consisting of a face name and an unordered
set of edge definitions for each face.
A single edge is described by the vertex numbers at its two ends, always in
ascending order.
All edges for the face are given, but in an undefined order.
For example face A could be described by the edges (1, 11), (7, 11), and (1, 7)
(The order of each vertex number in an edge is ascending, but the order in
which the edges are stated is arbitrary).
Similarly face B could be described by the edges (11, 23), (1, 17), (17, 23),
and (1, 11) in arbitrary order of the edges.
Let's call this second format the edge format.
Task
1. Write a routine to check if two perimeter formatted faces have the same perimeter. Use it to check if the following pairs of perimeters are the same:
Q: (8, 1, 3)
R: (1, 3, 8)
U: (18, 8, 14, 10, 12, 17, 19)
V: (8, 14, 10, 12, 17, 19, 18)
2. Write a routine and use it to transform the following faces from edge to perimeter format.
E: {(1, 11), (7, 11), (1, 7)}
F: {(11, 23), (1, 17), (17, 23), (1, 11)}
G: {(8, 14), (17, 19), (10, 12), (10, 14), (12, 17), (8, 18), (18, 19)}
H: {(1, 3), (9, 11), (3, 11), (1, 11)}
Show your output here.
| #11l | 11l | F perim_equal(p1, =p2)
I p1.len != p2.len | Set(p1) != Set(p2)
R 0B
I any((0 .< p1.len).map(n -> @p2 == (@p1[n ..] [+] @p1[0 .< n])))
R 1B
p2 = reversed(p2)
R any((0 .< p1.len).map(n -> @p2 == (@p1[n ..] [+] @p1[0 .< n])))
F edge_to_periphery(e)
V edges = sorted(e)
[Int] p
I !edges.empty
p = [edges[0][0], edges[0][1]]
edges.pop(0)
V last = I !p.empty {p.last} E -1
L !edges.empty
L(ij) edges
V (i, j) = ij
I i == last
p.append(j)
last = j
edges.pop(L.index)
L.break
E I j == last
p.append(i)
last = i
edges.pop(L.index)
L.break
L.was_no_break
R ‘>>>Error! Invalid edge format<<<’
R String(p[0 .< (len)-1])
print(‘Perimeter format equality checks:’)
L(eq_check) [(‘Q’, [8, 1, 3],
‘R’, [1, 3, 8]),
(‘U’, [18, 8, 14, 10, 12, 17, 19],
‘V’, [8, 14, 10, 12, 17, 19, 18])]
V (n1, p1, n2, p2) = eq_check
V eq = I perim_equal(p1, p2) {‘==’} E ‘!=’
print(‘ ’n1‘ ’eq‘ ’n2)
print("\nEdge to perimeter format translations:")
V edge_d = [‘E’ = [(1, 11), (7, 11), (1, 7)],
‘F’ = [(11, 23), (1, 17), (17, 23), (1, 11)],
‘G’ = [(8, 14), (17, 19), (10, 12), (10, 14), (12, 17), (8, 18), (18, 19)],
‘H’ = [(1, 3), (9, 11), (3, 11), (1, 11)]]
L(name, edges) edge_d
print(‘ ’name‘: ’edges"\n -> "edge_to_periphery(edges)) |
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
| #Oz | Oz | declare
fun {FizzBuzz X}
if X mod 15 == 0 then 'FizzBuzz'
elseif X mod 3 == 0 then 'Fizz'
elseif X mod 5 == 0 then 'Buzz'
else X
end
end
in
for I in 1..100 do
{Show {FizzBuzz I}}
end |
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
| #Euphoria | Euphoria |
function fibor(integer n)
if n<2 then return n end if
return fibor(n-1)+fibor(n-2)
end function
|
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
| #MUMPS | MUMPS | factors(num) New fctr,list,sep,sqrt
If num<1 Quit "Too small a number"
If num["." Quit "Not an integer"
Set sqrt=num**0.5\1
For fctr=1:1:sqrt Set:num/fctr'["." list(fctr)=1,list(num/fctr)=1
Set (list,fctr)="",sep="[" For Set fctr=$Order(list(fctr)) Quit:fctr="" Set list=list_sep_fctr,sep=","
Quit list_"]"
w $$factors(45) ; [1,3,5,9,15,45]
w $$factors(53) ; [1,53]
w $$factors(64) ; [1,2,4,8,16,32,64] |
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.
| #Rust | Rust | fn main() {
println!("new vec filtered: ");
let nums: Vec<i32> = (1..20).collect();
let evens: Vec<i32> = nums.iter().cloned().filter(|x| x % 2 == 0).collect();
println!("{:?}", evens);
// Filter an already existing vector
println!("original vec filtered: ");
let mut nums: Vec<i32> = (1..20).collect();
nums.retain(|x| x % 2 == 0);
println!("{:?}", nums);
} |
http://rosettacode.org/wiki/Faces_from_a_mesh | Faces from a mesh | A mesh defining a surface has uniquely numbered vertices, and named,
simple-polygonal faces described usually by an ordered list of edge numbers
going around the face,
For example:
External image of two faces
Rough textual version without edges:
1
17
7 A
B
11
23
A is the triangle (1, 11, 7), or equally (7, 11, 1), going anti-clockwise, or
any of all the rotations of those ordered vertices.
1
7 A
11
B is the four-sided face (1, 17, 23, 11), or equally (23, 17, 1, 11) or any
of their rotations.
1
17
B
11
23
Let's call the above the perimeter format as it traces around the perimeter.
A second format
A separate algorithm returns polygonal faces consisting of a face name and an unordered
set of edge definitions for each face.
A single edge is described by the vertex numbers at its two ends, always in
ascending order.
All edges for the face are given, but in an undefined order.
For example face A could be described by the edges (1, 11), (7, 11), and (1, 7)
(The order of each vertex number in an edge is ascending, but the order in
which the edges are stated is arbitrary).
Similarly face B could be described by the edges (11, 23), (1, 17), (17, 23),
and (1, 11) in arbitrary order of the edges.
Let's call this second format the edge format.
Task
1. Write a routine to check if two perimeter formatted faces have the same perimeter. Use it to check if the following pairs of perimeters are the same:
Q: (8, 1, 3)
R: (1, 3, 8)
U: (18, 8, 14, 10, 12, 17, 19)
V: (8, 14, 10, 12, 17, 19, 18)
2. Write a routine and use it to transform the following faces from edge to perimeter format.
E: {(1, 11), (7, 11), (1, 7)}
F: {(11, 23), (1, 17), (17, 23), (1, 11)}
G: {(8, 14), (17, 19), (10, 12), (10, 14), (12, 17), (8, 18), (18, 19)}
H: {(1, 3), (9, 11), (3, 11), (1, 11)}
Show your output here.
| #Go | Go | package main
import (
"fmt"
"sort"
)
// Check a slice contains a value.
func contains(s []int, f int) bool {
for _, e := range s {
if e == f {
return true
}
}
return false
}
// Assumes s1, s2 are of same length.
func sliceEqual(s1, s2 []int) bool {
for i := 0; i < len(s1); i++ {
if s1[i] != s2[i] {
return false
}
}
return true
}
// Reverses slice in place.
func reverse(s []int) {
for i, j := 0, len(s)-1; i < j; i, j = i+1, j-1 {
s[i], s[j] = s[j], s[i]
}
}
// Check two perimeters are equal.
func perimEqual(p1, p2 []int) bool {
le := len(p1)
if le != len(p2) {
return false
}
for _, p := range p1 {
if !contains(p2, p) {
return false
}
}
// use copy to avoid mutating 'p1'
c := make([]int, le)
copy(c, p1)
for r := 0; r < 2; r++ {
for i := 0; i < le; i++ {
if sliceEqual(c, p2) {
return true
}
// do circular shift to right
t := c[le-1]
copy(c[1:], c[0:le-1])
c[0] = t
}
// now process in opposite direction
reverse(c)
}
return false
}
type edge [2]int
// Translates a face to perimeter format.
func faceToPerim(face []edge) []int {
// use copy to avoid mutating 'face'
le := len(face)
if le == 0 {
return nil
}
edges := make([]edge, le)
for i := 0; i < le; i++ {
// check edge pairs are in correct order
if face[i][1] <= face[i][0] {
return nil
}
edges[i] = face[i]
}
// sort edges in ascending order
sort.Slice(edges, func(i, j int) bool {
if edges[i][0] != edges[j][0] {
return edges[i][0] < edges[j][0]
}
return edges[i][1] < edges[j][1]
})
var perim []int
first, last := edges[0][0], edges[0][1]
perim = append(perim, first, last)
// remove first edge
copy(edges, edges[1:])
edges = edges[0 : le-1]
le--
outer:
for le > 0 {
for i, e := range edges {
found := false
if e[0] == last {
perim = append(perim, e[1])
last, found = e[1], true
} else if e[1] == last {
perim = append(perim, e[0])
last, found = e[0], true
}
if found {
// remove i'th edge
copy(edges[i:], edges[i+1:])
edges = edges[0 : le-1]
le--
if last == first {
if le == 0 {
break outer
} else {
return nil
}
}
continue outer
}
}
}
return perim[0 : len(perim)-1]
}
func main() {
fmt.Println("Perimeter format equality checks:")
areEqual := perimEqual([]int{8, 1, 3}, []int{1, 3, 8})
fmt.Printf(" Q == R is %t\n", areEqual)
areEqual = perimEqual([]int{18, 8, 14, 10, 12, 17, 19}, []int{8, 14, 10, 12, 17, 19, 18})
fmt.Printf(" U == V is %t\n", areEqual)
e := []edge{{7, 11}, {1, 11}, {1, 7}}
f := []edge{{11, 23}, {1, 17}, {17, 23}, {1, 11}}
g := []edge{{8, 14}, {17, 19}, {10, 12}, {10, 14}, {12, 17}, {8, 18}, {18, 19}}
h := []edge{{1, 3}, {9, 11}, {3, 11}, {1, 11}}
fmt.Println("\nEdge to perimeter format translations:")
for i, face := range [][]edge{e, f, g, h} {
perim := faceToPerim(face)
if perim == nil {
fmt.Printf(" %c => Invalid edge format\n", i + 'E')
} else {
fmt.Printf(" %c => %v\n", i + 'E', perim)
}
}
} |
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
| #PARI.2FGP | PARI/GP | {for(n=1,100,
print(if(n%3,
if(n%5,
n
,
"Buzz"
)
,
if(n%5,
"Fizz"
,
"FizzBuzz"
)
))
)} |
http://rosettacode.org/wiki/Exponentiation_with_infix_operators_in_(or_operating_on)_the_base | Exponentiation with infix operators in (or operating on) the base | (Many programming languages, especially those with extended─precision integer arithmetic, usually
support one of **, ^, ↑ or some such for exponentiation.)
Some languages treat/honor infix operators when performing exponentiation (raising
numbers to some power by the language's exponentiation operator, if the computer
programming language has one).
Other programming languages may make use of the POW or some other BIF
(Built─In Ffunction), or some other library service.
If your language's exponentiation operator is not one of the usual ones, please comment on how to recognize it.
This task will deal with the case where there is some form of an infix operator operating
in (or operating on) the base.
Example
A negative five raised to the 3rd power could be specified as:
-5 ** 3 or as
-(5) ** 3 or as
(-5) ** 3 or as something else
(Not all computer programming languages have an exponential operator and/or support these syntax expression(s).
Task
compute and display exponentiation with a possible infix operator, whether specified and/or implied/inferred.
Raise the following numbers (integer or real):
-5 and
+5
to the following powers:
2nd and
3rd
using the following expressions (if applicable in your language):
-x**p
-(x)**p
(-x)**p
-(x**p)
Show here (on this page) the four (or more) types of symbolic expressions for each number and power.
Try to present the results in the same format/manner as the other programming entries to make any differences apparent.
The variables may be of any type(s) that is/are applicable in your language.
Related tasks
Exponentiation order
Exponentiation operator
Arbitrary-precision integers (included)
Parsing/RPN to infix conversion
Operator precedence
References
Wikipedia: Order of operations in Programming languages
| #Ada | Ada | logical_operator ::= and | or | xor
relational_operator ::= = | /= | < | <= | > | >=
binary_adding_operator ::= + | – | &
unary_adding_operator ::= + | –
multiplying_operator ::= * | / | mod | rem
highest_precedence_operator ::= ** | abs | not
|
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
| #Excel | Excel | FIBONACCI
=LAMBDA(n,
APPLYN(n - 2)(
LAMBDA(xs,
APPENDROWS(xs)(
SUM(
LASTNROWS(2)(xs)
)
)
)
)({1;1})
) |
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
| #Nanoquery | Nanoquery | n = int(input())
for i in range(1, n / 2)
if (n % i = 0)
print i + " "
end
end
println 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.
| #Salmon | Salmon | iterate(x; comprehend(y; [1...10]; y % 2 == 0) (y))
x!; |
http://rosettacode.org/wiki/Faces_from_a_mesh | Faces from a mesh | A mesh defining a surface has uniquely numbered vertices, and named,
simple-polygonal faces described usually by an ordered list of edge numbers
going around the face,
For example:
External image of two faces
Rough textual version without edges:
1
17
7 A
B
11
23
A is the triangle (1, 11, 7), or equally (7, 11, 1), going anti-clockwise, or
any of all the rotations of those ordered vertices.
1
7 A
11
B is the four-sided face (1, 17, 23, 11), or equally (23, 17, 1, 11) or any
of their rotations.
1
17
B
11
23
Let's call the above the perimeter format as it traces around the perimeter.
A second format
A separate algorithm returns polygonal faces consisting of a face name and an unordered
set of edge definitions for each face.
A single edge is described by the vertex numbers at its two ends, always in
ascending order.
All edges for the face are given, but in an undefined order.
For example face A could be described by the edges (1, 11), (7, 11), and (1, 7)
(The order of each vertex number in an edge is ascending, but the order in
which the edges are stated is arbitrary).
Similarly face B could be described by the edges (11, 23), (1, 17), (17, 23),
and (1, 11) in arbitrary order of the edges.
Let's call this second format the edge format.
Task
1. Write a routine to check if two perimeter formatted faces have the same perimeter. Use it to check if the following pairs of perimeters are the same:
Q: (8, 1, 3)
R: (1, 3, 8)
U: (18, 8, 14, 10, 12, 17, 19)
V: (8, 14, 10, 12, 17, 19, 18)
2. Write a routine and use it to transform the following faces from edge to perimeter format.
E: {(1, 11), (7, 11), (1, 7)}
F: {(11, 23), (1, 17), (17, 23), (1, 11)}
G: {(8, 14), (17, 19), (10, 12), (10, 14), (12, 17), (8, 18), (18, 19)}
H: {(1, 3), (9, 11), (3, 11), (1, 11)}
Show your output here.
| #Haskell | Haskell | import Data.List (find, delete, (\\))
import Control.Applicative ((<|>))
------------------------------------------------------------
newtype Perimeter a = Perimeter [a]
deriving Show
instance Eq a => Eq (Perimeter a) where
Perimeter p1 == Perimeter p2 =
null (p1 \\ p2)
&& ((p1 `elem` zipWith const (iterate rotate p2) p1)
|| Perimeter p1 == Perimeter (reverse p2))
rotate lst = zipWith const (tail (cycle lst)) lst
toEdges :: Ord a => Perimeter a -> Maybe (Edges a)
toEdges (Perimeter ps)
| allDifferent ps = Just . Edges $ zipWith ord ps (tail (cycle ps))
| otherwise = Nothing
where
ord a b = if a < b then (a, b) else (b, a)
allDifferent [] = True
allDifferent (x:xs) = all (x /=) xs && allDifferent xs
------------------------------------------------------------
newtype Edges a = Edges [(a, a)]
deriving Show
instance Eq a => Eq (Edges a) where
e1 == e2 = toPerimeter e1 == toPerimeter e2
toPerimeter :: Eq a => Edges a -> Maybe (Perimeter a)
toPerimeter (Edges ((a, b):es)) = Perimeter . (a :) <$> go b es
where
go x rs
| x == a = return []
| otherwise = do
p <- find ((x ==) . fst) rs <|> find ((x ==) . snd) rs
let next = if fst p == x then snd p else fst p
(x :) <$> go next (delete p rs) |
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
| #Pascal | Pascal | program fizzbuzz(output);
var
i: integer;
begin
for i := 1 to 100 do
if i mod 15 = 0 then
writeln('FizzBuzz')
else if i mod 3 = 0 then
writeln('Fizz')
else if i mod 5 = 0 then
writeln('Buzz')
else
writeln(i)
end. |
http://rosettacode.org/wiki/Exponentiation_with_infix_operators_in_(or_operating_on)_the_base | Exponentiation with infix operators in (or operating on) the base | (Many programming languages, especially those with extended─precision integer arithmetic, usually
support one of **, ^, ↑ or some such for exponentiation.)
Some languages treat/honor infix operators when performing exponentiation (raising
numbers to some power by the language's exponentiation operator, if the computer
programming language has one).
Other programming languages may make use of the POW or some other BIF
(Built─In Ffunction), or some other library service.
If your language's exponentiation operator is not one of the usual ones, please comment on how to recognize it.
This task will deal with the case where there is some form of an infix operator operating
in (or operating on) the base.
Example
A negative five raised to the 3rd power could be specified as:
-5 ** 3 or as
-(5) ** 3 or as
(-5) ** 3 or as something else
(Not all computer programming languages have an exponential operator and/or support these syntax expression(s).
Task
compute and display exponentiation with a possible infix operator, whether specified and/or implied/inferred.
Raise the following numbers (integer or real):
-5 and
+5
to the following powers:
2nd and
3rd
using the following expressions (if applicable in your language):
-x**p
-(x)**p
(-x)**p
-(x**p)
Show here (on this page) the four (or more) types of symbolic expressions for each number and power.
Try to present the results in the same format/manner as the other programming entries to make any differences apparent.
The variables may be of any type(s) that is/are applicable in your language.
Related tasks
Exponentiation order
Exponentiation operator
Arbitrary-precision integers (included)
Parsing/RPN to infix conversion
Operator precedence
References
Wikipedia: Order of operations in Programming languages
| #ALGOL_68 | ALGOL 68 | BEGIN
# show the results of exponentiation and unary minus in various combinations #
FOR x FROM -5 BY 10 TO 5 DO
FOR p FROM 2 TO 3 DO
print( ( "x = ", whole( x, -2 ), " p = ", whole( p, 0 ) ) );
print( ( " -x**p ", whole( -x**p, -4 ) ) );
print( ( " -(x)**p ", whole( -(x)**p, -4 ) ) );
print( ( " (-x)**p ", whole( (-x)**p, -4 ) ) );
print( ( " -(x**p) ", whole( -(x**p), -4 ) ) );
print( ( newline ) )
OD
OD
END |
http://rosettacode.org/wiki/Exponentiation_with_infix_operators_in_(or_operating_on)_the_base | Exponentiation with infix operators in (or operating on) the base | (Many programming languages, especially those with extended─precision integer arithmetic, usually
support one of **, ^, ↑ or some such for exponentiation.)
Some languages treat/honor infix operators when performing exponentiation (raising
numbers to some power by the language's exponentiation operator, if the computer
programming language has one).
Other programming languages may make use of the POW or some other BIF
(Built─In Ffunction), or some other library service.
If your language's exponentiation operator is not one of the usual ones, please comment on how to recognize it.
This task will deal with the case where there is some form of an infix operator operating
in (or operating on) the base.
Example
A negative five raised to the 3rd power could be specified as:
-5 ** 3 or as
-(5) ** 3 or as
(-5) ** 3 or as something else
(Not all computer programming languages have an exponential operator and/or support these syntax expression(s).
Task
compute and display exponentiation with a possible infix operator, whether specified and/or implied/inferred.
Raise the following numbers (integer or real):
-5 and
+5
to the following powers:
2nd and
3rd
using the following expressions (if applicable in your language):
-x**p
-(x)**p
(-x)**p
-(x**p)
Show here (on this page) the four (or more) types of symbolic expressions for each number and power.
Try to present the results in the same format/manner as the other programming entries to make any differences apparent.
The variables may be of any type(s) that is/are applicable in your language.
Related tasks
Exponentiation order
Exponentiation operator
Arbitrary-precision integers (included)
Parsing/RPN to infix conversion
Operator precedence
References
Wikipedia: Order of operations in Programming languages
| #AWK | AWK |
# syntax: GAWK -f EXPONENTIATION_WITH_INFIX_OPERATORS_IN_OR_OPERATING_ON_THE_BASE.AWK
# converted from FreeBASIC
BEGIN {
print(" x p | -x^p -(x)^p (-x)^p -(x^p)")
print("--------+------------------------------------")
for (x=-5; x<=5; x+=10) {
for (p=2; p<=3; p++) {
printf("%3d %3d | %8d %8d %8d %8d\n",x,p,(-x^p),(-(x)^p),((-x)^p),(-(x^p)))
}
}
exit(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
| #F.23 | F# |
let fibonacci n : bigint =
let rec f a b n =
match n with
| 0 -> a
| 1 -> b
| n -> (f b (a + b) (n - 1))
f (bigint 0) (bigint 1) n
> fibonacci 100;;
val it : bigint = 354224848179261915075I |
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
| #NetRexx | NetRexx | /* NetRexx ***********************************************************
* 21.04.2013 Walter Pachl
* 21.04.2013 add method main to accept argument(s)
*********************************************************************/
options replace format comments java crossref symbols nobinary
class divl
method main(argwords=String[]) static
arg=Rexx(argwords)
Parse arg a b
Say a b
If a='' Then Do
help='java divl low [high] shows'
help=help||' divisors of all numbers between low and high'
Say help
Return
End
If b='' Then b=a
loop x=a To b
say x '->' divs(x)
End
method divs(x) public static returns Rexx
if x==1 then return 1 /*handle special case of 1 */
lo=1
hi=x
odd=x//2 /* 1 if x is odd */
loop j=2+odd By 1+odd While j*j<x /*divide by numbers<sqrt(x) */
if x//j==0 then Do /*Divisible? Add two divisors:*/
lo=lo j /* list low divisors */
hi=x%j hi /* list high divisors */
End
End
If j*j=x Then /*for a square number as input */
lo=lo j /* add its square root */
return lo hi /* return both lists */ |
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.
| #Sather | Sather | class MARRAY{T} < $ARR{T} is
include ARRAY{T};
filter_by(r:ROUT{T}:BOOL):SAME is
o:MARRAY{T} := #;
loop e ::= elt!;
if r.call(e) then
o := o.append(#MARRAY{T}(|e|));
end;
end;
return o;
end;
end;
class MAIN is
main is
a ::= #MARRAY{INT}(|5, 6, 7, 8, 9, 10, 11|);
sel ::= a.filter_by( bind(_.is_even) );
loop #OUT + sel.elt! + " "; end;
#OUT + "\n";
end;
end; |
http://rosettacode.org/wiki/Faces_from_a_mesh | Faces from a mesh | A mesh defining a surface has uniquely numbered vertices, and named,
simple-polygonal faces described usually by an ordered list of edge numbers
going around the face,
For example:
External image of two faces
Rough textual version without edges:
1
17
7 A
B
11
23
A is the triangle (1, 11, 7), or equally (7, 11, 1), going anti-clockwise, or
any of all the rotations of those ordered vertices.
1
7 A
11
B is the four-sided face (1, 17, 23, 11), or equally (23, 17, 1, 11) or any
of their rotations.
1
17
B
11
23
Let's call the above the perimeter format as it traces around the perimeter.
A second format
A separate algorithm returns polygonal faces consisting of a face name and an unordered
set of edge definitions for each face.
A single edge is described by the vertex numbers at its two ends, always in
ascending order.
All edges for the face are given, but in an undefined order.
For example face A could be described by the edges (1, 11), (7, 11), and (1, 7)
(The order of each vertex number in an edge is ascending, but the order in
which the edges are stated is arbitrary).
Similarly face B could be described by the edges (11, 23), (1, 17), (17, 23),
and (1, 11) in arbitrary order of the edges.
Let's call this second format the edge format.
Task
1. Write a routine to check if two perimeter formatted faces have the same perimeter. Use it to check if the following pairs of perimeters are the same:
Q: (8, 1, 3)
R: (1, 3, 8)
U: (18, 8, 14, 10, 12, 17, 19)
V: (8, 14, 10, 12, 17, 19, 18)
2. Write a routine and use it to transform the following faces from edge to perimeter format.
E: {(1, 11), (7, 11), (1, 7)}
F: {(11, 23), (1, 17), (17, 23), (1, 11)}
G: {(8, 14), (17, 19), (10, 12), (10, 14), (12, 17), (8, 18), (18, 19)}
H: {(1, 3), (9, 11), (3, 11), (1, 11)}
Show your output here.
| #J | J | NB. construct a list of all rotations of one of the faces
NB. including all rotations of the reversed list.
NB. Find out if the other face is a member of this list.
NB. ,&:rotations -> append, but first enlist the rotations.
rotations=. |."0 1~ i.@#
reverse=: |.
same_perimeter=: e. (,&:rotations reverse)
(3, 1, 8)same_perimeter(8, 1, 3)
1
(18, 8, 14, 10, 12, 17, 19)same_perimeter(8, 14, 10, 12, 17, 19, 18)
1
|
http://rosettacode.org/wiki/Faces_from_a_mesh | Faces from a mesh | A mesh defining a surface has uniquely numbered vertices, and named,
simple-polygonal faces described usually by an ordered list of edge numbers
going around the face,
For example:
External image of two faces
Rough textual version without edges:
1
17
7 A
B
11
23
A is the triangle (1, 11, 7), or equally (7, 11, 1), going anti-clockwise, or
any of all the rotations of those ordered vertices.
1
7 A
11
B is the four-sided face (1, 17, 23, 11), or equally (23, 17, 1, 11) or any
of their rotations.
1
17
B
11
23
Let's call the above the perimeter format as it traces around the perimeter.
A second format
A separate algorithm returns polygonal faces consisting of a face name and an unordered
set of edge definitions for each face.
A single edge is described by the vertex numbers at its two ends, always in
ascending order.
All edges for the face are given, but in an undefined order.
For example face A could be described by the edges (1, 11), (7, 11), and (1, 7)
(The order of each vertex number in an edge is ascending, but the order in
which the edges are stated is arbitrary).
Similarly face B could be described by the edges (11, 23), (1, 17), (17, 23),
and (1, 11) in arbitrary order of the edges.
Let's call this second format the edge format.
Task
1. Write a routine to check if two perimeter formatted faces have the same perimeter. Use it to check if the following pairs of perimeters are the same:
Q: (8, 1, 3)
R: (1, 3, 8)
U: (18, 8, 14, 10, 12, 17, 19)
V: (8, 14, 10, 12, 17, 19, 18)
2. Write a routine and use it to transform the following faces from edge to perimeter format.
E: {(1, 11), (7, 11), (1, 7)}
F: {(11, 23), (1, 17), (17, 23), (1, 11)}
G: {(8, 14), (17, 19), (10, 12), (10, 14), (12, 17), (8, 18), (18, 19)}
H: {(1, 3), (9, 11), (3, 11), (1, 11)}
Show your output here.
| #Julia | Julia | iseq(f, g) = any(n -> f == circshift(g, n), 1:length(g))
function toface(evec)
try
ret, edges = collect(evec[1]), copy(evec[2:end])
while !isempty(edges)
i = findfirst(x -> ret[end] == x[1] || ret[end] == x[2], edges)
push!(ret, ret[end] == edges[i][1] ? edges[i][2] : edges[i][1])
deleteat!(edges, i)
end
return ret[1:end-1]
catch
println("Invalid edges vector: $evec")
exit(1)
end
end
const faces1 = [
[[8, 1, 3], [1, 3, 8]],
[[18, 8, 14, 10, 12, 17, 19], [8, 14, 10, 12, 17, 19, 18]]
]
const faces2 = [
[(1, 11), (7, 11), (1, 7)], [(11, 23), (1, 17), (17, 23), (1, 11)],
[(8, 14), (17, 19), (10, 12), (10, 14), (12, 17), (8, 18), (18, 19)],
[(1, 3), (9, 11), (3, 11), (1, 11)]
]
for faces in faces1
println("Faces are ", iseq(faces[1], faces[2]) ? "" : "not ", "equivalent.")
end
for face in faces2
println(toface(face))
end
|
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
| #PDP-8_Assembly | PDP-8 Assembly |
/--------------------------------------------------------
/THIS PROGRAM PRINTS THE INTEGERS FROM 1 TO 100 (INCL).
/WITH THE FOLLOWING RESTRICTIONS:
/ FOR MULTIPLES OF THREE, PRINT 'FIZZ'
/ FOR MULTIPLES OF FIVE, PRINT 'BUZZ'
/ FOR MULTIPLES OF BOTH THREE AND FIVE, PRINT 'FIZZBUZZ'
/--------------------------------------------------------
/--------------------------------------------------------
/DEFINES
/--------------------------------------------------------
SWBA=7447 /EAE MODE A INSTRUCTION
DVI=7407 /EAE DIVIDE INSTRUCTION
AIX0=0010 /AUTO INDEX REGISTER 0
CR=0215 /CARRIAGE RETURN
LF=0212 /LINEFEED
EOT=0000 /END OF TEXT NUL
FIZMOD=0003 /CONSTANT DECIMAL 3 (FIZZ)
BUZMOD=0005 /CONSTANT DECIMAL 5 (BUZZ)
K10=0012 /CONSTANT DECIMAL 10
LAST=0144 /FIZZBUZZ THE NUMBERS 1..LAST
/0144 OCTAL == 100 DECIMAL
/CAN BE ANY FROM [0001...7777]
/--------------------------------------------------------
/FIZZBUZZ START=0200
/--------------------------------------------------------
*200 /START IN MEMORY AT 0200 OCTAL
FZZBZZ, CLA /CLEAR AC
TAD (-LAST-1 /LOAD CONSTANT -(LAST+1)
DCA CNTR /SET UP MAIN COUNTER
TAD (-FIZMOD /SET UP FIZZ COUNTER
DCA FIZCTR /TO -3
TAD (-BUZMOD /SET UP BUZZ COUNTER
DCA BUZCTR /TO -5
LOOP, ISZ CNTR /READY?
SKP /NO: CONTINUE
JMP I [7600 /YES: RETURN TO OS/8, REPLACE BY
/'HLT' IF NOT ON OS/8
CHKFIZ, ISZ FIZCTR /MULTIPLE OF 3?
JMP CHKBUZ /NO: CONTINUE
TAD FIZSTR /YES: LOAD ADDRESS OF 'FIZZ'
JMS STROUT /PRINT IT TO TTY
TAD (-FIZMOD /RELOAD THE
DCA FIZCTR /MOD 3 COUNTER
CHKBUZ, ISZ BUZCTR /MULTIPLE OF 5?
JMP CHKNUM /NO: CONTINUE
TAD BUZSTR /YES: LOAD ADDRESS OF 'BUZZ'
JMS STROUT /PRINT IT TO TTY
TAD (-BUZMOD /RELOAD THE
DCA BUZCTR /MOD 5 COUNTER
JMP NXTLIN /PRINT NEWLINE AND CONTINUE
CHKNUM, TAD FIZCTR /CHECK WHETHER MOD 3 COUNTER
TAD (FIZMOD /IS RELOADED
SNA /DID WE JUST PRINT 'FIZZ'?
JMP NXTLIN /YES: PRINT NEWLINE AND CONTINUE
CLA /ZERO THE AC
NUM, TAD CNTR /LOAD THE MAIN NUMBER COUNTER
TAD (LAST+1 /OFFSET IT TO A POSITIVE VALUE
JMS NUMOUT /PRINT IT TO THE TTY
NXTLIN, TAD LINSTR /LOAD THE ADDRESS OF THE NEWLINE
JMS STROUT /PRINT IT TO TTY
JMP LOOP /CONTINUE WITH THE NEXT NUMBER
CNTR, 0 /MAIN COUNTER
FIZCTR, 0 /FIZZ COUNTER
BUZCTR, 0 /BUZZ COUNTER
/--------------------------------------------------------
/WRITE ASCII CHARACTER IN AC TO TTY
/PRE : AC=ASCII CHARACTER
/POST: AC=0
/--------------------------------------------------------
CHROUT, .-.
TLS /SEND CHARACTER TO TTY
TSF /IS TTY READY FOR NEXT CHARACTER?
JMP .-1 /NO TRY AGAIN
CLA /AC=0
JMP I CHROUT /RETURN
/--------------------------------------------------------
/WRITE NUL TERMINATED ASCII STRING TO TTY
/PRE : AC=ADDRESS OF STRING MINUS 1
/POST: AC=0
/--------------------------------------------------------
STROUT, .-.
DCA AIX0 /STORE POINTER IN AUTO INDEX 0
STRLOP, TAD I AIX0 /GET NEXT CHARACTER FROM STRING
SNA /SKIP IF NOT EOT
JMP I STROUT /RETURN
JMS CHROUT /PRINT CHARACTER
JMP STRLOP /GO GET NEXT CHARACTER
/--------------------------------------------------------
/WRITE NUMBER IN AC TO TTY AS DECIMAL
/PRE : AC=UNSIGNED NUMBER BETWEEN 0000 AND 7777
/POST: AC=0
/--------------------------------------------------------
NUMOUT, .-.
SWBA /SET EAE IN MODE A
MQL /MQ=NUM; AC=0
TAD BUFFER /LOAD END OF BUFFER
DCA BUFPTR /IN BUFPTR
SKP /NUM IS ALREADY IN MQ
NUMLOP, MQL /MQ=NUM; AC=0
DVI /MQ=NUM/10; AC=NUM-(NUM/10)*10
K10 /DECIMAL 10
TAD ("0 /ADD ASCII ZERO
DCA I BUFPTR /STORE CHAR BUFFER, BACK TO FRONT
CMA /AC=-1
TAD BUFPTR /DECREMENT
DCA BUFPTR /BUFFER POINTER
MQA /MQ -> AC
SZA /READY IF ZERO
JMP NUMLOP /GET NEXT DIGIT
TAD BUFPTR /LOAD START OF CONVERTED NUMBER
JMS STROUT /SEND IT TO TTY
JMP I NUMOUT /RETURN
BUFFER, .+4 /ADDRESS OF BUFFER
*.+4 /RESERVE 4 LOCATIONS (MAX=4095)
EOT /END OF BUFFER
BUFPTR, 0 /POINTER IN BUFFER
/--------------------------------------------------------
/STRINGS
/--------------------------------------------------------
FIZSTR, . /FIZZ STRING
"F; "I; "Z; "Z; EOT
BUZSTR, . /BUZZ STRING
"B; "U; "Z; "Z; EOT
LINSTR, . /NEWLINE STIRNG
CR; LF; EOT
$
|
http://rosettacode.org/wiki/Exponentiation_with_infix_operators_in_(or_operating_on)_the_base | Exponentiation with infix operators in (or operating on) the base | (Many programming languages, especially those with extended─precision integer arithmetic, usually
support one of **, ^, ↑ or some such for exponentiation.)
Some languages treat/honor infix operators when performing exponentiation (raising
numbers to some power by the language's exponentiation operator, if the computer
programming language has one).
Other programming languages may make use of the POW or some other BIF
(Built─In Ffunction), or some other library service.
If your language's exponentiation operator is not one of the usual ones, please comment on how to recognize it.
This task will deal with the case where there is some form of an infix operator operating
in (or operating on) the base.
Example
A negative five raised to the 3rd power could be specified as:
-5 ** 3 or as
-(5) ** 3 or as
(-5) ** 3 or as something else
(Not all computer programming languages have an exponential operator and/or support these syntax expression(s).
Task
compute and display exponentiation with a possible infix operator, whether specified and/or implied/inferred.
Raise the following numbers (integer or real):
-5 and
+5
to the following powers:
2nd and
3rd
using the following expressions (if applicable in your language):
-x**p
-(x)**p
(-x)**p
-(x**p)
Show here (on this page) the four (or more) types of symbolic expressions for each number and power.
Try to present the results in the same format/manner as the other programming entries to make any differences apparent.
The variables may be of any type(s) that is/are applicable in your language.
Related tasks
Exponentiation order
Exponentiation operator
Arbitrary-precision integers (included)
Parsing/RPN to infix conversion
Operator precedence
References
Wikipedia: Order of operations in Programming languages
| #BASIC | BASIC | S$=" ":M$=CHR$(13):?M$" X P -X^P -(X)^P (-X)^P -(X^P)":FORX=-5TO+5STEP10:FORP=2TO3:?M$MID$("+",1+(X<0));X" "PRIGHT$(S$+STR$(-X^P),8)RIGHT$(S$+STR$(-(X)^P),8)RIGHT$(S$+STR$((-X)^P),8)RIGHT$(S$+STR$(-(X^P)),8);:NEXTP,X |
http://rosettacode.org/wiki/Exponentiation_with_infix_operators_in_(or_operating_on)_the_base | Exponentiation with infix operators in (or operating on) the base | (Many programming languages, especially those with extended─precision integer arithmetic, usually
support one of **, ^, ↑ or some such for exponentiation.)
Some languages treat/honor infix operators when performing exponentiation (raising
numbers to some power by the language's exponentiation operator, if the computer
programming language has one).
Other programming languages may make use of the POW or some other BIF
(Built─In Ffunction), or some other library service.
If your language's exponentiation operator is not one of the usual ones, please comment on how to recognize it.
This task will deal with the case where there is some form of an infix operator operating
in (or operating on) the base.
Example
A negative five raised to the 3rd power could be specified as:
-5 ** 3 or as
-(5) ** 3 or as
(-5) ** 3 or as something else
(Not all computer programming languages have an exponential operator and/or support these syntax expression(s).
Task
compute and display exponentiation with a possible infix operator, whether specified and/or implied/inferred.
Raise the following numbers (integer or real):
-5 and
+5
to the following powers:
2nd and
3rd
using the following expressions (if applicable in your language):
-x**p
-(x)**p
(-x)**p
-(x**p)
Show here (on this page) the four (or more) types of symbolic expressions for each number and power.
Try to present the results in the same format/manner as the other programming entries to make any differences apparent.
The variables may be of any type(s) that is/are applicable in your language.
Related tasks
Exponentiation order
Exponentiation operator
Arbitrary-precision integers (included)
Parsing/RPN to infix conversion
Operator precedence
References
Wikipedia: Order of operations in Programming languages
| #F.23 | F# |
printfn "-5.0**2.0=%f; -(5.0**2.0)=%f" (-5.0**2.0) (-(5.0**2.0))
|
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
| #Factor | Factor | : fib ( n -- m )
dup 2 < [
[ 0 1 ] dip [ swap [ + ] keep ] times
drop
] unless ; |
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
| #Nim | Nim | import intsets, math, algorithm
proc factors(n: int): seq[int] =
var fs: IntSet
for x in 1 .. int(sqrt(float(n))):
if n mod x == 0:
fs.incl(x)
fs.incl(n div x)
for x in fs:
result.add(x)
result.sort()
echo factors(45) |
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.
| #Scala | Scala | (1 to 100).filter(_ % 2 == 0) |
http://rosettacode.org/wiki/Faces_from_a_mesh | Faces from a mesh | A mesh defining a surface has uniquely numbered vertices, and named,
simple-polygonal faces described usually by an ordered list of edge numbers
going around the face,
For example:
External image of two faces
Rough textual version without edges:
1
17
7 A
B
11
23
A is the triangle (1, 11, 7), or equally (7, 11, 1), going anti-clockwise, or
any of all the rotations of those ordered vertices.
1
7 A
11
B is the four-sided face (1, 17, 23, 11), or equally (23, 17, 1, 11) or any
of their rotations.
1
17
B
11
23
Let's call the above the perimeter format as it traces around the perimeter.
A second format
A separate algorithm returns polygonal faces consisting of a face name and an unordered
set of edge definitions for each face.
A single edge is described by the vertex numbers at its two ends, always in
ascending order.
All edges for the face are given, but in an undefined order.
For example face A could be described by the edges (1, 11), (7, 11), and (1, 7)
(The order of each vertex number in an edge is ascending, but the order in
which the edges are stated is arbitrary).
Similarly face B could be described by the edges (11, 23), (1, 17), (17, 23),
and (1, 11) in arbitrary order of the edges.
Let's call this second format the edge format.
Task
1. Write a routine to check if two perimeter formatted faces have the same perimeter. Use it to check if the following pairs of perimeters are the same:
Q: (8, 1, 3)
R: (1, 3, 8)
U: (18, 8, 14, 10, 12, 17, 19)
V: (8, 14, 10, 12, 17, 19, 18)
2. Write a routine and use it to transform the following faces from edge to perimeter format.
E: {(1, 11), (7, 11), (1, 7)}
F: {(11, 23), (1, 17), (17, 23), (1, 11)}
G: {(8, 14), (17, 19), (10, 12), (10, 14), (12, 17), (8, 18), (18, 19)}
H: {(1, 3), (9, 11), (3, 11), (1, 11)}
Show your output here.
| #Nim | Nim | import algorithm, strutils
type
Perimeter = seq[int]
Face = tuple[name: char; perimeter: Perimeter]
Edge = tuple[first, last: int]
const None = -1 # No point.
#---------------------------------------------------------------------------------------------------
func isSame(p1, p2: Perimeter): bool =
## Return "true" if "p1" and "p2" represent the same face.
if p1.len != p2.len: return false
for p in p1:
if p notin p2: return false
var start = p2.find(p1[0])
if p1 == p2[start..^1] & p2[0..<start]:
return true
let p3 = reversed(p2)
start = p3.find(p1[0])
if p1 == p3[start..^1] & p3[0..<start]:
return true
#---------------------------------------------------------------------------------------------------
func `$`(perimeter: Perimeter): string =
## Convert a perimeter to a string.
'(' & perimeter.join(", ") & ')'
func `$`(face: Face): string =
## Convert a perimeter formatted face to a string.
face.name & $face.perimeter
#---------------------------------------------------------------------------------------------------
func toPerimeter(edges: seq[Edge]): Perimeter =
## Convert a list of edges to perimeter representation.
## Return an empty perimeter if the list of edges doesn’t represent a face.
var edges = edges
let firstEdge = edges.pop() # First edge taken in account.
var nextPoint = firstEdge.first # Next point to search in remaining edges.
result.add(nextpoint)
while edges.len > 0:
# Search an edge.
var idx = None
for i, e in edges:
if e.first == nextPoint or e.last == nextPoint:
idx = i
nextPoint = if nextpoint == e.first: e.last else: e.first
break
if idx == None:
return @[] # No edge found containing "newPoint".
# Add next point to perimeter and remove the edge.
result.add(nextPoint)
edges.del(idx)
# Check that last added point is the expected one.
if nextPoint != firstEdge.last:
return @[]
#———————————————————————————————————————————————————————————————————————————————————————————————————
when isMainModule:
# List of pairs of perimeter formatted faces to compare.
const FP = [(('P', @[8, 1, 3]), ('R', @[1, 3, 8])),
(('U', @[18, 8, 14, 10, 12, 17, 19]), ('V', @[8, 14, 10, 12, 17, 19, 18]))]
echo "Perimeter comparison:"
for (p1, p2) in FP:
echo p1, if isSame(p1[1], p2[1]): " is same as " else: "is not same as", p2
# List of edge formatted faces.
const FE = {'E': @[(1, 11), (7, 11), (1, 7)],
'F': @[(11, 23), (1, 17), (17, 23), (1, 11)],
'G': @[(8, 14), (17, 19), (10, 12), (10, 14), (12, 17), (8, 18), (18, 19)],
'H': @[(1, 3), (9, 11), (3, 11), (1, 11)]}
echo ""
echo "Conversion from edge to perimeter format:"
for (faceName, edges) in FE:
let perimeter = edges.toPerimeter()
echo faceName, ": ", if perimeter.len == 0: "Invalid edge list" else: $perimeter |
http://rosettacode.org/wiki/Faces_from_a_mesh | Faces from a mesh | A mesh defining a surface has uniquely numbered vertices, and named,
simple-polygonal faces described usually by an ordered list of edge numbers
going around the face,
For example:
External image of two faces
Rough textual version without edges:
1
17
7 A
B
11
23
A is the triangle (1, 11, 7), or equally (7, 11, 1), going anti-clockwise, or
any of all the rotations of those ordered vertices.
1
7 A
11
B is the four-sided face (1, 17, 23, 11), or equally (23, 17, 1, 11) or any
of their rotations.
1
17
B
11
23
Let's call the above the perimeter format as it traces around the perimeter.
A second format
A separate algorithm returns polygonal faces consisting of a face name and an unordered
set of edge definitions for each face.
A single edge is described by the vertex numbers at its two ends, always in
ascending order.
All edges for the face are given, but in an undefined order.
For example face A could be described by the edges (1, 11), (7, 11), and (1, 7)
(The order of each vertex number in an edge is ascending, but the order in
which the edges are stated is arbitrary).
Similarly face B could be described by the edges (11, 23), (1, 17), (17, 23),
and (1, 11) in arbitrary order of the edges.
Let's call this second format the edge format.
Task
1. Write a routine to check if two perimeter formatted faces have the same perimeter. Use it to check if the following pairs of perimeters are the same:
Q: (8, 1, 3)
R: (1, 3, 8)
U: (18, 8, 14, 10, 12, 17, 19)
V: (8, 14, 10, 12, 17, 19, 18)
2. Write a routine and use it to transform the following faces from edge to perimeter format.
E: {(1, 11), (7, 11), (1, 7)}
F: {(11, 23), (1, 17), (17, 23), (1, 11)}
G: {(8, 14), (17, 19), (10, 12), (10, 14), (12, 17), (8, 18), (18, 19)}
H: {(1, 3), (9, 11), (3, 11), (1, 11)}
Show your output here.
| #Perl | Perl | use strict;
use warnings;
use feature 'say';
use Set::Scalar;
use Set::Bag;
use Storable qw(dclone);
sub show { my($pts) = @_; my $p='( '; $p .= '(' . join(' ',@$_) . ') ' for @$pts; $p.')' }
sub check_equivalence {
my($a, $b) = @_;
Set::Scalar->new(@$a) == Set::Scalar->new(@$b)
}
sub edge_to_periphery {
my $a = dclone \@_;
my $bag_a = Set::Bag->new;
for my $p (@$a) {
$bag_a->insert( @$p[0] => 1);
$bag_a->insert( @$p[1] => 1);
}
2 != @$bag_a{$_} and return 0 for keys %$bag_a;
my $b = shift @$a;
while ($#{$a} > 0) {
for my $k (0..$#{$a}) {
my $v = @$a[$k];
if (@$v[0] == @$b[-1]) {
push @$b, @$v[1];
splice(@$a,$k,1);
last
} elsif (@$v[1] == @$b[-1]) {
push @$b, @$v[0];
splice(@$a,$k,1);
last
}
}
}
@$b;
}
say 'Perimeter format equality checks:';
for ([[8, 1, 3], [1, 3, 8]],
[[18, 8, 14, 10, 12, 17, 19], [8, 14, 10, 12, 17, 19, 18]]) {
my($a, $b) = @$_;
say '(' . join(', ', @$a) . ') equivalent to (' . join(', ', @$b) . ')? ',
check_equivalence($a, $b) ? 'True' : 'False';
}
say "\nEdge to perimeter format translations:";
for ([[1, 11], [7, 11], [1, 7]],
[[11, 23], [1, 17], [17, 23], [1, 11]],
[[8, 14], [17, 19], [10, 12], [10, 14], [12, 17], [8, 18], [18, 19]],
[[1, 3], [9, 11], [3, 11], [1, 11]]) {
say show($_) . ' ==> (' . (join ' ', edge_to_periphery(@$_) or 'Invalid edge format') . ')'
} |
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
| #Peloton | Peloton | <# DEFINE USERDEFINEDROUTINE LITERAL>__FizzBuzz|<# SUPPRESSAUTOMATICWHITESPACE>
<# TEST ISITMODULUSZERO PARAMETER LITERAL>1|3</#>
<# TEST ISITMODULUSZERO PARAMETER LITERAL>1|5</#>
<# ONLYFIRSTOFLASTTWO><# SAY LITERAL>Fizz</#></#>
<# ONLYSECONDOFLASTTWO><# SAY LITERAL>Buzz</#></#>
<# BOTH><# SAY LITERAL>FizzBuzz</#></#>
<# NEITHER><# SAY PARAMETER>1</#></#>
</#></#>
<# ITERATE FORITERATION LITERAL LITERAL>100|<# ACT USERDEFINEDROUTINE POSITION FORITERATION>__FizzBuzz|...</#> </#> |
http://rosettacode.org/wiki/Exponentiation_with_infix_operators_in_(or_operating_on)_the_base | Exponentiation with infix operators in (or operating on) the base | (Many programming languages, especially those with extended─precision integer arithmetic, usually
support one of **, ^, ↑ or some such for exponentiation.)
Some languages treat/honor infix operators when performing exponentiation (raising
numbers to some power by the language's exponentiation operator, if the computer
programming language has one).
Other programming languages may make use of the POW or some other BIF
(Built─In Ffunction), or some other library service.
If your language's exponentiation operator is not one of the usual ones, please comment on how to recognize it.
This task will deal with the case where there is some form of an infix operator operating
in (or operating on) the base.
Example
A negative five raised to the 3rd power could be specified as:
-5 ** 3 or as
-(5) ** 3 or as
(-5) ** 3 or as something else
(Not all computer programming languages have an exponential operator and/or support these syntax expression(s).
Task
compute and display exponentiation with a possible infix operator, whether specified and/or implied/inferred.
Raise the following numbers (integer or real):
-5 and
+5
to the following powers:
2nd and
3rd
using the following expressions (if applicable in your language):
-x**p
-(x)**p
(-x)**p
-(x**p)
Show here (on this page) the four (or more) types of symbolic expressions for each number and power.
Try to present the results in the same format/manner as the other programming entries to make any differences apparent.
The variables may be of any type(s) that is/are applicable in your language.
Related tasks
Exponentiation order
Exponentiation operator
Arbitrary-precision integers (included)
Parsing/RPN to infix conversion
Operator precedence
References
Wikipedia: Order of operations in Programming languages
| #Factor | Factor | USING: infix locals prettyprint sequences
sequences.generalizations sequences.repeating ;
:: row ( x p -- seq )
x p "-x**p" [infix -x**p infix]
"-(x)**p" [infix -(x)**p infix]
"(-x)**p" [infix (-x)**p infix]
"-(x**p)" [infix -(x**p) infix] 10 narray ;
{ "x value" "p value" } { "expression" "result" } 8 cycle append
-5 2 row
-5 3 row
5 2 row
5 3 row
5 narray simple-table. |
http://rosettacode.org/wiki/Exponentiation_with_infix_operators_in_(or_operating_on)_the_base | Exponentiation with infix operators in (or operating on) the base | (Many programming languages, especially those with extended─precision integer arithmetic, usually
support one of **, ^, ↑ or some such for exponentiation.)
Some languages treat/honor infix operators when performing exponentiation (raising
numbers to some power by the language's exponentiation operator, if the computer
programming language has one).
Other programming languages may make use of the POW or some other BIF
(Built─In Ffunction), or some other library service.
If your language's exponentiation operator is not one of the usual ones, please comment on how to recognize it.
This task will deal with the case where there is some form of an infix operator operating
in (or operating on) the base.
Example
A negative five raised to the 3rd power could be specified as:
-5 ** 3 or as
-(5) ** 3 or as
(-5) ** 3 or as something else
(Not all computer programming languages have an exponential operator and/or support these syntax expression(s).
Task
compute and display exponentiation with a possible infix operator, whether specified and/or implied/inferred.
Raise the following numbers (integer or real):
-5 and
+5
to the following powers:
2nd and
3rd
using the following expressions (if applicable in your language):
-x**p
-(x)**p
(-x)**p
-(x**p)
Show here (on this page) the four (or more) types of symbolic expressions for each number and power.
Try to present the results in the same format/manner as the other programming entries to make any differences apparent.
The variables may be of any type(s) that is/are applicable in your language.
Related tasks
Exponentiation order
Exponentiation operator
Arbitrary-precision integers (included)
Parsing/RPN to infix conversion
Operator precedence
References
Wikipedia: Order of operations in Programming languages
| #FreeBASIC | FreeBASIC | print " x p | -x^p -(x)^p (-x)^p -(x^p)"
print "----------------+---------------------------------------------"
for x as integer = -5 to 5 step 10
for p as integer = 2 to 3
print using " ## ## | #### #### #### ####";_
x;p;(-x^p);(-(x)^p);((-x)^p);(-(x^p))
next p
next x |
http://rosettacode.org/wiki/Exponentiation_with_infix_operators_in_(or_operating_on)_the_base | Exponentiation with infix operators in (or operating on) the base | (Many programming languages, especially those with extended─precision integer arithmetic, usually
support one of **, ^, ↑ or some such for exponentiation.)
Some languages treat/honor infix operators when performing exponentiation (raising
numbers to some power by the language's exponentiation operator, if the computer
programming language has one).
Other programming languages may make use of the POW or some other BIF
(Built─In Ffunction), or some other library service.
If your language's exponentiation operator is not one of the usual ones, please comment on how to recognize it.
This task will deal with the case where there is some form of an infix operator operating
in (or operating on) the base.
Example
A negative five raised to the 3rd power could be specified as:
-5 ** 3 or as
-(5) ** 3 or as
(-5) ** 3 or as something else
(Not all computer programming languages have an exponential operator and/or support these syntax expression(s).
Task
compute and display exponentiation with a possible infix operator, whether specified and/or implied/inferred.
Raise the following numbers (integer or real):
-5 and
+5
to the following powers:
2nd and
3rd
using the following expressions (if applicable in your language):
-x**p
-(x)**p
(-x)**p
-(x**p)
Show here (on this page) the four (or more) types of symbolic expressions for each number and power.
Try to present the results in the same format/manner as the other programming entries to make any differences apparent.
The variables may be of any type(s) that is/are applicable in your language.
Related tasks
Exponentiation order
Exponentiation operator
Arbitrary-precision integers (included)
Parsing/RPN to infix conversion
Operator precedence
References
Wikipedia: Order of operations in Programming languages
| #Go | Go | package main
import (
"fmt"
"math"
)
type float float64
func (f float) p(e float) float { return float(math.Pow(float64(f), float64(e))) }
func main() {
ops := []string{"-x.p(e)", "-(x).p(e)", "(-x).p(e)", "-(x.p(e))"}
for _, x := range []float{float(-5), float(5)} {
for _, e := range []float{float(2), float(3)} {
fmt.Printf("x = %2.0f e = %0.0f | ", x, e)
fmt.Printf("%s = %4.0f | ", ops[0], -x.p(e))
fmt.Printf("%s = %4.0f | ", ops[1], -(x).p(e))
fmt.Printf("%s = %4.0f | ", ops[2], (-x).p(e))
fmt.Printf("%s = %4.0f\n", ops[3], -(x.p(e)))
}
}
} |
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
| #Falcon | Falcon | function fib_i(n)
if n < 2: return n
fibPrev = 1
fib = 1
for i in [2:n]
tmp = fib
fib += fibPrev
fibPrev = tmp
end
return fib
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
| #Niue | Niue | [ 'n ; [ negative-or-zero [ , ] if
[ n not-factor [ , ] when ] else ] n times n ] 'factors ;
[ dup 0 <= ] 'negative-or-zero ;
[ swap dup rot swap mod 0 = not ] 'not-factor ;
( tests )
100 factors .s .clr ( => 1 2 4 5 10 20 25 50 100 ) newline
53 factors .s .clr ( => 1 53 ) newline
64 factors .s .clr ( => 1 2 4 8 16 32 64 ) newline
12 factors .s .clr ( => 1 2 3 4 6 12 ) |
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
| #Ada | Ada |
subtype Consistent_Float is Float range Float'Range; -- No IEEE ideals
|
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
| #AppleScript | AppleScript | -- Permutate a list according to a given factorial base number.
on FBNShuffle(|Ω|, fbn)
set astid to AppleScript's text item delimiters
set AppleScript's text item delimiters to "."
set fbnDigits to fbn's text items
set AppleScript's text item delimiters to astid
repeat with m from 1 to (count fbnDigits)
set m_plus_g to m + (item m of fbnDigits)
set v to item m_plus_g of |Ω|
repeat with i from (m_plus_g - 1) to m by -1
set item (i + 1) of |Ω| to item i of |Ω|
end repeat
set item m of |Ω| to v
end repeat
end FBNShuffle
-- Generate all the factorial base numbers having a given number of digits.
on generateFBNs(numberOfDigits)
script o
property partials : {}
property permutations : {}
end script
if (numberOfDigits > 0) then
repeat with i from 0 to numberOfDigits
set end of o's permutations to (i as text)
end repeat
repeat with maxDigit from (numberOfDigits - 1) to 1 by -1
set o's partials to o's permutations
set o's permutations to {}
repeat with i from 1 to (count o's partials)
set thisPartial to item i of o's partials
repeat with j from 0 to maxDigit
set end of o's permutations to (thisPartial & ("." & j))
end repeat
end repeat
end repeat
end if
return o's permutations
end generateFBNs
-- Generate a random factorial base number having a given number of digits.
on generateRandomFBN(numberOfDigits)
set fbnDigits to {}
repeat with maxDigit from numberOfDigits to 1 by -1
set end of fbnDigits to (random number maxDigit)
end repeat
set astid to AppleScript's text item delimiters
set AppleScript's text item delimiters to "."
set fbn to fbnDigits as text
set AppleScript's text item delimiters to astid
return fbn
end generateRandomFBN
(* Test code *)
set astid to AppleScript's text item delimiters
set AppleScript's text item delimiters to ""
-- 1. Reproduce table of {0, 1, 2, 3} permutations
set output to {"1. Reproduce table of {0, 1, 2, 3} permutations:"}
set elements to {0, 1, 2, 3}
set listOfFBNs to generateFBNs((count elements) - 1)
repeat with fbn in listOfFBNs
copy elements to |Ω|
FBNShuffle(|Ω|, fbn)
set end of output to fbn's contents & " -> " & |Ω|
end repeat
-- 2. Generate and count all 11-digit permutations. No way!
set end of output to ""
set numberOfDigits to 11
set numberOfPermutations to 1
repeat with base from 2 to (numberOfDigits + 1)
set numberOfPermutations to numberOfPermutations * base
end repeat
set end of output to "2. " & numberOfDigits & "-digit factorial base numbers have " & (numberOfPermutations div 1) & " possible permutations!"
-- 3. Card shoe permutations with the given FBNs.
set end of output to ""
set shoe to {"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♣"}
copy shoe to shoe1
copy shoe to shoe2
set fbn1 to "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"
set fbn2 to "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"
FBNShuffle(shoe1, fbn1)
FBNShuffle(shoe2, fbn2)
set end of output to "3. Shuffle " & shoe
set end of output to "With " & fbn1 & (linefeed & " -> " & shoe1)
set end of output to "With " & fbn2 & (linefeed & " -> " & shoe2)
-- 4. Card shoe permutation with randomly generated FBN.
set end of output to ""
set fbn3 to generateRandomFBN(51)
FBNShuffle(shoe, fbn3)
set end of output to "4. With randomly generated " & fbn3 & (linefeed & " -> " & shoe)
set AppleScript's text item delimiters to linefeed
set output to output as text
set AppleScript's text item delimiters to astid
return output |
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
| #11l | 11l | V fact = [1]
L(n) 1..11
fact.append(fact[n-1] * n)
L(b) 9..12
print(‘The factorions for base ’b‘ are:’)
L(i) 1..1'499'999
V fact_sum = 0
V j = i
L j > 0
V d = j % b
fact_sum += fact[d]
j I/= b
I 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.
| #Scheme | Scheme |
(define filter
(lambda (fn lst)
(let iter ((lst lst) (result '()))
(if (null? lst)
(reverse result)
(let ((item (car lst))
(rest (cdr lst)))
(if (fn item)
(iter rest (cons item result))
(iter rest result)))))))
|
http://rosettacode.org/wiki/Faces_from_a_mesh | Faces from a mesh | A mesh defining a surface has uniquely numbered vertices, and named,
simple-polygonal faces described usually by an ordered list of edge numbers
going around the face,
For example:
External image of two faces
Rough textual version without edges:
1
17
7 A
B
11
23
A is the triangle (1, 11, 7), or equally (7, 11, 1), going anti-clockwise, or
any of all the rotations of those ordered vertices.
1
7 A
11
B is the four-sided face (1, 17, 23, 11), or equally (23, 17, 1, 11) or any
of their rotations.
1
17
B
11
23
Let's call the above the perimeter format as it traces around the perimeter.
A second format
A separate algorithm returns polygonal faces consisting of a face name and an unordered
set of edge definitions for each face.
A single edge is described by the vertex numbers at its two ends, always in
ascending order.
All edges for the face are given, but in an undefined order.
For example face A could be described by the edges (1, 11), (7, 11), and (1, 7)
(The order of each vertex number in an edge is ascending, but the order in
which the edges are stated is arbitrary).
Similarly face B could be described by the edges (11, 23), (1, 17), (17, 23),
and (1, 11) in arbitrary order of the edges.
Let's call this second format the edge format.
Task
1. Write a routine to check if two perimeter formatted faces have the same perimeter. Use it to check if the following pairs of perimeters are the same:
Q: (8, 1, 3)
R: (1, 3, 8)
U: (18, 8, 14, 10, 12, 17, 19)
V: (8, 14, 10, 12, 17, 19, 18)
2. Write a routine and use it to transform the following faces from edge to perimeter format.
E: {(1, 11), (7, 11), (1, 7)}
F: {(11, 23), (1, 17), (17, 23), (1, 11)}
G: {(8, 14), (17, 19), (10, 12), (10, 14), (12, 17), (8, 18), (18, 19)}
H: {(1, 3), (9, 11), (3, 11), (1, 11)}
Show your output here.
| #Phix | Phix | with javascript_semantics
function perequiv(sequence a, b)
-- Works by aligning and rotating in one step, so theoretically much faster on massive sets.
-- (ahem, faster than multiple rotates, index-only loops would obviously be even faster...)
bool res = (length(a)==length(b))
if res and length(a)>0 then
integer k = find(a[1],b)
if k=0 then
res = false
else
-- align with a (ie make b[1]==a[1], by
-- rotating b k places in one operation)
b = b[k..$]&b[1..k-1]
if a!=b then
-- eg {8,3,4,5} <=> {8,5,4,3}, ie
-- rotate *and* keep in alignment.
b[2..$] = reverse(b[2..$])
res = (a==b)
end if
end if
end if
-- return res
return {"false","true"}[res+1]
end function
function edge2peri(sequence edges)
sequence was = edges, res = {}
string error = ""
integer lnk = 0
if length(edges)<2 then
error = "too short"
else
-- edges = sort(deep_copy(edges)) -- (see note below)
res = deep_copy(edges[1])
edges = edges[2..$]
lnk = res[2]
while length(edges) and error="" do
bool found = false
for i=1 to length(edges) do
integer k = find(lnk,edges[i])
if k then
lnk = edges[i][3-k]
edges[i..i] = {}
if edges={} then
if lnk!=res[1] then error = "oh dear" end if
else
if find(lnk,res) then error = "oops" end if
res &= lnk
end if
found = true
exit
end if
end for
if not found then error = "bad link" exit end if
end while
end if
if length(error) then res = {error,res,lnk,edges,was} end if
return res
end function
constant ptests = {{{8, 1, 3}, {1, 3, 8}},
{{18, 8, 14, 10, 12, 17, 19}, {8, 14, 10, 12, 17, 19, 18}},
-- (check our results below against Go etc)
{{1,11,7},{1,7,11}},
{{11,23,17,1},{1,11,23,17}}}
for i=1 to length(ptests) do
sequence {p,q} = ptests[i]
printf(1,"%v equivalent to %v: %s\n",{p,q,perequiv(p,q)})
end for
constant etests = {{{1, 11}, {7, 11}, {1, 7}},
{{11, 23}, {1, 17}, {17, 23}, {1, 11}},
{{8, 14}, {17, 19}, {10, 12}, {10, 14}, {12, 17}, {8, 18}, {18, 19}},
{{1, 3}, {9, 11}, {3, 11}, {1, 11}}}
for i=1 to length(etests) do
printf(1,"%v\n",{edge2peri(etests[i])})
end for
|
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