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http://rosettacode.org/wiki/File_input/output | File input/output | File input/output is part of Short Circuit's Console Program Basics selection.
Task
Create a file called "output.txt", and place in it the contents of the file "input.txt", via an intermediate variable.
In other words, your program will demonstrate:
how to read from a file into a variable
how to write a variable's contents into a file
Oneliners that skip the intermediate variable are of secondary interest — operating systems have copy commands for that.
| #Pascal | Pascal | #!/usr/bin/perl
open my $fh_in, '<', 'input.txt' or die "could not open <input.txt> for reading: $!";
open my $fh_out, '>', 'output.txt' or die "could not open <output.txt> for writing: $!";
# '>' overwrites file, '>>' appends to file, just like in the shell
binmode $fh_out; # marks filehandle for binary content on systems where that matters
print $fh_out $_ while <$fh_in>;
# prints current line to file associated with $fh_out filehandle
# the same, less concise
#while (<$fh_in>) {
# print $fh_out $_;
#};
close $fh_in;
close $fh_out; |
http://rosettacode.org/wiki/File_input/output | File input/output | File input/output is part of Short Circuit's Console Program Basics selection.
Task
Create a file called "output.txt", and place in it the contents of the file "input.txt", via an intermediate variable.
In other words, your program will demonstrate:
how to read from a file into a variable
how to write a variable's contents into a file
Oneliners that skip the intermediate variable are of secondary interest — operating systems have copy commands for that.
| #Perl | Perl | #!/usr/bin/perl
open my $fh_in, '<', 'input.txt' or die "could not open <input.txt> for reading: $!";
open my $fh_out, '>', 'output.txt' or die "could not open <output.txt> for writing: $!";
# '>' overwrites file, '>>' appends to file, just like in the shell
binmode $fh_out; # marks filehandle for binary content on systems where that matters
print $fh_out $_ while <$fh_in>;
# prints current line to file associated with $fh_out filehandle
# the same, less concise
#while (<$fh_in>) {
# print $fh_out $_;
#};
close $fh_in;
close $fh_out; |
http://rosettacode.org/wiki/Fibonacci_word | Fibonacci word | The Fibonacci Word may be created in a manner analogous to the Fibonacci Sequence as described here:
Define F_Word1 as 1
Define F_Word2 as 0
Form F_Word3 as F_Word2 concatenated with F_Word1 i.e.: 01
Form F_Wordn as F_Wordn-1 concatenated with F_wordn-2
Task
Perform the above steps for n = 37.
You may display the first few but not the larger values of n.
{Doing so will get the task's author into trouble with them what be (again!).}
Instead, create a table for F_Words 1 to 37 which shows:
The number of characters in the word
The word's Entropy
Related tasks
Fibonacci word/fractal
Entropy
Entropy/Narcissist
| #Wren | Wren | import "/fmt" for Fmt
var entropy = Fn.new { |s|
var m = {}
for (c in s) {
var d = m[c]
m[c] = (d) ? d + 1 : 1
}
var hm = 0
for (k in m.keys) {
var c = m[k]
hm = hm + c * c.log2
}
var l = s.count
return l.log2 - hm/l
}
var fibWord = Fn.new { |n|
if (n < 2) return n.toString
var a = "1"
var b = "0"
var i = 3
while (i <= n) {
var c = b + a
a = b
b = c
i = i + 1
}
return b
}
Fmt.print("$2s $10s $10m $s", "n", "Length", "Entropy", "Fib word")
for (i in 1..37) {
var fw = fibWord.call(i)
if (i < 10) {
Fmt.print("$2d $,10d $0.8f $s", i, fw.count, entropy.call(fw), fw)
} else {
Fmt.print("$2d $,10d $0.8f $s", i, fw.count, entropy.call(fw), Fmt.abbreviate(20, fw))
}
} |
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
| #APL | APL |
fib←{⍵≤1:⍵ ⋄ (∇ ⍵-1)+∇ ⍵-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
| #bc | bc | /* Calculate the factors of n and return their count.
* This function mutates the global array f[] which will
* contain all factors of n in ascending order after the call!
*/
define f(n) {
auto i, d, h, h[], l, o
/* Local variables:
* i: Loop variable.
* d: Complementary (higher) factor to i.
* h: Will always point to the last element of h[].
* h[]: Array to hold the greater factor of the pair (x, y), where
* x * y == n. The factors are stored in descending order.
* l: Will always point to the next free spot in f[].
* o: For saving the value of scale.
*/
/* Use integer arithmetic */
o = scale
scale = 0
/* Two factors are 1 and n (if n != 1) */
f[l++] = 1
if (n == 1) return(1)
h[0] = n
/* Main loop */
for (i = 2; i < h[h]; i++) {
if (n % i == 0) {
d = n / i
if (d != i) {
h[++h] = d
}
f[l++] = i
}
}
/* Append the values in h[] to f[] */
while (h >= 0) {
f[l++] = h[h--]
}
scale = o
return(l)
} |
http://rosettacode.org/wiki/Fast_Fourier_transform | Fast Fourier transform | Task
Calculate the FFT (Fast Fourier Transform) of an input sequence.
The most general case allows for complex numbers at the input
and results in a sequence of equal length, again of complex numbers.
If you need to restrict yourself to real numbers, the output should
be the magnitude (i.e.: sqrt(re2 + im2)) of the complex result.
The classic version is the recursive Cooley–Tukey FFT. Wikipedia has pseudo-code for that.
Further optimizations are possible but not required.
| #Klong | Klong | fft::{ff2::{[n e o p t k];n::#x;
f::{p::2:#x;e::ff2(*'p);o::ff2({x@1}'p);k::-1;
t::{k::k+1;cmul(cexp(cdiv(cmul([0 -2];(k*pi),0);n,0));x)}'o;
(e cadd't),e csub't};
:[n<2;x;f(x)]};
n::#x;k::{(2^x)<n}{1+x}:~1;n#ff2({x,0}'x,&(2^k)-n)} |
http://rosettacode.org/wiki/Factors_of_a_Mersenne_number | Factors of a Mersenne number | A Mersenne number is a number in the form of 2P-1.
If P is prime, the Mersenne number may be a Mersenne prime
(if P is not prime, the Mersenne number is also not prime).
In the search for Mersenne prime numbers it is advantageous to eliminate exponents by finding a small factor before starting a, potentially lengthy, Lucas-Lehmer test.
There are very efficient algorithms for determining if a number divides 2P-1 (or equivalently, if 2P mod (the number) = 1).
Some languages already have built-in implementations of this exponent-and-mod operation (called modPow or similar).
The following is how to implement this modPow yourself:
For example, let's compute 223 mod 47.
Convert the exponent 23 to binary, you get 10111. Starting with square = 1, repeatedly square it.
Remove the top bit of the exponent, and if it's 1 multiply square by the base of the exponentiation (2), then compute square modulo 47.
Use the result of the modulo from the last step as the initial value of square in the next step:
remove optional
square top bit multiply by 2 mod 47
──────────── ─────── ───────────── ──────
1*1 = 1 1 0111 1*2 = 2 2
2*2 = 4 0 111 no 4
4*4 = 16 1 11 16*2 = 32 32
32*32 = 1024 1 1 1024*2 = 2048 27
27*27 = 729 1 729*2 = 1458 1
Since 223 mod 47 = 1, 47 is a factor of 2P-1.
(To see this, subtract 1 from both sides: 223-1 = 0 mod 47.)
Since we've shown that 47 is a factor, 223-1 is not prime.
Further properties of Mersenne numbers allow us to refine the process even more.
Any factor q of 2P-1 must be of the form 2kP+1, k being a positive integer or zero. Furthermore, q must be 1 or 7 mod 8.
Finally any potential factor q must be prime.
As in other trial division algorithms, the algorithm stops when 2kP+1 > sqrt(N).
These primality tests only work on Mersenne numbers where P is prime. For example, M4=15 yields no factors using these techniques, but factors into 3 and 5, neither of which fit 2kP+1.
Task
Using the above method find a factor of 2929-1 (aka M929)
Related tasks
count in factors
prime decomposition
factors of an integer
Sieve of Eratosthenes
primality by trial division
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
See also
Computers in 1948: 2127 - 1
(Note: This video is no longer available because the YouTube account associated with this video has been terminated.)
| #Maxima | Maxima | mersenne_fac(p) := block([m: 2^p - 1, k: 1],
while mod(m, 2 * k * p + 1) # 0 do k: k + 1,
2 * k * p + 1
)$
mersenne_fac(929);
/* 13007 */ |
http://rosettacode.org/wiki/Factors_of_a_Mersenne_number | Factors of a Mersenne number | A Mersenne number is a number in the form of 2P-1.
If P is prime, the Mersenne number may be a Mersenne prime
(if P is not prime, the Mersenne number is also not prime).
In the search for Mersenne prime numbers it is advantageous to eliminate exponents by finding a small factor before starting a, potentially lengthy, Lucas-Lehmer test.
There are very efficient algorithms for determining if a number divides 2P-1 (or equivalently, if 2P mod (the number) = 1).
Some languages already have built-in implementations of this exponent-and-mod operation (called modPow or similar).
The following is how to implement this modPow yourself:
For example, let's compute 223 mod 47.
Convert the exponent 23 to binary, you get 10111. Starting with square = 1, repeatedly square it.
Remove the top bit of the exponent, and if it's 1 multiply square by the base of the exponentiation (2), then compute square modulo 47.
Use the result of the modulo from the last step as the initial value of square in the next step:
remove optional
square top bit multiply by 2 mod 47
──────────── ─────── ───────────── ──────
1*1 = 1 1 0111 1*2 = 2 2
2*2 = 4 0 111 no 4
4*4 = 16 1 11 16*2 = 32 32
32*32 = 1024 1 1 1024*2 = 2048 27
27*27 = 729 1 729*2 = 1458 1
Since 223 mod 47 = 1, 47 is a factor of 2P-1.
(To see this, subtract 1 from both sides: 223-1 = 0 mod 47.)
Since we've shown that 47 is a factor, 223-1 is not prime.
Further properties of Mersenne numbers allow us to refine the process even more.
Any factor q of 2P-1 must be of the form 2kP+1, k being a positive integer or zero. Furthermore, q must be 1 or 7 mod 8.
Finally any potential factor q must be prime.
As in other trial division algorithms, the algorithm stops when 2kP+1 > sqrt(N).
These primality tests only work on Mersenne numbers where P is prime. For example, M4=15 yields no factors using these techniques, but factors into 3 and 5, neither of which fit 2kP+1.
Task
Using the above method find a factor of 2929-1 (aka M929)
Related tasks
count in factors
prime decomposition
factors of an integer
Sieve of Eratosthenes
primality by trial division
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
See also
Computers in 1948: 2127 - 1
(Note: This video is no longer available because the YouTube account associated with this video has been terminated.)
| #Nim | Nim | import math
proc isPrime(a: int): bool =
if a == 2: return true
if a < 2 or a mod 2 == 0: return false
for i in countup(3, int sqrt(float a), 2):
if a mod i == 0:
return false
return true
const q = 929
if not isPrime q: quit 1
var r = q
while r > 0: r = r shl 1
var d = 2 * q + 1
while true:
var i = 1
var p = r
while p != 0:
i = (i * i) mod d
if p < 0: i *= 2
if i > d: i -= d
p = p shl 1
if i != 1: d += 2 * q
else: break
echo "2^",q," - 1 = 0 (mod ",d,")" |
http://rosettacode.org/wiki/Farey_sequence | Farey sequence | The Farey sequence Fn of order n is the sequence of completely reduced fractions between 0 and 1 which, when in lowest terms, have denominators less than or equal to n, arranged in order of increasing size.
The Farey sequence is sometimes incorrectly called a Farey series.
Each Farey sequence:
starts with the value 0 (zero), denoted by the fraction
0
1
{\displaystyle {\frac {0}{1}}}
ends with the value 1 (unity), denoted by the fraction
1
1
{\displaystyle {\frac {1}{1}}}
.
The Farey sequences of orders 1 to 5 are:
F
1
=
0
1
,
1
1
{\displaystyle {\bf {\it {F}}}_{1}={\frac {0}{1}},{\frac {1}{1}}}
F
2
=
0
1
,
1
2
,
1
1
{\displaystyle {\bf {\it {F}}}_{2}={\frac {0}{1}},{\frac {1}{2}},{\frac {1}{1}}}
F
3
=
0
1
,
1
3
,
1
2
,
2
3
,
1
1
{\displaystyle {\bf {\it {F}}}_{3}={\frac {0}{1}},{\frac {1}{3}},{\frac {1}{2}},{\frac {2}{3}},{\frac {1}{1}}}
F
4
=
0
1
,
1
4
,
1
3
,
1
2
,
2
3
,
3
4
,
1
1
{\displaystyle {\bf {\it {F}}}_{4}={\frac {0}{1}},{\frac {1}{4}},{\frac {1}{3}},{\frac {1}{2}},{\frac {2}{3}},{\frac {3}{4}},{\frac {1}{1}}}
F
5
=
0
1
,
1
5
,
1
4
,
1
3
,
2
5
,
1
2
,
3
5
,
2
3
,
3
4
,
4
5
,
1
1
{\displaystyle {\bf {\it {F}}}_{5}={\frac {0}{1}},{\frac {1}{5}},{\frac {1}{4}},{\frac {1}{3}},{\frac {2}{5}},{\frac {1}{2}},{\frac {3}{5}},{\frac {2}{3}},{\frac {3}{4}},{\frac {4}{5}},{\frac {1}{1}}}
Task
Compute and show the Farey sequence for orders 1 through 11 (inclusive).
Compute and display the number of fractions in the Farey sequence for order 100 through 1,000 (inclusive) by hundreds.
Show the fractions as n/d (using the solidus [or slash] to separate the numerator from the denominator).
The length (the number of fractions) of a Farey sequence asymptotically approaches:
3 × n2 ÷
π
{\displaystyle \pi }
2
See also
OEIS sequence A006842 numerators of Farey series of order 1, 2, ···
OEIS sequence A006843 denominators of Farey series of order 1, 2, ···
OEIS sequence A005728 number of fractions in Farey series of order n
MathWorld entry Farey sequence
Wikipedia entry Farey sequence
| #Ring | Ring |
# Project : Farey sequence
for i = 1 to 11
count = 0
see "F" + string(i) + " = "
farey(i, false)
next
see nl
for x = 100 to 1000 step 100
count = 0
see "F" + string(x) + " = "
see farey(x, false)
see nl
next
func farey(n, descending)
a = 0
b = 1
c = 1
d = n
if descending = true
a = 1
c = n -1
ok
count = count + 1
if n < 12
see string(a) + "/" + string(b) + " "
ok
while ((c <= n) and not descending) or ((a > 0) and descending)
aa = a
bb = b
cc = c
dd = d
k = floor((n + b) / d)
a = cc
b = dd
c = k * cc - aa
d = k * dd - bb
count = count + 1
if n < 12
see string(a) + "/" + string(b) + " "
ok
end
if n < 12
see nl
ok
return count
|
http://rosettacode.org/wiki/Farey_sequence | Farey sequence | The Farey sequence Fn of order n is the sequence of completely reduced fractions between 0 and 1 which, when in lowest terms, have denominators less than or equal to n, arranged in order of increasing size.
The Farey sequence is sometimes incorrectly called a Farey series.
Each Farey sequence:
starts with the value 0 (zero), denoted by the fraction
0
1
{\displaystyle {\frac {0}{1}}}
ends with the value 1 (unity), denoted by the fraction
1
1
{\displaystyle {\frac {1}{1}}}
.
The Farey sequences of orders 1 to 5 are:
F
1
=
0
1
,
1
1
{\displaystyle {\bf {\it {F}}}_{1}={\frac {0}{1}},{\frac {1}{1}}}
F
2
=
0
1
,
1
2
,
1
1
{\displaystyle {\bf {\it {F}}}_{2}={\frac {0}{1}},{\frac {1}{2}},{\frac {1}{1}}}
F
3
=
0
1
,
1
3
,
1
2
,
2
3
,
1
1
{\displaystyle {\bf {\it {F}}}_{3}={\frac {0}{1}},{\frac {1}{3}},{\frac {1}{2}},{\frac {2}{3}},{\frac {1}{1}}}
F
4
=
0
1
,
1
4
,
1
3
,
1
2
,
2
3
,
3
4
,
1
1
{\displaystyle {\bf {\it {F}}}_{4}={\frac {0}{1}},{\frac {1}{4}},{\frac {1}{3}},{\frac {1}{2}},{\frac {2}{3}},{\frac {3}{4}},{\frac {1}{1}}}
F
5
=
0
1
,
1
5
,
1
4
,
1
3
,
2
5
,
1
2
,
3
5
,
2
3
,
3
4
,
4
5
,
1
1
{\displaystyle {\bf {\it {F}}}_{5}={\frac {0}{1}},{\frac {1}{5}},{\frac {1}{4}},{\frac {1}{3}},{\frac {2}{5}},{\frac {1}{2}},{\frac {3}{5}},{\frac {2}{3}},{\frac {3}{4}},{\frac {4}{5}},{\frac {1}{1}}}
Task
Compute and show the Farey sequence for orders 1 through 11 (inclusive).
Compute and display the number of fractions in the Farey sequence for order 100 through 1,000 (inclusive) by hundreds.
Show the fractions as n/d (using the solidus [or slash] to separate the numerator from the denominator).
The length (the number of fractions) of a Farey sequence asymptotically approaches:
3 × n2 ÷
π
{\displaystyle \pi }
2
See also
OEIS sequence A006842 numerators of Farey series of order 1, 2, ···
OEIS sequence A006843 denominators of Farey series of order 1, 2, ···
OEIS sequence A005728 number of fractions in Farey series of order n
MathWorld entry Farey sequence
Wikipedia entry Farey sequence
| #Ruby | Ruby | def farey(n, length=false)
if length
(n*(n+3))/2 - (2..n).sum{|k| farey(n/k, true)}
else
(1..n).each_with_object([]){|k,a|(0..k).each{|m|a << Rational(m,k)}}.uniq.sort
end
end
puts 'Farey sequence for order 1 through 11 (inclusive):'
for n in 1..11
puts "F(#{n}): " + farey(n).join(", ")
end
puts 'Number of fractions in the Farey sequence:'
for i in (100..1000).step(100)
puts "F(%4d) =%7d" % [i, farey(i, true)]
end |
http://rosettacode.org/wiki/Fibonacci_n-step_number_sequences | Fibonacci n-step number sequences | These number series are an expansion of the ordinary Fibonacci sequence where:
For
n
=
2
{\displaystyle n=2}
we have the Fibonacci sequence; with initial values
[
1
,
1
]
{\displaystyle [1,1]}
and
F
k
2
=
F
k
−
1
2
+
F
k
−
2
2
{\displaystyle F_{k}^{2}=F_{k-1}^{2}+F_{k-2}^{2}}
For
n
=
3
{\displaystyle n=3}
we have the tribonacci sequence; with initial values
[
1
,
1
,
2
]
{\displaystyle [1,1,2]}
and
F
k
3
=
F
k
−
1
3
+
F
k
−
2
3
+
F
k
−
3
3
{\displaystyle F_{k}^{3}=F_{k-1}^{3}+F_{k-2}^{3}+F_{k-3}^{3}}
For
n
=
4
{\displaystyle n=4}
we have the tetranacci sequence; with initial values
[
1
,
1
,
2
,
4
]
{\displaystyle [1,1,2,4]}
and
F
k
4
=
F
k
−
1
4
+
F
k
−
2
4
+
F
k
−
3
4
+
F
k
−
4
4
{\displaystyle F_{k}^{4}=F_{k-1}^{4}+F_{k-2}^{4}+F_{k-3}^{4}+F_{k-4}^{4}}
...
For general
n
>
2
{\displaystyle n>2}
we have the Fibonacci
n
{\displaystyle n}
-step sequence -
F
k
n
{\displaystyle F_{k}^{n}}
; with initial values of the first
n
{\displaystyle n}
values of the
(
n
−
1
)
{\displaystyle (n-1)}
'th Fibonacci
n
{\displaystyle n}
-step sequence
F
k
n
−
1
{\displaystyle F_{k}^{n-1}}
; and
k
{\displaystyle k}
'th value of this
n
{\displaystyle n}
'th sequence being
F
k
n
=
∑
i
=
1
(
n
)
F
k
−
i
(
n
)
{\displaystyle F_{k}^{n}=\sum _{i=1}^{(n)}{F_{k-i}^{(n)}}}
For small values of
n
{\displaystyle n}
, Greek numeric prefixes are sometimes used to individually name each series.
Fibonacci
n
{\displaystyle n}
-step sequences
n
{\displaystyle n}
Series name
Values
2
fibonacci
1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 ...
3
tribonacci
1 1 2 4 7 13 24 44 81 149 274 504 927 1705 3136 ...
4
tetranacci
1 1 2 4 8 15 29 56 108 208 401 773 1490 2872 5536 ...
5
pentanacci
1 1 2 4 8 16 31 61 120 236 464 912 1793 3525 6930 ...
6
hexanacci
1 1 2 4 8 16 32 63 125 248 492 976 1936 3840 7617 ...
7
heptanacci
1 1 2 4 8 16 32 64 127 253 504 1004 2000 3984 7936 ...
8
octonacci
1 1 2 4 8 16 32 64 128 255 509 1016 2028 4048 8080 ...
9
nonanacci
1 1 2 4 8 16 32 64 128 256 511 1021 2040 4076 8144 ...
10
decanacci
1 1 2 4 8 16 32 64 128 256 512 1023 2045 4088 8172 ...
Allied sequences can be generated where the initial values are changed:
The Lucas series sums the two preceding values like the fibonacci series for
n
=
2
{\displaystyle n=2}
but uses
[
2
,
1
]
{\displaystyle [2,1]}
as its initial values.
Task
Write a function to generate Fibonacci
n
{\displaystyle n}
-step number sequences given its initial values and assuming the number of initial values determines how many previous values are summed to make the next number of the series.
Use this to print and show here at least the first ten members of the Fibo/tribo/tetra-nacci and Lucas sequences.
Related tasks
Fibonacci sequence
Wolfram Mathworld
Hofstadter Q sequence
Leonardo numbers
Also see
Lucas Numbers - Numberphile (Video)
Tribonacci Numbers (and the Rauzy Fractal) - Numberphile (Video)
Wikipedia, Lucas number
MathWorld, Fibonacci Number
Some identities for r-Fibonacci numbers
OEIS Fibonacci numbers
OEIS Lucas numbers
| #J | J | nacci =: (] , +/@{.)^:(-@#@]`(-#)`]) |
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.
| #IDL | IDL | result = array[where(NOT array AND 1)] |
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
| #K | K | `0:\:{:[0=#a:{,/$(:[0=x!3;"Fizz"];:[0=x!5;"Buzz"])}@x;$x;a]}'1_!101 |
http://rosettacode.org/wiki/File_input/output | File input/output | File input/output is part of Short Circuit's Console Program Basics selection.
Task
Create a file called "output.txt", and place in it the contents of the file "input.txt", via an intermediate variable.
In other words, your program will demonstrate:
how to read from a file into a variable
how to write a variable's contents into a file
Oneliners that skip the intermediate variable are of secondary interest — operating systems have copy commands for that.
| #Phix | Phix | integer fn = open("input.txt","rb")
string txt = get_text(fn)
close(fn)
fn = open("output.txt","wb")
puts(fn,txt)
close(fn)
|
http://rosettacode.org/wiki/File_input/output | File input/output | File input/output is part of Short Circuit's Console Program Basics selection.
Task
Create a file called "output.txt", and place in it the contents of the file "input.txt", via an intermediate variable.
In other words, your program will demonstrate:
how to read from a file into a variable
how to write a variable's contents into a file
Oneliners that skip the intermediate variable are of secondary interest — operating systems have copy commands for that.
| #PHP | PHP | <?php
if (!$in = fopen('input.txt', 'r')) {
die('Could not open input file.');
}
if (!$out = fopen('output.txt', 'w')) {
die('Could not open output file.');
}
while (!feof($in)) {
$data = fread($in, 512);
fwrite($out, $data);
}
fclose($out);
fclose($in);
?> |
http://rosettacode.org/wiki/Fibonacci_word | Fibonacci word | The Fibonacci Word may be created in a manner analogous to the Fibonacci Sequence as described here:
Define F_Word1 as 1
Define F_Word2 as 0
Form F_Word3 as F_Word2 concatenated with F_Word1 i.e.: 01
Form F_Wordn as F_Wordn-1 concatenated with F_wordn-2
Task
Perform the above steps for n = 37.
You may display the first few but not the larger values of n.
{Doing so will get the task's author into trouble with them what be (again!).}
Instead, create a table for F_Words 1 to 37 which shows:
The number of characters in the word
The word's Entropy
Related tasks
Fibonacci word/fractal
Entropy
Entropy/Narcissist
| #zkl | zkl | fcn entropy(bs){ //binary String-->Float
len:=bs.len(); num1s:=(bs-"0").len();
T(num1s,len-num1s).filter().apply('wrap(p){ p=p.toFloat()/len; -p*p.log() })
.sum(0.0) / (2.0).log();
}
" N Length Entropy Fibword".println();
ws:=L("1","0");
foreach n in ([1..37]){
if(n>2) ws.append(ws[-1]+ws[-2]);
w:=ws[-1];
"%3d %10d %2.10f %s".fmt(n,w.len(),entropy(w),
w.len()<50 and w or "<too long>").println();
} |
http://rosettacode.org/wiki/Fibonacci_word | Fibonacci word | The Fibonacci Word may be created in a manner analogous to the Fibonacci Sequence as described here:
Define F_Word1 as 1
Define F_Word2 as 0
Form F_Word3 as F_Word2 concatenated with F_Word1 i.e.: 01
Form F_Wordn as F_Wordn-1 concatenated with F_wordn-2
Task
Perform the above steps for n = 37.
You may display the first few but not the larger values of n.
{Doing so will get the task's author into trouble with them what be (again!).}
Instead, create a table for F_Words 1 to 37 which shows:
The number of characters in the word
The word's Entropy
Related tasks
Fibonacci word/fractal
Entropy
Entropy/Narcissist
| #ZX_Spectrum_Basic | ZX Spectrum Basic | 10 LET x$="1": LET y$="0": LET z$=""
20 PRINT "N, Length, Entropy, Word"
30 LET n=1
40 PRINT n;" ";LEN x$;" ";
50 LET s$=x$: LET base=2: GO SUB 1000
60 PRINT entropy
70 PRINT x$
80 LET n=2
90 PRINT n;" ";LEN y$;" ";
100 LET s$=y$: GO SUB 1000
110 PRINT entropy
120 PRINT y$
130 FOR n=1 TO 18
140 LET x$="1": LET y$="0"
150 FOR i=1 TO n
160 LET z$=y$+x$
170 LET p$=x$: LET x$=y$: LET y$=p$
180 LET p$=y$: LET y$=z$: LET z$=p$
190 NEXT i
200 LET x$="": LET z$=""
210 LET s$=y$: GO SUB 1000
220 PRINT n+2;" ";LEN y$;" ";entropy
230 PRINT y$ AND (LEN y$<32)
240 NEXT n
250 STOP
1000 REM Calculate entropy
1010 LET sourcelen=LEN s$: LET entropy=0
1020 DIM t(255)
1030 FOR j=1 TO sourcelen
1040 LET digit=VAL s$(j)+1: LET t(digit)=t(digit)+1
1050 NEXT j
1060 FOR j=1 TO 255
1070 IF t(j)>0 THEN LET prop=t(j)/sourcelen: LET entropy=entropy-(prop*LN (prop)/LN (base))
1080 NEXT j
1090 RETURN |
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
| #AppleScript | AppleScript | set fibs to {}
set x to (text returned of (display dialog "What fibbonaci number do you want?" default answer "3"))
set x to x as integer
repeat with y from 1 to x
if (y = 1 or y = 2) then
copy 1 to the end of fibs
else
copy ((item (y - 1) of fibs) + (item (y - 2) of fibs)) to the end of fibs
end if
end repeat
return item x of fibs |
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
| #Befunge | Befunge | 10:p&v: >:0:g%#v_0:g\:0:g/\v
>:0:g:*`| > >0:g1+0:p
>:0:g:*-#v_0:g\>$>:!#@_.v
> ^ ^ ," "< |
http://rosettacode.org/wiki/Fast_Fourier_transform | Fast Fourier transform | Task
Calculate the FFT (Fast Fourier Transform) of an input sequence.
The most general case allows for complex numbers at the input
and results in a sequence of equal length, again of complex numbers.
If you need to restrict yourself to real numbers, the output should
be the magnitude (i.e.: sqrt(re2 + im2)) of the complex result.
The classic version is the recursive Cooley–Tukey FFT. Wikipedia has pseudo-code for that.
Further optimizations are possible but not required.
| #Kotlin | Kotlin | import java.lang.Math.*
class Complex(val re: Double, val im: Double) {
operator infix fun plus(x: Complex) = Complex(re + x.re, im + x.im)
operator infix fun minus(x: Complex) = Complex(re - x.re, im - x.im)
operator infix fun times(x: Double) = Complex(re * x, im * x)
operator infix fun times(x: Complex) = Complex(re * x.re - im * x.im, re * x.im + im * x.re)
operator infix fun div(x: Double) = Complex(re / x, im / x)
val exp: Complex by lazy { Complex(cos(im), sin(im)) * (cosh(re) + sinh(re)) }
override fun toString() = when {
b == "0.000" -> a
a == "0.000" -> b + 'i'
im > 0 -> a + " + " + b + 'i'
else -> a + " - " + b + 'i'
}
private val a = "%1.3f".format(re)
private val b = "%1.3f".format(abs(im))
} |
http://rosettacode.org/wiki/Factors_of_a_Mersenne_number | Factors of a Mersenne number | A Mersenne number is a number in the form of 2P-1.
If P is prime, the Mersenne number may be a Mersenne prime
(if P is not prime, the Mersenne number is also not prime).
In the search for Mersenne prime numbers it is advantageous to eliminate exponents by finding a small factor before starting a, potentially lengthy, Lucas-Lehmer test.
There are very efficient algorithms for determining if a number divides 2P-1 (or equivalently, if 2P mod (the number) = 1).
Some languages already have built-in implementations of this exponent-and-mod operation (called modPow or similar).
The following is how to implement this modPow yourself:
For example, let's compute 223 mod 47.
Convert the exponent 23 to binary, you get 10111. Starting with square = 1, repeatedly square it.
Remove the top bit of the exponent, and if it's 1 multiply square by the base of the exponentiation (2), then compute square modulo 47.
Use the result of the modulo from the last step as the initial value of square in the next step:
remove optional
square top bit multiply by 2 mod 47
──────────── ─────── ───────────── ──────
1*1 = 1 1 0111 1*2 = 2 2
2*2 = 4 0 111 no 4
4*4 = 16 1 11 16*2 = 32 32
32*32 = 1024 1 1 1024*2 = 2048 27
27*27 = 729 1 729*2 = 1458 1
Since 223 mod 47 = 1, 47 is a factor of 2P-1.
(To see this, subtract 1 from both sides: 223-1 = 0 mod 47.)
Since we've shown that 47 is a factor, 223-1 is not prime.
Further properties of Mersenne numbers allow us to refine the process even more.
Any factor q of 2P-1 must be of the form 2kP+1, k being a positive integer or zero. Furthermore, q must be 1 or 7 mod 8.
Finally any potential factor q must be prime.
As in other trial division algorithms, the algorithm stops when 2kP+1 > sqrt(N).
These primality tests only work on Mersenne numbers where P is prime. For example, M4=15 yields no factors using these techniques, but factors into 3 and 5, neither of which fit 2kP+1.
Task
Using the above method find a factor of 2929-1 (aka M929)
Related tasks
count in factors
prime decomposition
factors of an integer
Sieve of Eratosthenes
primality by trial division
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
See also
Computers in 1948: 2127 - 1
(Note: This video is no longer available because the YouTube account associated with this video has been terminated.)
| #Octave | Octave | % test a bit; lsb is 1 (like built-in bit* ops)
function b = bittst(n, p)
b = bitand(n, 2^(p-1)) > 0;
endfunction
function f = Mfactor(p)
% msb is the index of the first non-zero bit
[b, msb] = max(bitand(p, 2 .^ [32:-1:1]) > 0);
maxk = floor(sqrt(intmax()) / p);
for k = 1 : maxk
q = 2*p*k + 1;
if ( ! isprime(q) )
continue;
endif
if ( (mod(q, 8) != 1) && ( mod(q, 8) != 7) )
continue;
endif
n = 1;
for i = msb:-1:1
if ( bittst(p, i) )
n = mod(n*n*2, q);
else
n = mod(n*n, q);
endif
endfor
if ( n==1 )
f = q;
return
endif
endfor
f = 0;
endfunction
printf("%d\n", Mfactor(929)); |
http://rosettacode.org/wiki/Farey_sequence | Farey sequence | The Farey sequence Fn of order n is the sequence of completely reduced fractions between 0 and 1 which, when in lowest terms, have denominators less than or equal to n, arranged in order of increasing size.
The Farey sequence is sometimes incorrectly called a Farey series.
Each Farey sequence:
starts with the value 0 (zero), denoted by the fraction
0
1
{\displaystyle {\frac {0}{1}}}
ends with the value 1 (unity), denoted by the fraction
1
1
{\displaystyle {\frac {1}{1}}}
.
The Farey sequences of orders 1 to 5 are:
F
1
=
0
1
,
1
1
{\displaystyle {\bf {\it {F}}}_{1}={\frac {0}{1}},{\frac {1}{1}}}
F
2
=
0
1
,
1
2
,
1
1
{\displaystyle {\bf {\it {F}}}_{2}={\frac {0}{1}},{\frac {1}{2}},{\frac {1}{1}}}
F
3
=
0
1
,
1
3
,
1
2
,
2
3
,
1
1
{\displaystyle {\bf {\it {F}}}_{3}={\frac {0}{1}},{\frac {1}{3}},{\frac {1}{2}},{\frac {2}{3}},{\frac {1}{1}}}
F
4
=
0
1
,
1
4
,
1
3
,
1
2
,
2
3
,
3
4
,
1
1
{\displaystyle {\bf {\it {F}}}_{4}={\frac {0}{1}},{\frac {1}{4}},{\frac {1}{3}},{\frac {1}{2}},{\frac {2}{3}},{\frac {3}{4}},{\frac {1}{1}}}
F
5
=
0
1
,
1
5
,
1
4
,
1
3
,
2
5
,
1
2
,
3
5
,
2
3
,
3
4
,
4
5
,
1
1
{\displaystyle {\bf {\it {F}}}_{5}={\frac {0}{1}},{\frac {1}{5}},{\frac {1}{4}},{\frac {1}{3}},{\frac {2}{5}},{\frac {1}{2}},{\frac {3}{5}},{\frac {2}{3}},{\frac {3}{4}},{\frac {4}{5}},{\frac {1}{1}}}
Task
Compute and show the Farey sequence for orders 1 through 11 (inclusive).
Compute and display the number of fractions in the Farey sequence for order 100 through 1,000 (inclusive) by hundreds.
Show the fractions as n/d (using the solidus [or slash] to separate the numerator from the denominator).
The length (the number of fractions) of a Farey sequence asymptotically approaches:
3 × n2 ÷
π
{\displaystyle \pi }
2
See also
OEIS sequence A006842 numerators of Farey series of order 1, 2, ···
OEIS sequence A006843 denominators of Farey series of order 1, 2, ···
OEIS sequence A005728 number of fractions in Farey series of order n
MathWorld entry Farey sequence
Wikipedia entry Farey sequence
| #Rust | Rust | #[derive(Copy, Clone)]
struct Fraction {
numerator: u32,
denominator: u32,
}
use std::fmt;
impl fmt::Display for Fraction {
fn fmt(&self, f: &mut fmt::Formatter<'_>) -> fmt::Result {
write!(f, "{}/{}", self.numerator, self.denominator)
}
}
impl Fraction {
fn new(n: u32, d: u32) -> Fraction {
Fraction {
numerator: n,
denominator: d,
}
}
}
fn farey_sequence(n: u32) -> impl std::iter::Iterator<Item = Fraction> {
let mut a = 0;
let mut b = 1;
let mut c = 1;
let mut d = n;
std::iter::from_fn(move || {
if a > n {
return None;
}
let result = Fraction::new(a, b);
let k = (n + b) / d;
let next_c = k * c - a;
let next_d = k * d - b;
a = c;
b = d;
c = next_c;
d = next_d;
Some(result)
})
}
fn main() {
for n in 1..=11 {
print!("{}:", n);
for f in farey_sequence(n) {
print!(" {}", f);
}
println!();
}
for n in (100..=1000).step_by(100) {
println!("{}: {}", n, farey_sequence(n).count());
}
} |
http://rosettacode.org/wiki/Farey_sequence | Farey sequence | The Farey sequence Fn of order n is the sequence of completely reduced fractions between 0 and 1 which, when in lowest terms, have denominators less than or equal to n, arranged in order of increasing size.
The Farey sequence is sometimes incorrectly called a Farey series.
Each Farey sequence:
starts with the value 0 (zero), denoted by the fraction
0
1
{\displaystyle {\frac {0}{1}}}
ends with the value 1 (unity), denoted by the fraction
1
1
{\displaystyle {\frac {1}{1}}}
.
The Farey sequences of orders 1 to 5 are:
F
1
=
0
1
,
1
1
{\displaystyle {\bf {\it {F}}}_{1}={\frac {0}{1}},{\frac {1}{1}}}
F
2
=
0
1
,
1
2
,
1
1
{\displaystyle {\bf {\it {F}}}_{2}={\frac {0}{1}},{\frac {1}{2}},{\frac {1}{1}}}
F
3
=
0
1
,
1
3
,
1
2
,
2
3
,
1
1
{\displaystyle {\bf {\it {F}}}_{3}={\frac {0}{1}},{\frac {1}{3}},{\frac {1}{2}},{\frac {2}{3}},{\frac {1}{1}}}
F
4
=
0
1
,
1
4
,
1
3
,
1
2
,
2
3
,
3
4
,
1
1
{\displaystyle {\bf {\it {F}}}_{4}={\frac {0}{1}},{\frac {1}{4}},{\frac {1}{3}},{\frac {1}{2}},{\frac {2}{3}},{\frac {3}{4}},{\frac {1}{1}}}
F
5
=
0
1
,
1
5
,
1
4
,
1
3
,
2
5
,
1
2
,
3
5
,
2
3
,
3
4
,
4
5
,
1
1
{\displaystyle {\bf {\it {F}}}_{5}={\frac {0}{1}},{\frac {1}{5}},{\frac {1}{4}},{\frac {1}{3}},{\frac {2}{5}},{\frac {1}{2}},{\frac {3}{5}},{\frac {2}{3}},{\frac {3}{4}},{\frac {4}{5}},{\frac {1}{1}}}
Task
Compute and show the Farey sequence for orders 1 through 11 (inclusive).
Compute and display the number of fractions in the Farey sequence for order 100 through 1,000 (inclusive) by hundreds.
Show the fractions as n/d (using the solidus [or slash] to separate the numerator from the denominator).
The length (the number of fractions) of a Farey sequence asymptotically approaches:
3 × n2 ÷
π
{\displaystyle \pi }
2
See also
OEIS sequence A006842 numerators of Farey series of order 1, 2, ···
OEIS sequence A006843 denominators of Farey series of order 1, 2, ···
OEIS sequence A005728 number of fractions in Farey series of order n
MathWorld entry Farey sequence
Wikipedia entry Farey sequence
| #Scala | Scala |
object FareySequence {
def fareySequence(n: Int, start: (Int, Int), stop: (Int, Int)): LazyList[(Int, Int)] = {
val (nominator_l, denominator_l) = start
val (nominator_r, denominator_r) = stop
val mediant = ((nominator_l + nominator_r), (denominator_l + denominator_r))
if (mediant._2 <= n) fareySequence(n, start, mediant) ++ mediant #:: fareySequence(n, mediant, stop)
else LazyList.empty
}
def farey(n: Int, start: (Int, Int) = (0, 1), stop: (Int, Int) = (1, 1)): LazyList[(Int, Int)] = {
start #:: fareySequence(n, start, stop) ++ stop #:: LazyList.empty[(Int, Int)]
}
def main(args: Array[String]): Unit = {
for (i <- 1 to 11) {
println(s"$i: " + farey(i).map(e => s"${e._1}/${e._2}").mkString(", "))
}
println
for (i <- 100 to 1000 by 100) {
println(s"$i: " + farey(i).length + " elements")
}
}
}
|
http://rosettacode.org/wiki/Fibonacci_n-step_number_sequences | Fibonacci n-step number sequences | These number series are an expansion of the ordinary Fibonacci sequence where:
For
n
=
2
{\displaystyle n=2}
we have the Fibonacci sequence; with initial values
[
1
,
1
]
{\displaystyle [1,1]}
and
F
k
2
=
F
k
−
1
2
+
F
k
−
2
2
{\displaystyle F_{k}^{2}=F_{k-1}^{2}+F_{k-2}^{2}}
For
n
=
3
{\displaystyle n=3}
we have the tribonacci sequence; with initial values
[
1
,
1
,
2
]
{\displaystyle [1,1,2]}
and
F
k
3
=
F
k
−
1
3
+
F
k
−
2
3
+
F
k
−
3
3
{\displaystyle F_{k}^{3}=F_{k-1}^{3}+F_{k-2}^{3}+F_{k-3}^{3}}
For
n
=
4
{\displaystyle n=4}
we have the tetranacci sequence; with initial values
[
1
,
1
,
2
,
4
]
{\displaystyle [1,1,2,4]}
and
F
k
4
=
F
k
−
1
4
+
F
k
−
2
4
+
F
k
−
3
4
+
F
k
−
4
4
{\displaystyle F_{k}^{4}=F_{k-1}^{4}+F_{k-2}^{4}+F_{k-3}^{4}+F_{k-4}^{4}}
...
For general
n
>
2
{\displaystyle n>2}
we have the Fibonacci
n
{\displaystyle n}
-step sequence -
F
k
n
{\displaystyle F_{k}^{n}}
; with initial values of the first
n
{\displaystyle n}
values of the
(
n
−
1
)
{\displaystyle (n-1)}
'th Fibonacci
n
{\displaystyle n}
-step sequence
F
k
n
−
1
{\displaystyle F_{k}^{n-1}}
; and
k
{\displaystyle k}
'th value of this
n
{\displaystyle n}
'th sequence being
F
k
n
=
∑
i
=
1
(
n
)
F
k
−
i
(
n
)
{\displaystyle F_{k}^{n}=\sum _{i=1}^{(n)}{F_{k-i}^{(n)}}}
For small values of
n
{\displaystyle n}
, Greek numeric prefixes are sometimes used to individually name each series.
Fibonacci
n
{\displaystyle n}
-step sequences
n
{\displaystyle n}
Series name
Values
2
fibonacci
1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 ...
3
tribonacci
1 1 2 4 7 13 24 44 81 149 274 504 927 1705 3136 ...
4
tetranacci
1 1 2 4 8 15 29 56 108 208 401 773 1490 2872 5536 ...
5
pentanacci
1 1 2 4 8 16 31 61 120 236 464 912 1793 3525 6930 ...
6
hexanacci
1 1 2 4 8 16 32 63 125 248 492 976 1936 3840 7617 ...
7
heptanacci
1 1 2 4 8 16 32 64 127 253 504 1004 2000 3984 7936 ...
8
octonacci
1 1 2 4 8 16 32 64 128 255 509 1016 2028 4048 8080 ...
9
nonanacci
1 1 2 4 8 16 32 64 128 256 511 1021 2040 4076 8144 ...
10
decanacci
1 1 2 4 8 16 32 64 128 256 512 1023 2045 4088 8172 ...
Allied sequences can be generated where the initial values are changed:
The Lucas series sums the two preceding values like the fibonacci series for
n
=
2
{\displaystyle n=2}
but uses
[
2
,
1
]
{\displaystyle [2,1]}
as its initial values.
Task
Write a function to generate Fibonacci
n
{\displaystyle n}
-step number sequences given its initial values and assuming the number of initial values determines how many previous values are summed to make the next number of the series.
Use this to print and show here at least the first ten members of the Fibo/tribo/tetra-nacci and Lucas sequences.
Related tasks
Fibonacci sequence
Wolfram Mathworld
Hofstadter Q sequence
Leonardo numbers
Also see
Lucas Numbers - Numberphile (Video)
Tribonacci Numbers (and the Rauzy Fractal) - Numberphile (Video)
Wikipedia, Lucas number
MathWorld, Fibonacci Number
Some identities for r-Fibonacci numbers
OEIS Fibonacci numbers
OEIS Lucas numbers
| #Java | Java | class Fibonacci
{
public static int[] lucas(int n, int numRequested)
{
if (n < 2)
throw new IllegalArgumentException("Fibonacci value must be at least 2");
return fibonacci((n == 2) ? new int[] { 2, 1 } : lucas(n - 1, n), numRequested);
}
public static int[] fibonacci(int n, int numRequested)
{
if (n < 2)
throw new IllegalArgumentException("Fibonacci value must be at least 2");
return fibonacci((n == 2) ? new int[] { 1, 1 } : fibonacci(n - 1, n), numRequested);
}
public static int[] fibonacci(int[] startingValues, int numRequested)
{
int[] output = new int[numRequested];
int n = startingValues.length;
System.arraycopy(startingValues, 0, output, 0, n);
for (int i = n; i < numRequested; i++)
for (int j = 1; j <= n; j++)
output[i] += output[i - j];
return output;
}
public static void main(String[] args)
{
for (int n = 2; n <= 10; n++)
{
System.out.print("nacci(" + n + "):");
for (int value : fibonacci(n, 15))
System.out.print(" " + value);
System.out.println();
}
for (int n = 2; n <= 10; n++)
{
System.out.print("lucas(" + n + "):");
for (int value : lucas(n, 15))
System.out.print(" " + value);
System.out.println();
}
}
} |
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.
| #J | J | (#~ f) v |
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
| #Kamailio_Script | Kamailio Script | # FizzBuzz
log_stderror=yes
loadmodule "pv"
loadmodule "xlog"
route {
$var(i) = 1;
while ($var(i) <= 1000) {
if ($var(i) mod 15 == 0) {
xlog("FizzBuzz\n");
} else if ($var(i) mod 5 == 0) {
xlog("Buzz\n");
} else if ($var(i) mod 3 == 0) {
xlog("Fizz\n");
} else {
xlog("$var(i)\n");
}
$var(i) = $var(i) + 1;
}
} |
http://rosettacode.org/wiki/File_input/output | File input/output | File input/output is part of Short Circuit's Console Program Basics selection.
Task
Create a file called "output.txt", and place in it the contents of the file "input.txt", via an intermediate variable.
In other words, your program will demonstrate:
how to read from a file into a variable
how to write a variable's contents into a file
Oneliners that skip the intermediate variable are of secondary interest — operating systems have copy commands for that.
| #PicoLisp | PicoLisp | (let V (in "input.txt" (till))
(out "output.txt" (prin V)) ) |
http://rosettacode.org/wiki/File_input/output | File input/output | File input/output is part of Short Circuit's Console Program Basics selection.
Task
Create a file called "output.txt", and place in it the contents of the file "input.txt", via an intermediate variable.
In other words, your program will demonstrate:
how to read from a file into a variable
how to write a variable's contents into a file
Oneliners that skip the intermediate variable are of secondary interest — operating systems have copy commands for that.
| #Pike | Pike |
object lines = Stdio.File("input.txt")->line_iterator();
object out = Stdio.File("output.txt", "cw");
foreach(lines; int line_number; string line)
out->write(line + "\n");
|
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
| #Arendelle | Arendelle | ( fibonacci , 1; 1 )
[ 98 , // 100 numbers of fibonacci
( fibonacci[ @fibonacci? ] ,
@fibonacci[ @fibonacci - 1 ] + @fibonacci[ @fibonacci - 2 ]
)
"Index: | @fibonacci? | => | @fibonacci[ @fibonacci? - 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
| #BQN | BQN | Factors ← (1+↕)⊸(⊣/˜0=|)
•Show Factors 12345
•Show Factors 729 |
http://rosettacode.org/wiki/Fast_Fourier_transform | Fast Fourier transform | Task
Calculate the FFT (Fast Fourier Transform) of an input sequence.
The most general case allows for complex numbers at the input
and results in a sequence of equal length, again of complex numbers.
If you need to restrict yourself to real numbers, the output should
be the magnitude (i.e.: sqrt(re2 + im2)) of the complex result.
The classic version is the recursive Cooley–Tukey FFT. Wikipedia has pseudo-code for that.
Further optimizations are possible but not required.
| #Lambdatalk | Lambdatalk |
1) the function fft
{def fft
{lambda {:s :x}
{if {= {list.length :x} 1}
then :x
else {let { {:s :s}
{:ev {fft :s {evens :x}} }
{:od {fft :s {odds :x}} } }
{let { {:ev :ev} {:t {rotate :s :od 0 {list.length :od}}} }
{list.append {list.map Cadd :ev :t}
{list.map Csub :ev :t}} }}}}}
{def rotate
{lambda {:s :f :k :N}
{if {list.null? :f}
then nil
else {cons {Cmul {car :f} {Cexp {Cnew 0 {/ {* :s {PI} :k} :N}}}}
{rotate :s {cdr :f} {+ :k 1} :N}}}}}
2) functions for lists
We add to the existing {lambda talk}'s list primitives a small set of functions required by the function fft.
{def evens
{lambda {:l}
{if {list.null? :l}
then nil
else {cons {car :l} {evens {cdr {cdr :l}}}}}}}
{def odds
{lambda {:l}
{if {list.null? {cdr :l}}
then nil
else {cons {car {cdr :l}} {odds {cdr {cdr :l}}}}}}}
{def list.map
{def list.map.r
{lambda {:f :a :b :c}
{if {list.null? :a}
then :c
else {list.map.r :f {cdr :a} {cdr :b}
{cons {:f {car :a} {car :b}} :c}} }}}
{lambda {:f :a :b}
{list.map.r :f {list.reverse :a} {list.reverse :b} nil}}}
{def list.append
{def list.append.r
{lambda {:a :b}
{if {list.null? :b}
then :a
else {list.append.r {cons {car :b} :a} {cdr :b}}}}}
{lambda {:a :b}
{list.append.r :b {list.reverse :a}} }}
3) functions for Cnumbers
{lambda talk} has no primitive functions working on complex numbers. We add the minimal set required by the function fft.
{def Cnew
{lambda {:x :y}
{cons :x :y} }}
{def Cnorm
{lambda {:c}
{sqrt {+ {* {car :c} {car :c}}
{* {cdr :c} {cdr :c}}}} }}
{def Cadd
{lambda {:x :y}
{cons {+ {car :x} {car :y}}
{+ {cdr :x} {cdr :y}}} }}
{def Csub
{lambda {:x :y}
{cons {- {car :x} {car :y}}
{- {cdr :x} {cdr :y}}} }}
{def Cmul
{lambda {:x :y}
{cons {- {* {car :x} {car :y}} {* {cdr :x} {cdr :y}}}
{+ {* {car :x} {cdr :y}} {* {cdr :x} {car :y}}}} }}
{def Cexp
{lambda {:x}
{cons {* {exp {car :x}} {cos {cdr :x}}}
{* {exp {car :x}} {sin {cdr :x}}}} }}
{def Clist
{lambda {:s}
{list.new {map {lambda {:i} {cons :i 0}} :s}}}}
4) testing
Applying the fft function on such a sample (1 1 1 1 0 0 0 0) where numbers have been promoted as complex
{list.disp {fft -1 {Clist 1 1 1 1 0 0 0 0}}} ->
(4 0)
(1 -2.414213562373095)
(0 0)
(1 -0.4142135623730949)
(0 0)
(0.9999999999999999 0.4142135623730949)
(0 0)
(0.9999999999999997 2.414213562373095)
A more usefull example can be seen in http://lambdaway.free.fr/lambdaspeech/?view=zorg
|
http://rosettacode.org/wiki/Factors_of_a_Mersenne_number | Factors of a Mersenne number | A Mersenne number is a number in the form of 2P-1.
If P is prime, the Mersenne number may be a Mersenne prime
(if P is not prime, the Mersenne number is also not prime).
In the search for Mersenne prime numbers it is advantageous to eliminate exponents by finding a small factor before starting a, potentially lengthy, Lucas-Lehmer test.
There are very efficient algorithms for determining if a number divides 2P-1 (or equivalently, if 2P mod (the number) = 1).
Some languages already have built-in implementations of this exponent-and-mod operation (called modPow or similar).
The following is how to implement this modPow yourself:
For example, let's compute 223 mod 47.
Convert the exponent 23 to binary, you get 10111. Starting with square = 1, repeatedly square it.
Remove the top bit of the exponent, and if it's 1 multiply square by the base of the exponentiation (2), then compute square modulo 47.
Use the result of the modulo from the last step as the initial value of square in the next step:
remove optional
square top bit multiply by 2 mod 47
──────────── ─────── ───────────── ──────
1*1 = 1 1 0111 1*2 = 2 2
2*2 = 4 0 111 no 4
4*4 = 16 1 11 16*2 = 32 32
32*32 = 1024 1 1 1024*2 = 2048 27
27*27 = 729 1 729*2 = 1458 1
Since 223 mod 47 = 1, 47 is a factor of 2P-1.
(To see this, subtract 1 from both sides: 223-1 = 0 mod 47.)
Since we've shown that 47 is a factor, 223-1 is not prime.
Further properties of Mersenne numbers allow us to refine the process even more.
Any factor q of 2P-1 must be of the form 2kP+1, k being a positive integer or zero. Furthermore, q must be 1 or 7 mod 8.
Finally any potential factor q must be prime.
As in other trial division algorithms, the algorithm stops when 2kP+1 > sqrt(N).
These primality tests only work on Mersenne numbers where P is prime. For example, M4=15 yields no factors using these techniques, but factors into 3 and 5, neither of which fit 2kP+1.
Task
Using the above method find a factor of 2929-1 (aka M929)
Related tasks
count in factors
prime decomposition
factors of an integer
Sieve of Eratosthenes
primality by trial division
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
See also
Computers in 1948: 2127 - 1
(Note: This video is no longer available because the YouTube account associated with this video has been terminated.)
| #PARI.2FGP | PARI/GP | factorMersenne(p)={
forstep(q=2*p+1,sqrt(2)<<(p\2),2*p,
[1,0,0,0,0,0,1][q%8] && Mod(2, q)^p==1 && return(q)
);
1<<p-1
};
factorMersenne(929) |
http://rosettacode.org/wiki/Factors_of_a_Mersenne_number | Factors of a Mersenne number | A Mersenne number is a number in the form of 2P-1.
If P is prime, the Mersenne number may be a Mersenne prime
(if P is not prime, the Mersenne number is also not prime).
In the search for Mersenne prime numbers it is advantageous to eliminate exponents by finding a small factor before starting a, potentially lengthy, Lucas-Lehmer test.
There are very efficient algorithms for determining if a number divides 2P-1 (or equivalently, if 2P mod (the number) = 1).
Some languages already have built-in implementations of this exponent-and-mod operation (called modPow or similar).
The following is how to implement this modPow yourself:
For example, let's compute 223 mod 47.
Convert the exponent 23 to binary, you get 10111. Starting with square = 1, repeatedly square it.
Remove the top bit of the exponent, and if it's 1 multiply square by the base of the exponentiation (2), then compute square modulo 47.
Use the result of the modulo from the last step as the initial value of square in the next step:
remove optional
square top bit multiply by 2 mod 47
──────────── ─────── ───────────── ──────
1*1 = 1 1 0111 1*2 = 2 2
2*2 = 4 0 111 no 4
4*4 = 16 1 11 16*2 = 32 32
32*32 = 1024 1 1 1024*2 = 2048 27
27*27 = 729 1 729*2 = 1458 1
Since 223 mod 47 = 1, 47 is a factor of 2P-1.
(To see this, subtract 1 from both sides: 223-1 = 0 mod 47.)
Since we've shown that 47 is a factor, 223-1 is not prime.
Further properties of Mersenne numbers allow us to refine the process even more.
Any factor q of 2P-1 must be of the form 2kP+1, k being a positive integer or zero. Furthermore, q must be 1 or 7 mod 8.
Finally any potential factor q must be prime.
As in other trial division algorithms, the algorithm stops when 2kP+1 > sqrt(N).
These primality tests only work on Mersenne numbers where P is prime. For example, M4=15 yields no factors using these techniques, but factors into 3 and 5, neither of which fit 2kP+1.
Task
Using the above method find a factor of 2929-1 (aka M929)
Related tasks
count in factors
prime decomposition
factors of an integer
Sieve of Eratosthenes
primality by trial division
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
See also
Computers in 1948: 2127 - 1
(Note: This video is no longer available because the YouTube account associated with this video has been terminated.)
| #Pascal | Pascal | program FactorsMersenneNumber(input, output);
function isPrime(n: longint): boolean;
var
d: longint;
begin
isPrime := true;
if (n mod 2) = 0 then
begin
isPrime := (n = 2);
exit;
end;
if (n mod 3) = 0 then
begin
isPrime := (n = 3);
exit;
end;
d := 5;
while d*d <= n do
begin
if (n mod d) = 0 then
begin
isPrime := false;
exit;
end;
d := d + 2;
end;
end;
function btest(n, pos: longint): boolean;
begin
btest := (n shr pos) mod 2 = 1;
end;
function MFactor(p: longint): longint;
var
i, k, maxk, msb, n, q: longint;
begin
for i := 30 downto 0 do
if btest(p, i) then
begin
msb := i;
break;
end;
maxk := 16384 div p; // limit for k to prevent overflow of 32 bit signed integer
for k := 1 to maxk do
begin
q := 2*p*k + 1;
if not isprime(q) then
continue;
if ((q mod 8) <> 1) and ((q mod 8) <> 7) then
continue;
n := 1;
for i := msb downto 0 do
if btest(p, i) then
n := (n*n*2) mod q
else
n := (n*n) mod q;
if n = 1 then
begin
mfactor := q;
exit;
end;
end;
mfactor := 0;
end;
var
exponent, factor: longint;
begin
write('Enter the exponent of the Mersenne number (suggestion: 929): ');
readln(exponent);
if not isPrime(exponent) then
begin
writeln('M', exponent, ' (2**', exponent, ' - 1) is not prime.');
exit;
end;
factor := MFactor(exponent);
if factor = 0 then
writeln('M', exponent, ' (2**', exponent, ' - 1) has no factor.')
else
writeln('M', exponent, ' (2**', exponent, ' - 1) has the factor: ', factor);
end. |
http://rosettacode.org/wiki/Farey_sequence | Farey sequence | The Farey sequence Fn of order n is the sequence of completely reduced fractions between 0 and 1 which, when in lowest terms, have denominators less than or equal to n, arranged in order of increasing size.
The Farey sequence is sometimes incorrectly called a Farey series.
Each Farey sequence:
starts with the value 0 (zero), denoted by the fraction
0
1
{\displaystyle {\frac {0}{1}}}
ends with the value 1 (unity), denoted by the fraction
1
1
{\displaystyle {\frac {1}{1}}}
.
The Farey sequences of orders 1 to 5 are:
F
1
=
0
1
,
1
1
{\displaystyle {\bf {\it {F}}}_{1}={\frac {0}{1}},{\frac {1}{1}}}
F
2
=
0
1
,
1
2
,
1
1
{\displaystyle {\bf {\it {F}}}_{2}={\frac {0}{1}},{\frac {1}{2}},{\frac {1}{1}}}
F
3
=
0
1
,
1
3
,
1
2
,
2
3
,
1
1
{\displaystyle {\bf {\it {F}}}_{3}={\frac {0}{1}},{\frac {1}{3}},{\frac {1}{2}},{\frac {2}{3}},{\frac {1}{1}}}
F
4
=
0
1
,
1
4
,
1
3
,
1
2
,
2
3
,
3
4
,
1
1
{\displaystyle {\bf {\it {F}}}_{4}={\frac {0}{1}},{\frac {1}{4}},{\frac {1}{3}},{\frac {1}{2}},{\frac {2}{3}},{\frac {3}{4}},{\frac {1}{1}}}
F
5
=
0
1
,
1
5
,
1
4
,
1
3
,
2
5
,
1
2
,
3
5
,
2
3
,
3
4
,
4
5
,
1
1
{\displaystyle {\bf {\it {F}}}_{5}={\frac {0}{1}},{\frac {1}{5}},{\frac {1}{4}},{\frac {1}{3}},{\frac {2}{5}},{\frac {1}{2}},{\frac {3}{5}},{\frac {2}{3}},{\frac {3}{4}},{\frac {4}{5}},{\frac {1}{1}}}
Task
Compute and show the Farey sequence for orders 1 through 11 (inclusive).
Compute and display the number of fractions in the Farey sequence for order 100 through 1,000 (inclusive) by hundreds.
Show the fractions as n/d (using the solidus [or slash] to separate the numerator from the denominator).
The length (the number of fractions) of a Farey sequence asymptotically approaches:
3 × n2 ÷
π
{\displaystyle \pi }
2
See also
OEIS sequence A006842 numerators of Farey series of order 1, 2, ···
OEIS sequence A006843 denominators of Farey series of order 1, 2, ···
OEIS sequence A005728 number of fractions in Farey series of order n
MathWorld entry Farey sequence
Wikipedia entry Farey sequence
| #Scheme | Scheme |
(import (scheme base)
(scheme write))
;; create a generator for Farey sequence n
;; using next term formula from https://en.wikipedia.org/wiki/Farey_sequence
(define (farey-generator n)
(let ((a #f) (b 1) (c #f) (d n))
(lambda ()
(cond ((not a) ; first item in sequence
(set! a 0)
(/ a b))
((not c) ; second item in sequence
(set! c 1)
(/ c d))
((= c d) ; return #f when finished sequence
#f)
(else ; compute next term
(let* ((f (floor (/ (+ n b) d)))
(p (- (* f c) a))
(q (- (* f d) b)))
(set! a c)
(set! b d)
(set! c p)
(set! d q)
(/ p q)))))))
(define (farey-sequence n display?)
(define (display-rat n) ; ensure 0,1 show /1
(display n)
(when (= 1 (denominator n))
(display "/1"))
(display " "))
;
(let ((gen (farey-generator n)))
(do ((res (gen) (gen))
(count 0 (+ 1 count)))
((not res) (when display? (newline))
count)
(when display? (display-rat res)))))
;;
(display "Farey sequence for order 1 through 11 (inclusive):\n")
(do ((i 1 (+ i 1)))
((> i 11) )
(display (string-append "F(" (number->string i) "): "))
(farey-sequence i #t))
(display "\nNumber of fractions in the Farey sequence:\n")
(do ((i 100 (+ i 100)))
((> i 1000) )
(display
(string-append "F(" (number->string i) ") = "
(number->string (farey-sequence i #f))))
(newline))
|
http://rosettacode.org/wiki/Fibonacci_n-step_number_sequences | Fibonacci n-step number sequences | These number series are an expansion of the ordinary Fibonacci sequence where:
For
n
=
2
{\displaystyle n=2}
we have the Fibonacci sequence; with initial values
[
1
,
1
]
{\displaystyle [1,1]}
and
F
k
2
=
F
k
−
1
2
+
F
k
−
2
2
{\displaystyle F_{k}^{2}=F_{k-1}^{2}+F_{k-2}^{2}}
For
n
=
3
{\displaystyle n=3}
we have the tribonacci sequence; with initial values
[
1
,
1
,
2
]
{\displaystyle [1,1,2]}
and
F
k
3
=
F
k
−
1
3
+
F
k
−
2
3
+
F
k
−
3
3
{\displaystyle F_{k}^{3}=F_{k-1}^{3}+F_{k-2}^{3}+F_{k-3}^{3}}
For
n
=
4
{\displaystyle n=4}
we have the tetranacci sequence; with initial values
[
1
,
1
,
2
,
4
]
{\displaystyle [1,1,2,4]}
and
F
k
4
=
F
k
−
1
4
+
F
k
−
2
4
+
F
k
−
3
4
+
F
k
−
4
4
{\displaystyle F_{k}^{4}=F_{k-1}^{4}+F_{k-2}^{4}+F_{k-3}^{4}+F_{k-4}^{4}}
...
For general
n
>
2
{\displaystyle n>2}
we have the Fibonacci
n
{\displaystyle n}
-step sequence -
F
k
n
{\displaystyle F_{k}^{n}}
; with initial values of the first
n
{\displaystyle n}
values of the
(
n
−
1
)
{\displaystyle (n-1)}
'th Fibonacci
n
{\displaystyle n}
-step sequence
F
k
n
−
1
{\displaystyle F_{k}^{n-1}}
; and
k
{\displaystyle k}
'th value of this
n
{\displaystyle n}
'th sequence being
F
k
n
=
∑
i
=
1
(
n
)
F
k
−
i
(
n
)
{\displaystyle F_{k}^{n}=\sum _{i=1}^{(n)}{F_{k-i}^{(n)}}}
For small values of
n
{\displaystyle n}
, Greek numeric prefixes are sometimes used to individually name each series.
Fibonacci
n
{\displaystyle n}
-step sequences
n
{\displaystyle n}
Series name
Values
2
fibonacci
1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 ...
3
tribonacci
1 1 2 4 7 13 24 44 81 149 274 504 927 1705 3136 ...
4
tetranacci
1 1 2 4 8 15 29 56 108 208 401 773 1490 2872 5536 ...
5
pentanacci
1 1 2 4 8 16 31 61 120 236 464 912 1793 3525 6930 ...
6
hexanacci
1 1 2 4 8 16 32 63 125 248 492 976 1936 3840 7617 ...
7
heptanacci
1 1 2 4 8 16 32 64 127 253 504 1004 2000 3984 7936 ...
8
octonacci
1 1 2 4 8 16 32 64 128 255 509 1016 2028 4048 8080 ...
9
nonanacci
1 1 2 4 8 16 32 64 128 256 511 1021 2040 4076 8144 ...
10
decanacci
1 1 2 4 8 16 32 64 128 256 512 1023 2045 4088 8172 ...
Allied sequences can be generated where the initial values are changed:
The Lucas series sums the two preceding values like the fibonacci series for
n
=
2
{\displaystyle n=2}
but uses
[
2
,
1
]
{\displaystyle [2,1]}
as its initial values.
Task
Write a function to generate Fibonacci
n
{\displaystyle n}
-step number sequences given its initial values and assuming the number of initial values determines how many previous values are summed to make the next number of the series.
Use this to print and show here at least the first ten members of the Fibo/tribo/tetra-nacci and Lucas sequences.
Related tasks
Fibonacci sequence
Wolfram Mathworld
Hofstadter Q sequence
Leonardo numbers
Also see
Lucas Numbers - Numberphile (Video)
Tribonacci Numbers (and the Rauzy Fractal) - Numberphile (Video)
Wikipedia, Lucas number
MathWorld, Fibonacci Number
Some identities for r-Fibonacci numbers
OEIS Fibonacci numbers
OEIS Lucas numbers
| #JavaScript | JavaScript | function fib(arity, len) {
return nacci(nacci([1,1], arity, arity), arity, len);
}
function lucas(arity, len) {
return nacci(nacci([2,1], arity, arity), arity, len);
}
function nacci(a, arity, len) {
while (a.length < len) {
var sum = 0;
for (var i = Math.max(0, a.length - arity); i < a.length; i++)
sum += a[i];
a.push(sum);
}
return a;
}
function main() {
for (var arity = 2; arity <= 10; arity++)
console.log("fib(" + arity + "): " + fib(arity, 15));
for (var arity = 2; arity <= 10; arity++)
console.log("lucas(" + arity + "): " + lucas(arity, 15));
}
main(); |
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.
| #Java | Java | int[] array = {1, 2, 3, 4, 5 };
List<Integer> evensList = new ArrayList<Integer>();
for (int i: array) {
if (i % 2 == 0) evensList.add(i);
}
int[] evens = evensList.toArray(new int[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
| #Kaya | Kaya | // fizzbuzz in Kaya
program fizzbuzz;
Void fizzbuzz(Int size) {
for i in [1..size] {
if (i % 15 == 0) {
putStrLn("FizzBuzz");
} else if (i % 5 == 0) {
putStrLn("Buzz");
} else if (i % 3 == 0) {
putStrLn("Fizz");
} else {
putStrLn( string(i) );
}
}
}
Void main() {
fizzbuzz(100);
} |
http://rosettacode.org/wiki/File_input/output | File input/output | File input/output is part of Short Circuit's Console Program Basics selection.
Task
Create a file called "output.txt", and place in it the contents of the file "input.txt", via an intermediate variable.
In other words, your program will demonstrate:
how to read from a file into a variable
how to write a variable's contents into a file
Oneliners that skip the intermediate variable are of secondary interest — operating systems have copy commands for that.
| #PL.2FI | PL/I |
declare in file, out file;
open file (in) title ('/INPUT.TXT,type(text),recsize(100)') input;
open file (out) title ('/OUTPUT.TXT,type(text),recsize(100') output;
do forever;
get file (in) edit (line) (L);
put file (out) edit (line) (A);
end;
|
http://rosettacode.org/wiki/File_input/output | File input/output | File input/output is part of Short Circuit's Console Program Basics selection.
Task
Create a file called "output.txt", and place in it the contents of the file "input.txt", via an intermediate variable.
In other words, your program will demonstrate:
how to read from a file into a variable
how to write a variable's contents into a file
Oneliners that skip the intermediate variable are of secondary interest — operating systems have copy commands for that.
| #Pop11 | Pop11 | lvars i_stream = discin('input.txt');
lvars o_stream = discout('output.txt');
lvars c;
while (i_stream() ->> c) /= termin do
o_stream(c);
endwhile; |
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
| #ARM_Assembly | ARM Assembly | fibonacci:
push {r1-r3}
mov r1, #0
mov r2, #1
fibloop:
mov r3, r2
add r2, r1, r2
mov r1, r3
sub r0, r0, #1
cmp r0, #1
bne fibloop
mov r0, r2
pop {r1-r3}
mov pc, lr |
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
| #Burlesque | Burlesque | blsq ) 32767 fc
{1 7 31 151 217 1057 4681 32767} |
http://rosettacode.org/wiki/Fast_Fourier_transform | Fast Fourier transform | Task
Calculate the FFT (Fast Fourier Transform) of an input sequence.
The most general case allows for complex numbers at the input
and results in a sequence of equal length, again of complex numbers.
If you need to restrict yourself to real numbers, the output should
be the magnitude (i.e.: sqrt(re2 + im2)) of the complex result.
The classic version is the recursive Cooley–Tukey FFT. Wikipedia has pseudo-code for that.
Further optimizations are possible but not required.
| #Liberty_BASIC | Liberty BASIC |
P =8
S =int( log( P) /log( 2) +0.9999)
Pi =3.14159265
R1 =2^S
R =R1 -1
R2 =div( R1, 2)
R4 =div( R1, 4)
R3 =R4 +R2
Dim Re( R1), Im( R1), Co( R3)
for N =0 to P -1
read dummy: Re( N) =dummy
read dummy: Im( N) =dummy
next N
data 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0
S2 =div( S, 2)
S1 =S -S2
P1 =2^S1
P2 =2^S2
dim V( P1 -1)
V( 0) =0
DV =1
DP =P1
for J =1 to S1
HA =div( DP, 2)
PT =P1 -HA
for I =HA to PT step DP
V( I) =V( I -HA) +DV
next I
DV =DV +DV
DP =HA
next J
K =2 *Pi /R1
for X =0 to R4
COX =cos( K *X)
Co( X) =COX
Co( R2 -X) =0 -COX
Co( R2 +X) =0 -COX
next X
print "FFT: bit reversal"
for I =0 to P1 -1
IP =I *P2
for J =0 to P2 -1
H =IP +J
G =V( J) *P2 +V( I)
if G >H then temp =Re( G): Re( G) =Re( H): Re( H) =temp
if G >H then temp =Im( G): Im( G) =Im( H): Im( H) =temp
next J
next I
T =1
for stage =0 to S -1
print " Stage:- "; stage
D =div( R2, T)
for Z =0 to T -1
L =D *Z
LS =L +R4
for I =0 to D -1
A =2 *I *T +Z
B =A +T
F1 =Re( A)
F2 =Im( A)
P1 =Co( L) *Re( B)
P2 =Co( LS) *Im( B)
P3 =Co( LS) *Re( B)
P4 =Co( L) *Im( B)
Re( A) =F1 +P1 -P2
Im( A) =F2 +P3 +P4
Re( B) =F1 -P1 +P2
Im( B) =F2 -P3 -P4
next I
next Z
T =T +T
next stage
print " M Re( M) Im( M)"
for M =0 to R
if abs( Re( M)) <10^-5 then Re( M) =0
if abs( Im( M)) <10^-5 then Im( M) =0
print " "; M, Re( M), Im( M)
next M
end
wait
function div( a, b)
div =int( a /b)
end function
end
|
http://rosettacode.org/wiki/Factors_of_a_Mersenne_number | Factors of a Mersenne number | A Mersenne number is a number in the form of 2P-1.
If P is prime, the Mersenne number may be a Mersenne prime
(if P is not prime, the Mersenne number is also not prime).
In the search for Mersenne prime numbers it is advantageous to eliminate exponents by finding a small factor before starting a, potentially lengthy, Lucas-Lehmer test.
There are very efficient algorithms for determining if a number divides 2P-1 (or equivalently, if 2P mod (the number) = 1).
Some languages already have built-in implementations of this exponent-and-mod operation (called modPow or similar).
The following is how to implement this modPow yourself:
For example, let's compute 223 mod 47.
Convert the exponent 23 to binary, you get 10111. Starting with square = 1, repeatedly square it.
Remove the top bit of the exponent, and if it's 1 multiply square by the base of the exponentiation (2), then compute square modulo 47.
Use the result of the modulo from the last step as the initial value of square in the next step:
remove optional
square top bit multiply by 2 mod 47
──────────── ─────── ───────────── ──────
1*1 = 1 1 0111 1*2 = 2 2
2*2 = 4 0 111 no 4
4*4 = 16 1 11 16*2 = 32 32
32*32 = 1024 1 1 1024*2 = 2048 27
27*27 = 729 1 729*2 = 1458 1
Since 223 mod 47 = 1, 47 is a factor of 2P-1.
(To see this, subtract 1 from both sides: 223-1 = 0 mod 47.)
Since we've shown that 47 is a factor, 223-1 is not prime.
Further properties of Mersenne numbers allow us to refine the process even more.
Any factor q of 2P-1 must be of the form 2kP+1, k being a positive integer or zero. Furthermore, q must be 1 or 7 mod 8.
Finally any potential factor q must be prime.
As in other trial division algorithms, the algorithm stops when 2kP+1 > sqrt(N).
These primality tests only work on Mersenne numbers where P is prime. For example, M4=15 yields no factors using these techniques, but factors into 3 and 5, neither of which fit 2kP+1.
Task
Using the above method find a factor of 2929-1 (aka M929)
Related tasks
count in factors
prime decomposition
factors of an integer
Sieve of Eratosthenes
primality by trial division
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
See also
Computers in 1948: 2127 - 1
(Note: This video is no longer available because the YouTube account associated with this video has been terminated.)
| #Perl | Perl | use strict;
use utf8;
sub factors {
my $n = shift;
my $p = 2;
my @out;
while ($n >= $p * $p) {
while ($n % $p == 0) {
push @out, $p;
$n /= $p;
}
$p = next_prime($p);
}
push @out, $n if $n > 1 || !@out;
@out;
}
sub next_prime {
my $p = shift;
do { $p = $p == 2 ? 3 : $p + 2 } until is_prime($p);
$p;
}
my %pcache;
sub is_prime {
my $x = shift;
$pcache{$x} //= (factors($x) == 1)
}
sub mtest {
my @bits = split "", sprintf("%b", shift);
my $p = shift;
my $sq = 1;
while (@bits) {
$sq = $sq * $sq;
$sq *= 2 if shift @bits;
$sq %= $p;
}
$sq == 1;
}
for my $m (2 .. 60, 929) {
next unless is_prime($m);
use bigint;
my ($f, $k, $x) = (0, 0, 2**$m - 1);
my $q;
while (++$k) {
$q = 2 * $k * $m + 1;
next if (($q & 7) != 1 && ($q & 7) != 7);
next unless is_prime($q);
last if $q * $q > $x;
last if $f = mtest($m, $q);
}
print $f? "M$m = $x = $q × @{[$x / $q]}\n"
: "M$m = $x is prime\n";
} |
http://rosettacode.org/wiki/Farey_sequence | Farey sequence | The Farey sequence Fn of order n is the sequence of completely reduced fractions between 0 and 1 which, when in lowest terms, have denominators less than or equal to n, arranged in order of increasing size.
The Farey sequence is sometimes incorrectly called a Farey series.
Each Farey sequence:
starts with the value 0 (zero), denoted by the fraction
0
1
{\displaystyle {\frac {0}{1}}}
ends with the value 1 (unity), denoted by the fraction
1
1
{\displaystyle {\frac {1}{1}}}
.
The Farey sequences of orders 1 to 5 are:
F
1
=
0
1
,
1
1
{\displaystyle {\bf {\it {F}}}_{1}={\frac {0}{1}},{\frac {1}{1}}}
F
2
=
0
1
,
1
2
,
1
1
{\displaystyle {\bf {\it {F}}}_{2}={\frac {0}{1}},{\frac {1}{2}},{\frac {1}{1}}}
F
3
=
0
1
,
1
3
,
1
2
,
2
3
,
1
1
{\displaystyle {\bf {\it {F}}}_{3}={\frac {0}{1}},{\frac {1}{3}},{\frac {1}{2}},{\frac {2}{3}},{\frac {1}{1}}}
F
4
=
0
1
,
1
4
,
1
3
,
1
2
,
2
3
,
3
4
,
1
1
{\displaystyle {\bf {\it {F}}}_{4}={\frac {0}{1}},{\frac {1}{4}},{\frac {1}{3}},{\frac {1}{2}},{\frac {2}{3}},{\frac {3}{4}},{\frac {1}{1}}}
F
5
=
0
1
,
1
5
,
1
4
,
1
3
,
2
5
,
1
2
,
3
5
,
2
3
,
3
4
,
4
5
,
1
1
{\displaystyle {\bf {\it {F}}}_{5}={\frac {0}{1}},{\frac {1}{5}},{\frac {1}{4}},{\frac {1}{3}},{\frac {2}{5}},{\frac {1}{2}},{\frac {3}{5}},{\frac {2}{3}},{\frac {3}{4}},{\frac {4}{5}},{\frac {1}{1}}}
Task
Compute and show the Farey sequence for orders 1 through 11 (inclusive).
Compute and display the number of fractions in the Farey sequence for order 100 through 1,000 (inclusive) by hundreds.
Show the fractions as n/d (using the solidus [or slash] to separate the numerator from the denominator).
The length (the number of fractions) of a Farey sequence asymptotically approaches:
3 × n2 ÷
π
{\displaystyle \pi }
2
See also
OEIS sequence A006842 numerators of Farey series of order 1, 2, ···
OEIS sequence A006843 denominators of Farey series of order 1, 2, ···
OEIS sequence A005728 number of fractions in Farey series of order n
MathWorld entry Farey sequence
Wikipedia entry Farey sequence
| #Sidef | Sidef | func farey_count(n) { # A005728
1 + sum(1..n, {|k| euler_phi(k) })
}
func farey(n) {
var seq = [0]
var (a,b,c,d) = (0,1,1,n)
while (c <= n) {
var k = (n+b)//d
(a,b,c,d) = (c, d, k*c - a, k*d - b)
seq << a/b
}
return seq
}
say "Farey sequence for order 1 through 11 (inclusive):"
for n in (1..11) {
say("F(%2d): %s" % (n, farey(n).map{.as_frac}.join(" ")))
}
say "\nNumber of fractions in the Farey sequence:"
for n in (100..1000 -> by(100)) {
say ("F(%4d) =%7d" % (n, farey_count(n)))
} |
http://rosettacode.org/wiki/Fibonacci_n-step_number_sequences | Fibonacci n-step number sequences | These number series are an expansion of the ordinary Fibonacci sequence where:
For
n
=
2
{\displaystyle n=2}
we have the Fibonacci sequence; with initial values
[
1
,
1
]
{\displaystyle [1,1]}
and
F
k
2
=
F
k
−
1
2
+
F
k
−
2
2
{\displaystyle F_{k}^{2}=F_{k-1}^{2}+F_{k-2}^{2}}
For
n
=
3
{\displaystyle n=3}
we have the tribonacci sequence; with initial values
[
1
,
1
,
2
]
{\displaystyle [1,1,2]}
and
F
k
3
=
F
k
−
1
3
+
F
k
−
2
3
+
F
k
−
3
3
{\displaystyle F_{k}^{3}=F_{k-1}^{3}+F_{k-2}^{3}+F_{k-3}^{3}}
For
n
=
4
{\displaystyle n=4}
we have the tetranacci sequence; with initial values
[
1
,
1
,
2
,
4
]
{\displaystyle [1,1,2,4]}
and
F
k
4
=
F
k
−
1
4
+
F
k
−
2
4
+
F
k
−
3
4
+
F
k
−
4
4
{\displaystyle F_{k}^{4}=F_{k-1}^{4}+F_{k-2}^{4}+F_{k-3}^{4}+F_{k-4}^{4}}
...
For general
n
>
2
{\displaystyle n>2}
we have the Fibonacci
n
{\displaystyle n}
-step sequence -
F
k
n
{\displaystyle F_{k}^{n}}
; with initial values of the first
n
{\displaystyle n}
values of the
(
n
−
1
)
{\displaystyle (n-1)}
'th Fibonacci
n
{\displaystyle n}
-step sequence
F
k
n
−
1
{\displaystyle F_{k}^{n-1}}
; and
k
{\displaystyle k}
'th value of this
n
{\displaystyle n}
'th sequence being
F
k
n
=
∑
i
=
1
(
n
)
F
k
−
i
(
n
)
{\displaystyle F_{k}^{n}=\sum _{i=1}^{(n)}{F_{k-i}^{(n)}}}
For small values of
n
{\displaystyle n}
, Greek numeric prefixes are sometimes used to individually name each series.
Fibonacci
n
{\displaystyle n}
-step sequences
n
{\displaystyle n}
Series name
Values
2
fibonacci
1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 ...
3
tribonacci
1 1 2 4 7 13 24 44 81 149 274 504 927 1705 3136 ...
4
tetranacci
1 1 2 4 8 15 29 56 108 208 401 773 1490 2872 5536 ...
5
pentanacci
1 1 2 4 8 16 31 61 120 236 464 912 1793 3525 6930 ...
6
hexanacci
1 1 2 4 8 16 32 63 125 248 492 976 1936 3840 7617 ...
7
heptanacci
1 1 2 4 8 16 32 64 127 253 504 1004 2000 3984 7936 ...
8
octonacci
1 1 2 4 8 16 32 64 128 255 509 1016 2028 4048 8080 ...
9
nonanacci
1 1 2 4 8 16 32 64 128 256 511 1021 2040 4076 8144 ...
10
decanacci
1 1 2 4 8 16 32 64 128 256 512 1023 2045 4088 8172 ...
Allied sequences can be generated where the initial values are changed:
The Lucas series sums the two preceding values like the fibonacci series for
n
=
2
{\displaystyle n=2}
but uses
[
2
,
1
]
{\displaystyle [2,1]}
as its initial values.
Task
Write a function to generate Fibonacci
n
{\displaystyle n}
-step number sequences given its initial values and assuming the number of initial values determines how many previous values are summed to make the next number of the series.
Use this to print and show here at least the first ten members of the Fibo/tribo/tetra-nacci and Lucas sequences.
Related tasks
Fibonacci sequence
Wolfram Mathworld
Hofstadter Q sequence
Leonardo numbers
Also see
Lucas Numbers - Numberphile (Video)
Tribonacci Numbers (and the Rauzy Fractal) - Numberphile (Video)
Wikipedia, Lucas number
MathWorld, Fibonacci Number
Some identities for r-Fibonacci numbers
OEIS Fibonacci numbers
OEIS Lucas numbers
| #jq | jq | # Input: the initial array
def nacci(arity; len):
arity as $arity | len as $len
| reduce range(length; $len) as $i
(.;
([0, (length - $arity)] | max ) as $lower
| . + [ .[ ($lower) : length] | add] ) ;
def fib(arity; len):
arity as $arity | len as $len
| [1,1] | nacci($arity; $arity) | nacci($arity; $len) ;
def lucas(arity; len):
arity as $arity | len as $len
| [2,1] | nacci($arity; $arity) | nacci($arity; $len) ; |
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.
| #JavaFX_Script | JavaFX Script | def array = [1..100];
def evens = array[n | n mod 2 == 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
| #KL1 | KL1 |
:- module main.
main :-
nats(100, Nats),
fizzbuzz(Nats, Output),
display(Output).
nats(Max, Out) :-
nats(Max, 1, Out).
nats(Max, Count, Out) :- Count =< Max |
Out = [Count|NewOut],
NewCount := Count + 1,
nats(Max, NewCount, NewOut).
nats(Max, Count, Out) :- Count > Max |
Out = [].
fizzbuzz([N|Rest], Out) :- N mod 3 =:= 0, N mod 5 =:= 0 |
Out = ['FizzBuzz' | NewOut],
fizzbuzz(Rest, NewOut).
fizzbuzz([], Out) :-
Out = [].
alternatively.
fizzbuzz([N|Rest], Out) :- N mod 3 =:= 0 |
Out = ['Fizz' | NewOut],
fizzbuzz(Rest, NewOut).
fizzbuzz([N|Rest], Out) :- N mod 5 =:= 0 |
Out = ['Buzz' | NewOut],
fizzbuzz(Rest, NewOut).
alternatively.
fizzbuzz([N|Rest], Out) :-
Out = [N | NewOut],
fizzbuzz(Rest, NewOut).
display([Message|Rest]) :-
io:outstream([print(Message), nl]),
display(Rest).
display([]).
|
http://rosettacode.org/wiki/File_input/output | File input/output | File input/output is part of Short Circuit's Console Program Basics selection.
Task
Create a file called "output.txt", and place in it the contents of the file "input.txt", via an intermediate variable.
In other words, your program will demonstrate:
how to read from a file into a variable
how to write a variable's contents into a file
Oneliners that skip the intermediate variable are of secondary interest — operating systems have copy commands for that.
| #PowerShell | PowerShell | Get-Content $PWD\input.txt | Out-File $PWD\output.txt |
http://rosettacode.org/wiki/File_input/output | File input/output | File input/output is part of Short Circuit's Console Program Basics selection.
Task
Create a file called "output.txt", and place in it the contents of the file "input.txt", via an intermediate variable.
In other words, your program will demonstrate:
how to read from a file into a variable
how to write a variable's contents into a file
Oneliners that skip the intermediate variable are of secondary interest — operating systems have copy commands for that.
| #PureBasic | PureBasic | CopyFile("input.txt","output.txt") |
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
| #ArnoldC | ArnoldC | IT'S SHOWTIME
HEY CHRISTMAS TREE f1
YOU SET US UP @I LIED
TALK TO THE HAND f1
HEY CHRISTMAS TREE f2
YOU SET US UP @NO PROBLEMO
HEY CHRISTMAS TREE f3
YOU SET US UP @I LIED
STICK AROUND @NO PROBLEMO
GET TO THE CHOPPER f3
HERE IS MY INVITATION f1
GET UP f2
ENOUGH TALK
TALK TO THE HAND f3
GET TO THE CHOPPER f1
HERE IS MY INVITATION f2
ENOUGH TALK
GET TO THE CHOPPER f2
HERE IS MY INVITATION f3
ENOUGH TALK
CHILL
YOU HAVE BEEN TERMINATED |
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
| #C | C | #include <stdio.h>
#include <stdlib.h>
typedef struct {
int *list;
short count;
} Factors;
void xferFactors( Factors *fctrs, int *flist, int flix )
{
int ix, ij;
int newSize = fctrs->count + flix;
if (newSize > flix) {
fctrs->list = realloc( fctrs->list, newSize * sizeof(int));
}
else {
fctrs->list = malloc( newSize * sizeof(int));
}
for (ij=0,ix=fctrs->count; ix<newSize; ij++,ix++) {
fctrs->list[ix] = flist[ij];
}
fctrs->count = newSize;
}
Factors *factor( int num, Factors *fctrs)
{
int flist[301], flix;
int dvsr;
flix = 0;
fctrs->count = 0;
free(fctrs->list);
fctrs->list = NULL;
for (dvsr=1; dvsr*dvsr < num; dvsr++) {
if (num % dvsr != 0) continue;
if ( flix == 300) {
xferFactors( fctrs, flist, flix );
flix = 0;
}
flist[flix++] = dvsr;
flist[flix++] = num/dvsr;
}
if (dvsr*dvsr == num)
flist[flix++] = dvsr;
if (flix > 0)
xferFactors( fctrs, flist, flix );
return fctrs;
}
int main(int argc, char*argv[])
{
int nums2factor[] = { 2059, 223092870, 3135, 45 };
Factors ftors = { NULL, 0};
char sep;
int i,j;
for (i=0; i<4; i++) {
factor( nums2factor[i], &ftors );
printf("\nfactors of %d are:\n ", nums2factor[i]);
sep = ' ';
for (j=0; j<ftors.count; j++) {
printf("%c %d", sep, ftors.list[j]);
sep = ',';
}
printf("\n");
}
return 0;
} |
http://rosettacode.org/wiki/Fast_Fourier_transform | Fast Fourier transform | Task
Calculate the FFT (Fast Fourier Transform) of an input sequence.
The most general case allows for complex numbers at the input
and results in a sequence of equal length, again of complex numbers.
If you need to restrict yourself to real numbers, the output should
be the magnitude (i.e.: sqrt(re2 + im2)) of the complex result.
The classic version is the recursive Cooley–Tukey FFT. Wikipedia has pseudo-code for that.
Further optimizations are possible but not required.
| #Lua | Lua | -- operations on complex number
complex = {__mt={} }
function complex.new (r, i)
local new={r=r, i=i or 0}
setmetatable(new,complex.__mt)
return new
end
function complex.__mt.__add (c1, c2)
return complex.new(c1.r + c2.r, c1.i + c2.i)
end
function complex.__mt.__sub (c1, c2)
return complex.new(c1.r - c2.r, c1.i - c2.i)
end
function complex.__mt.__mul (c1, c2)
return complex.new(c1.r*c2.r - c1.i*c2.i,
c1.r*c2.i + c1.i*c2.r)
end
function complex.expi (i)
return complex.new(math.cos(i),math.sin(i))
end
function complex.__mt.__tostring(c)
return "("..c.r..","..c.i..")"
end
-- Cooley–Tukey FFT (in-place, divide-and-conquer)
-- Higher memory requirements and redundancy although more intuitive
function fft(vect)
local n=#vect
if n<=1 then return vect end
-- divide
local odd,even={},{}
for i=1,n,2 do
odd[#odd+1]=vect[i]
even[#even+1]=vect[i+1]
end
-- conquer
fft(even);
fft(odd);
-- combine
for k=1,n/2 do
local t=even[k] * complex.expi(-2*math.pi*(k-1)/n)
vect[k] = odd[k] + t;
vect[k+n/2] = odd[k] - t;
end
return vect
end
function toComplex(vectr)
vect={}
for i,r in ipairs(vectr) do
vect[i]=complex.new(r)
end
return vect
end
-- test
data = toComplex{1, 1, 1, 1, 0, 0, 0, 0};
-- this works for old lua versions & luaJIT (depends on version!)
-- print("orig:", unpack(data))
-- print("fft:", unpack(fft(data)))
-- Beginning with Lua 5.2 you have to write
print("orig:", table.unpack(data))
print("fft:", table.unpack(fft(data))) |
http://rosettacode.org/wiki/Factors_of_a_Mersenne_number | Factors of a Mersenne number | A Mersenne number is a number in the form of 2P-1.
If P is prime, the Mersenne number may be a Mersenne prime
(if P is not prime, the Mersenne number is also not prime).
In the search for Mersenne prime numbers it is advantageous to eliminate exponents by finding a small factor before starting a, potentially lengthy, Lucas-Lehmer test.
There are very efficient algorithms for determining if a number divides 2P-1 (or equivalently, if 2P mod (the number) = 1).
Some languages already have built-in implementations of this exponent-and-mod operation (called modPow or similar).
The following is how to implement this modPow yourself:
For example, let's compute 223 mod 47.
Convert the exponent 23 to binary, you get 10111. Starting with square = 1, repeatedly square it.
Remove the top bit of the exponent, and if it's 1 multiply square by the base of the exponentiation (2), then compute square modulo 47.
Use the result of the modulo from the last step as the initial value of square in the next step:
remove optional
square top bit multiply by 2 mod 47
──────────── ─────── ───────────── ──────
1*1 = 1 1 0111 1*2 = 2 2
2*2 = 4 0 111 no 4
4*4 = 16 1 11 16*2 = 32 32
32*32 = 1024 1 1 1024*2 = 2048 27
27*27 = 729 1 729*2 = 1458 1
Since 223 mod 47 = 1, 47 is a factor of 2P-1.
(To see this, subtract 1 from both sides: 223-1 = 0 mod 47.)
Since we've shown that 47 is a factor, 223-1 is not prime.
Further properties of Mersenne numbers allow us to refine the process even more.
Any factor q of 2P-1 must be of the form 2kP+1, k being a positive integer or zero. Furthermore, q must be 1 or 7 mod 8.
Finally any potential factor q must be prime.
As in other trial division algorithms, the algorithm stops when 2kP+1 > sqrt(N).
These primality tests only work on Mersenne numbers where P is prime. For example, M4=15 yields no factors using these techniques, but factors into 3 and 5, neither of which fit 2kP+1.
Task
Using the above method find a factor of 2929-1 (aka M929)
Related tasks
count in factors
prime decomposition
factors of an integer
Sieve of Eratosthenes
primality by trial division
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
See also
Computers in 1948: 2127 - 1
(Note: This video is no longer available because the YouTube account associated with this video has been terminated.)
| #Phix | Phix | with javascript_semantics
function modpow(atom x, atom n, atom m)
atom i = n,
y = 1,
z = x
while i do
if and_bits(i,1) then
y = mod(y*z,m)
end if
z = mod(z*z,m)
i = floor(i/2)
end while
return y
end function
function mersenne_factor(integer p)
if not is_prime(p) then return -1 end if
atom limit = sqrt(power(2,p))-1
integer k = 1
while 1 do
atom q = 2*p*k + 1
if q>=limit then exit end if
if find(mod(q,8),{1,7})
and is_prime(q)
and modpow(2,p,q)=1 then
return q
end if
k += 1
end while
return 0
end function
sequence tests = {11, 23, 29, 37, 41, 43, 47, 53, 59, 67, 71, 73, 79, 83, 97, 929, 937}
for i=1 to length(tests) do
integer ti = tests[i]
printf(1,"A factor of M%d is %d\n",{ti,mersenne_factor(ti)})
end for
|
http://rosettacode.org/wiki/Farey_sequence | Farey sequence | The Farey sequence Fn of order n is the sequence of completely reduced fractions between 0 and 1 which, when in lowest terms, have denominators less than or equal to n, arranged in order of increasing size.
The Farey sequence is sometimes incorrectly called a Farey series.
Each Farey sequence:
starts with the value 0 (zero), denoted by the fraction
0
1
{\displaystyle {\frac {0}{1}}}
ends with the value 1 (unity), denoted by the fraction
1
1
{\displaystyle {\frac {1}{1}}}
.
The Farey sequences of orders 1 to 5 are:
F
1
=
0
1
,
1
1
{\displaystyle {\bf {\it {F}}}_{1}={\frac {0}{1}},{\frac {1}{1}}}
F
2
=
0
1
,
1
2
,
1
1
{\displaystyle {\bf {\it {F}}}_{2}={\frac {0}{1}},{\frac {1}{2}},{\frac {1}{1}}}
F
3
=
0
1
,
1
3
,
1
2
,
2
3
,
1
1
{\displaystyle {\bf {\it {F}}}_{3}={\frac {0}{1}},{\frac {1}{3}},{\frac {1}{2}},{\frac {2}{3}},{\frac {1}{1}}}
F
4
=
0
1
,
1
4
,
1
3
,
1
2
,
2
3
,
3
4
,
1
1
{\displaystyle {\bf {\it {F}}}_{4}={\frac {0}{1}},{\frac {1}{4}},{\frac {1}{3}},{\frac {1}{2}},{\frac {2}{3}},{\frac {3}{4}},{\frac {1}{1}}}
F
5
=
0
1
,
1
5
,
1
4
,
1
3
,
2
5
,
1
2
,
3
5
,
2
3
,
3
4
,
4
5
,
1
1
{\displaystyle {\bf {\it {F}}}_{5}={\frac {0}{1}},{\frac {1}{5}},{\frac {1}{4}},{\frac {1}{3}},{\frac {2}{5}},{\frac {1}{2}},{\frac {3}{5}},{\frac {2}{3}},{\frac {3}{4}},{\frac {4}{5}},{\frac {1}{1}}}
Task
Compute and show the Farey sequence for orders 1 through 11 (inclusive).
Compute and display the number of fractions in the Farey sequence for order 100 through 1,000 (inclusive) by hundreds.
Show the fractions as n/d (using the solidus [or slash] to separate the numerator from the denominator).
The length (the number of fractions) of a Farey sequence asymptotically approaches:
3 × n2 ÷
π
{\displaystyle \pi }
2
See also
OEIS sequence A006842 numerators of Farey series of order 1, 2, ···
OEIS sequence A006843 denominators of Farey series of order 1, 2, ···
OEIS sequence A005728 number of fractions in Farey series of order n
MathWorld entry Farey sequence
Wikipedia entry Farey sequence
| #Stata | Stata | mata
function totient(n_) {
n = n_
if (n<4) {
if (n<1) return(.)
else if (n>1) return(n-1)
else return(1)
}
else {
r = 1
if (mod(n,2)==0) {
n = floor(n/2)
while (mod(n,2)==0) {
n = floor(n/2)
r = r*2
}
}
for (k=3; k*k<=n; k=k+2) {
if (mod(n,k)==0) {
r = r*(k-1)
n = floor(n/k)
while (mod(n,k)==0) {
n = floor(n/k)
r = r*k
}
}
}
if (n>1) r = r*(n-1)
return(r)
}
}
function map(f,a) {
n = rows(a)
p = cols(a)
b = J(n,p,.)
for (i=1; i<=n; i++) {
for (j=1; j<=p; j++) {
b[i,j] = (*f)(a[i,j])
}
}
return(b)
}
function farey_length(n) {
return(1+sum(map(&totient(),1::n)))
}
function farey(n) {
m = 1+sum(map(&totient(),1::n))
r = J(m,2,.)
r[1,.] = 0,1
a = 0
b = 1
c = 1
d = n
i = 1
while (c<=n) {
k = floor((n+b)/d)
a = k*c-a
b = k*d-b
swap(a,c)
swap(b,d)
r[++i,.] = a,b
}
return(r)
}
for (n=1; n<=11; n++) {
a = farey(n)
m = rows(a)
for (i=1; i<=m; i++) printf("%f/%f ",a[i,1],a[i,2])
printf("\n")
}
map(&farey_length(),100*(1..10))
end |
http://rosettacode.org/wiki/Farey_sequence | Farey sequence | The Farey sequence Fn of order n is the sequence of completely reduced fractions between 0 and 1 which, when in lowest terms, have denominators less than or equal to n, arranged in order of increasing size.
The Farey sequence is sometimes incorrectly called a Farey series.
Each Farey sequence:
starts with the value 0 (zero), denoted by the fraction
0
1
{\displaystyle {\frac {0}{1}}}
ends with the value 1 (unity), denoted by the fraction
1
1
{\displaystyle {\frac {1}{1}}}
.
The Farey sequences of orders 1 to 5 are:
F
1
=
0
1
,
1
1
{\displaystyle {\bf {\it {F}}}_{1}={\frac {0}{1}},{\frac {1}{1}}}
F
2
=
0
1
,
1
2
,
1
1
{\displaystyle {\bf {\it {F}}}_{2}={\frac {0}{1}},{\frac {1}{2}},{\frac {1}{1}}}
F
3
=
0
1
,
1
3
,
1
2
,
2
3
,
1
1
{\displaystyle {\bf {\it {F}}}_{3}={\frac {0}{1}},{\frac {1}{3}},{\frac {1}{2}},{\frac {2}{3}},{\frac {1}{1}}}
F
4
=
0
1
,
1
4
,
1
3
,
1
2
,
2
3
,
3
4
,
1
1
{\displaystyle {\bf {\it {F}}}_{4}={\frac {0}{1}},{\frac {1}{4}},{\frac {1}{3}},{\frac {1}{2}},{\frac {2}{3}},{\frac {3}{4}},{\frac {1}{1}}}
F
5
=
0
1
,
1
5
,
1
4
,
1
3
,
2
5
,
1
2
,
3
5
,
2
3
,
3
4
,
4
5
,
1
1
{\displaystyle {\bf {\it {F}}}_{5}={\frac {0}{1}},{\frac {1}{5}},{\frac {1}{4}},{\frac {1}{3}},{\frac {2}{5}},{\frac {1}{2}},{\frac {3}{5}},{\frac {2}{3}},{\frac {3}{4}},{\frac {4}{5}},{\frac {1}{1}}}
Task
Compute and show the Farey sequence for orders 1 through 11 (inclusive).
Compute and display the number of fractions in the Farey sequence for order 100 through 1,000 (inclusive) by hundreds.
Show the fractions as n/d (using the solidus [or slash] to separate the numerator from the denominator).
The length (the number of fractions) of a Farey sequence asymptotically approaches:
3 × n2 ÷
π
{\displaystyle \pi }
2
See also
OEIS sequence A006842 numerators of Farey series of order 1, 2, ···
OEIS sequence A006843 denominators of Farey series of order 1, 2, ···
OEIS sequence A005728 number of fractions in Farey series of order n
MathWorld entry Farey sequence
Wikipedia entry Farey sequence
| #Swift | Swift | class Farey {
let n: Int
init(_ x: Int) {
n = x
}
//using algorithm from wikipedia
var sequence: [(Int,Int)] {
var a = 0
var b = 1
var c = 1
var d = n
var results = [(a, b)]
while c <= n {
let k = (n + b) / d
let oldA = a
let oldB = b
a = c
b = d
c = k * c - oldA
d = k * d - oldB
results += [(a, b)]
}
return results
}
var formattedSequence: String {
var s = "\(n):"
for pair in sequence {
s += " \(pair.0)/\(pair.1)"
}
return s
}
}
print("Sequences\n")
for n in 1...11 {
print(Farey(n).formattedSequence)
}
print("\nSequence Lengths\n")
for n in 1...10 {
let m = n * 100
print("\(m): \(Farey(m).sequence.count)")
} |
http://rosettacode.org/wiki/Fibonacci_n-step_number_sequences | Fibonacci n-step number sequences | These number series are an expansion of the ordinary Fibonacci sequence where:
For
n
=
2
{\displaystyle n=2}
we have the Fibonacci sequence; with initial values
[
1
,
1
]
{\displaystyle [1,1]}
and
F
k
2
=
F
k
−
1
2
+
F
k
−
2
2
{\displaystyle F_{k}^{2}=F_{k-1}^{2}+F_{k-2}^{2}}
For
n
=
3
{\displaystyle n=3}
we have the tribonacci sequence; with initial values
[
1
,
1
,
2
]
{\displaystyle [1,1,2]}
and
F
k
3
=
F
k
−
1
3
+
F
k
−
2
3
+
F
k
−
3
3
{\displaystyle F_{k}^{3}=F_{k-1}^{3}+F_{k-2}^{3}+F_{k-3}^{3}}
For
n
=
4
{\displaystyle n=4}
we have the tetranacci sequence; with initial values
[
1
,
1
,
2
,
4
]
{\displaystyle [1,1,2,4]}
and
F
k
4
=
F
k
−
1
4
+
F
k
−
2
4
+
F
k
−
3
4
+
F
k
−
4
4
{\displaystyle F_{k}^{4}=F_{k-1}^{4}+F_{k-2}^{4}+F_{k-3}^{4}+F_{k-4}^{4}}
...
For general
n
>
2
{\displaystyle n>2}
we have the Fibonacci
n
{\displaystyle n}
-step sequence -
F
k
n
{\displaystyle F_{k}^{n}}
; with initial values of the first
n
{\displaystyle n}
values of the
(
n
−
1
)
{\displaystyle (n-1)}
'th Fibonacci
n
{\displaystyle n}
-step sequence
F
k
n
−
1
{\displaystyle F_{k}^{n-1}}
; and
k
{\displaystyle k}
'th value of this
n
{\displaystyle n}
'th sequence being
F
k
n
=
∑
i
=
1
(
n
)
F
k
−
i
(
n
)
{\displaystyle F_{k}^{n}=\sum _{i=1}^{(n)}{F_{k-i}^{(n)}}}
For small values of
n
{\displaystyle n}
, Greek numeric prefixes are sometimes used to individually name each series.
Fibonacci
n
{\displaystyle n}
-step sequences
n
{\displaystyle n}
Series name
Values
2
fibonacci
1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 ...
3
tribonacci
1 1 2 4 7 13 24 44 81 149 274 504 927 1705 3136 ...
4
tetranacci
1 1 2 4 8 15 29 56 108 208 401 773 1490 2872 5536 ...
5
pentanacci
1 1 2 4 8 16 31 61 120 236 464 912 1793 3525 6930 ...
6
hexanacci
1 1 2 4 8 16 32 63 125 248 492 976 1936 3840 7617 ...
7
heptanacci
1 1 2 4 8 16 32 64 127 253 504 1004 2000 3984 7936 ...
8
octonacci
1 1 2 4 8 16 32 64 128 255 509 1016 2028 4048 8080 ...
9
nonanacci
1 1 2 4 8 16 32 64 128 256 511 1021 2040 4076 8144 ...
10
decanacci
1 1 2 4 8 16 32 64 128 256 512 1023 2045 4088 8172 ...
Allied sequences can be generated where the initial values are changed:
The Lucas series sums the two preceding values like the fibonacci series for
n
=
2
{\displaystyle n=2}
but uses
[
2
,
1
]
{\displaystyle [2,1]}
as its initial values.
Task
Write a function to generate Fibonacci
n
{\displaystyle n}
-step number sequences given its initial values and assuming the number of initial values determines how many previous values are summed to make the next number of the series.
Use this to print and show here at least the first ten members of the Fibo/tribo/tetra-nacci and Lucas sequences.
Related tasks
Fibonacci sequence
Wolfram Mathworld
Hofstadter Q sequence
Leonardo numbers
Also see
Lucas Numbers - Numberphile (Video)
Tribonacci Numbers (and the Rauzy Fractal) - Numberphile (Video)
Wikipedia, Lucas number
MathWorld, Fibonacci Number
Some identities for r-Fibonacci numbers
OEIS Fibonacci numbers
OEIS Lucas numbers
| #Julia | Julia |
type NFib{T<:Integer}
n::T
klim::T
seeder::Function
end
type FState
a::Array{BigInt,1}
adex::Integer
k::Integer
end
function Base.start{T<:Integer}(nf::NFib{T})
a = nf.seeder(nf.n)
adex = 1
k = 1
return FState(a, adex, k)
end
function Base.done{T<:Integer}(nf::NFib{T}, fs::FState)
fs.k > nf.klim
end
function Base.next{T<:Integer}(nf::NFib{T}, fs::FState)
f = sum(fs.a)
fs.a[fs.adex] = f
fs.adex = rem1(fs.adex+1, nf.n)
fs.k += 1
return (f, fs)
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.
| #JavaScript | JavaScript | var arr = [1,2,3,4,5];
var evens = arr.filter(function(a) {return a % 2 == 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
| #Klong | Klong |
{:[0=x!15;:FizzBuzz:|0=x!5;:Buzz:|0=x!3;:Fizz;x]}'1+!100
|
http://rosettacode.org/wiki/File_input/output | File input/output | File input/output is part of Short Circuit's Console Program Basics selection.
Task
Create a file called "output.txt", and place in it the contents of the file "input.txt", via an intermediate variable.
In other words, your program will demonstrate:
how to read from a file into a variable
how to write a variable's contents into a file
Oneliners that skip the intermediate variable are of secondary interest — operating systems have copy commands for that.
| #Python | Python | import shutil
shutil.copyfile('input.txt', 'output.txt') |
http://rosettacode.org/wiki/File_input/output | File input/output | File input/output is part of Short Circuit's Console Program Basics selection.
Task
Create a file called "output.txt", and place in it the contents of the file "input.txt", via an intermediate variable.
In other words, your program will demonstrate:
how to read from a file into a variable
how to write a variable's contents into a file
Oneliners that skip the intermediate variable are of secondary interest — operating systems have copy commands for that.
| #Quackery | Quackery | $ "input.txt" sharefile drop
temp put
temp share
$ "output.txt" putfile drop
|
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
| #Arturo | Arturo | fib: $[x][
if? x<2 [1]
else [(fib x-1) + (fib x-2)]
]
loop 1..25 [x][
print ["Fibonacci of" x "=" fib x]
] |
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
| #C.23 | C# | static void Main (string[] args) {
do {
Console.WriteLine ("Number:");
Int64 p = 0;
do {
try {
p = Convert.ToInt64 (Console.ReadLine ());
break;
} catch (Exception) { }
} while (true);
Console.WriteLine ("For 1 through " + ((int) Math.Sqrt (p)).ToString () + "");
for (int x = 1; x <= (int) Math.Sqrt (p); x++) {
if (p % x == 0)
Console.WriteLine ("Found: " + x.ToString () + ". " + p.ToString () + " / " + x.ToString () + " = " + (p / x).ToString ());
}
Console.WriteLine ("Done.");
} while (true);
} |
http://rosettacode.org/wiki/Fast_Fourier_transform | Fast Fourier transform | Task
Calculate the FFT (Fast Fourier Transform) of an input sequence.
The most general case allows for complex numbers at the input
and results in a sequence of equal length, again of complex numbers.
If you need to restrict yourself to real numbers, the output should
be the magnitude (i.e.: sqrt(re2 + im2)) of the complex result.
The classic version is the recursive Cooley–Tukey FFT. Wikipedia has pseudo-code for that.
Further optimizations are possible but not required.
| #Maple | Maple |
with( DiscreteTransforms ):
FourierTransform( <1,1,1,1,0,0,0,0>, normalization=none );
|
http://rosettacode.org/wiki/Factors_of_a_Mersenne_number | Factors of a Mersenne number | A Mersenne number is a number in the form of 2P-1.
If P is prime, the Mersenne number may be a Mersenne prime
(if P is not prime, the Mersenne number is also not prime).
In the search for Mersenne prime numbers it is advantageous to eliminate exponents by finding a small factor before starting a, potentially lengthy, Lucas-Lehmer test.
There are very efficient algorithms for determining if a number divides 2P-1 (or equivalently, if 2P mod (the number) = 1).
Some languages already have built-in implementations of this exponent-and-mod operation (called modPow or similar).
The following is how to implement this modPow yourself:
For example, let's compute 223 mod 47.
Convert the exponent 23 to binary, you get 10111. Starting with square = 1, repeatedly square it.
Remove the top bit of the exponent, and if it's 1 multiply square by the base of the exponentiation (2), then compute square modulo 47.
Use the result of the modulo from the last step as the initial value of square in the next step:
remove optional
square top bit multiply by 2 mod 47
──────────── ─────── ───────────── ──────
1*1 = 1 1 0111 1*2 = 2 2
2*2 = 4 0 111 no 4
4*4 = 16 1 11 16*2 = 32 32
32*32 = 1024 1 1 1024*2 = 2048 27
27*27 = 729 1 729*2 = 1458 1
Since 223 mod 47 = 1, 47 is a factor of 2P-1.
(To see this, subtract 1 from both sides: 223-1 = 0 mod 47.)
Since we've shown that 47 is a factor, 223-1 is not prime.
Further properties of Mersenne numbers allow us to refine the process even more.
Any factor q of 2P-1 must be of the form 2kP+1, k being a positive integer or zero. Furthermore, q must be 1 or 7 mod 8.
Finally any potential factor q must be prime.
As in other trial division algorithms, the algorithm stops when 2kP+1 > sqrt(N).
These primality tests only work on Mersenne numbers where P is prime. For example, M4=15 yields no factors using these techniques, but factors into 3 and 5, neither of which fit 2kP+1.
Task
Using the above method find a factor of 2929-1 (aka M929)
Related tasks
count in factors
prime decomposition
factors of an integer
Sieve of Eratosthenes
primality by trial division
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
See also
Computers in 1948: 2127 - 1
(Note: This video is no longer available because the YouTube account associated with this video has been terminated.)
| #PHP | PHP | echo 'M929 has a factor: ', mersenneFactor(929), '</br>';
function mersenneFactor($p) {
$limit = sqrt(pow(2, $p) - 1);
for ($k = 1; 2 * $p * $k - 1 < $limit; $k++) {
$q = 2 * $p * $k + 1;
if (isPrime($q) && ($q % 8 == 1 || $q % 8 == 7) && bcpowmod("2", "$p", "$q") == "1") {
return $q;
}
}
return 0;
}
function isPrime($n) {
if ($n < 2 || $n % 2 == 0) return $n == 2;
for ($i = 3; $i * $i <= $n; $i += 2) {
if ($n % $i == 0) {
return false;
}
}
return true;
} |
http://rosettacode.org/wiki/Farey_sequence | Farey sequence | The Farey sequence Fn of order n is the sequence of completely reduced fractions between 0 and 1 which, when in lowest terms, have denominators less than or equal to n, arranged in order of increasing size.
The Farey sequence is sometimes incorrectly called a Farey series.
Each Farey sequence:
starts with the value 0 (zero), denoted by the fraction
0
1
{\displaystyle {\frac {0}{1}}}
ends with the value 1 (unity), denoted by the fraction
1
1
{\displaystyle {\frac {1}{1}}}
.
The Farey sequences of orders 1 to 5 are:
F
1
=
0
1
,
1
1
{\displaystyle {\bf {\it {F}}}_{1}={\frac {0}{1}},{\frac {1}{1}}}
F
2
=
0
1
,
1
2
,
1
1
{\displaystyle {\bf {\it {F}}}_{2}={\frac {0}{1}},{\frac {1}{2}},{\frac {1}{1}}}
F
3
=
0
1
,
1
3
,
1
2
,
2
3
,
1
1
{\displaystyle {\bf {\it {F}}}_{3}={\frac {0}{1}},{\frac {1}{3}},{\frac {1}{2}},{\frac {2}{3}},{\frac {1}{1}}}
F
4
=
0
1
,
1
4
,
1
3
,
1
2
,
2
3
,
3
4
,
1
1
{\displaystyle {\bf {\it {F}}}_{4}={\frac {0}{1}},{\frac {1}{4}},{\frac {1}{3}},{\frac {1}{2}},{\frac {2}{3}},{\frac {3}{4}},{\frac {1}{1}}}
F
5
=
0
1
,
1
5
,
1
4
,
1
3
,
2
5
,
1
2
,
3
5
,
2
3
,
3
4
,
4
5
,
1
1
{\displaystyle {\bf {\it {F}}}_{5}={\frac {0}{1}},{\frac {1}{5}},{\frac {1}{4}},{\frac {1}{3}},{\frac {2}{5}},{\frac {1}{2}},{\frac {3}{5}},{\frac {2}{3}},{\frac {3}{4}},{\frac {4}{5}},{\frac {1}{1}}}
Task
Compute and show the Farey sequence for orders 1 through 11 (inclusive).
Compute and display the number of fractions in the Farey sequence for order 100 through 1,000 (inclusive) by hundreds.
Show the fractions as n/d (using the solidus [or slash] to separate the numerator from the denominator).
The length (the number of fractions) of a Farey sequence asymptotically approaches:
3 × n2 ÷
π
{\displaystyle \pi }
2
See also
OEIS sequence A006842 numerators of Farey series of order 1, 2, ···
OEIS sequence A006843 denominators of Farey series of order 1, 2, ···
OEIS sequence A005728 number of fractions in Farey series of order n
MathWorld entry Farey sequence
Wikipedia entry Farey sequence
| #Tcl | Tcl | package require Tcl 8.6
proc farey {n} {
set nums [lrepeat [expr {$n+1}] 1]
set result {{0 1}}
for {set found 1} {$found} {} {
set nj [lindex $nums [set j 1]]
for {set found 0;set i 1} {$i <= $n} {incr i} {
if {[lindex $nums $i]*$j < $nj*$i} {
set nj [lindex $nums [set j $i]]
set found 1
}
}
lappend result [list $nj $j]
for {set i $j} {$i <= $n} {incr i $j} {
lset nums $i [expr {[lindex $nums $i] + 1}]
}
}
return $result
}
for {set i 1} {$i <= 11} {incr i} {
puts F($i):\x20[lmap n [farey $i] {join $n /}]
}
for {set i 100} {$i <= 1000} {incr i 100} {
puts |F($i)|\x20=\x20[llength [farey $i]]
} |
http://rosettacode.org/wiki/Fibonacci_n-step_number_sequences | Fibonacci n-step number sequences | These number series are an expansion of the ordinary Fibonacci sequence where:
For
n
=
2
{\displaystyle n=2}
we have the Fibonacci sequence; with initial values
[
1
,
1
]
{\displaystyle [1,1]}
and
F
k
2
=
F
k
−
1
2
+
F
k
−
2
2
{\displaystyle F_{k}^{2}=F_{k-1}^{2}+F_{k-2}^{2}}
For
n
=
3
{\displaystyle n=3}
we have the tribonacci sequence; with initial values
[
1
,
1
,
2
]
{\displaystyle [1,1,2]}
and
F
k
3
=
F
k
−
1
3
+
F
k
−
2
3
+
F
k
−
3
3
{\displaystyle F_{k}^{3}=F_{k-1}^{3}+F_{k-2}^{3}+F_{k-3}^{3}}
For
n
=
4
{\displaystyle n=4}
we have the tetranacci sequence; with initial values
[
1
,
1
,
2
,
4
]
{\displaystyle [1,1,2,4]}
and
F
k
4
=
F
k
−
1
4
+
F
k
−
2
4
+
F
k
−
3
4
+
F
k
−
4
4
{\displaystyle F_{k}^{4}=F_{k-1}^{4}+F_{k-2}^{4}+F_{k-3}^{4}+F_{k-4}^{4}}
...
For general
n
>
2
{\displaystyle n>2}
we have the Fibonacci
n
{\displaystyle n}
-step sequence -
F
k
n
{\displaystyle F_{k}^{n}}
; with initial values of the first
n
{\displaystyle n}
values of the
(
n
−
1
)
{\displaystyle (n-1)}
'th Fibonacci
n
{\displaystyle n}
-step sequence
F
k
n
−
1
{\displaystyle F_{k}^{n-1}}
; and
k
{\displaystyle k}
'th value of this
n
{\displaystyle n}
'th sequence being
F
k
n
=
∑
i
=
1
(
n
)
F
k
−
i
(
n
)
{\displaystyle F_{k}^{n}=\sum _{i=1}^{(n)}{F_{k-i}^{(n)}}}
For small values of
n
{\displaystyle n}
, Greek numeric prefixes are sometimes used to individually name each series.
Fibonacci
n
{\displaystyle n}
-step sequences
n
{\displaystyle n}
Series name
Values
2
fibonacci
1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 ...
3
tribonacci
1 1 2 4 7 13 24 44 81 149 274 504 927 1705 3136 ...
4
tetranacci
1 1 2 4 8 15 29 56 108 208 401 773 1490 2872 5536 ...
5
pentanacci
1 1 2 4 8 16 31 61 120 236 464 912 1793 3525 6930 ...
6
hexanacci
1 1 2 4 8 16 32 63 125 248 492 976 1936 3840 7617 ...
7
heptanacci
1 1 2 4 8 16 32 64 127 253 504 1004 2000 3984 7936 ...
8
octonacci
1 1 2 4 8 16 32 64 128 255 509 1016 2028 4048 8080 ...
9
nonanacci
1 1 2 4 8 16 32 64 128 256 511 1021 2040 4076 8144 ...
10
decanacci
1 1 2 4 8 16 32 64 128 256 512 1023 2045 4088 8172 ...
Allied sequences can be generated where the initial values are changed:
The Lucas series sums the two preceding values like the fibonacci series for
n
=
2
{\displaystyle n=2}
but uses
[
2
,
1
]
{\displaystyle [2,1]}
as its initial values.
Task
Write a function to generate Fibonacci
n
{\displaystyle n}
-step number sequences given its initial values and assuming the number of initial values determines how many previous values are summed to make the next number of the series.
Use this to print and show here at least the first ten members of the Fibo/tribo/tetra-nacci and Lucas sequences.
Related tasks
Fibonacci sequence
Wolfram Mathworld
Hofstadter Q sequence
Leonardo numbers
Also see
Lucas Numbers - Numberphile (Video)
Tribonacci Numbers (and the Rauzy Fractal) - Numberphile (Video)
Wikipedia, Lucas number
MathWorld, Fibonacci Number
Some identities for r-Fibonacci numbers
OEIS Fibonacci numbers
OEIS Lucas numbers
| #Kotlin | Kotlin | // version 1.1.2
fun fibN(initial: IntArray, numTerms: Int) : IntArray {
val n = initial.size
require(n >= 2 && numTerms >= 0)
val fibs = initial.copyOf(numTerms)
if (numTerms <= n) return fibs
for (i in n until numTerms) {
var sum = 0
for (j in i - n until i) sum += fibs[j]
fibs[i] = sum
}
return fibs
}
fun main(args: Array<String>) {
val names = arrayOf("fibonacci", "tribonacci", "tetranacci", "pentanacci", "hexanacci",
"heptanacci", "octonacci", "nonanacci", "decanacci")
val initial = intArrayOf(1, 1, 2, 4, 8, 16, 32, 64, 128, 256)
println(" n name values")
var values = fibN(intArrayOf(2, 1), 15).joinToString(", ")
println("%2d %-10s %s".format(2, "lucas", values))
for (i in 0..8) {
values = fibN(initial.sliceArray(0 until i + 2), 15).joinToString(", ")
println("%2d %-10s %s".format(i + 2, names[i], values))
}
} |
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.
| #jq | jq | (1,2,3,4,5,6,7,8,9) | select(. % 2 == 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
| #Kotlin | Kotlin | fun fizzBuzz() {
for (number in 1..100) {
println(
when {
number % 15 == 0 -> "FizzBuzz"
number % 3 == 0 -> "Fizz"
number % 5 == 0 -> "Buzz"
else -> number
}
)
}
} |
http://rosettacode.org/wiki/File_input/output | File input/output | File input/output is part of Short Circuit's Console Program Basics selection.
Task
Create a file called "output.txt", and place in it the contents of the file "input.txt", via an intermediate variable.
In other words, your program will demonstrate:
how to read from a file into a variable
how to write a variable's contents into a file
Oneliners that skip the intermediate variable are of secondary interest — operating systems have copy commands for that.
| #R | R | src <- file("input.txt", "r")
dest <- file("output.txt", "w")
fc <- readLines(src, -1)
writeLines(fc, dest)
close(src); close(dest) |
http://rosettacode.org/wiki/File_input/output | File input/output | File input/output is part of Short Circuit's Console Program Basics selection.
Task
Create a file called "output.txt", and place in it the contents of the file "input.txt", via an intermediate variable.
In other words, your program will demonstrate:
how to read from a file into a variable
how to write a variable's contents into a file
Oneliners that skip the intermediate variable are of secondary interest — operating systems have copy commands for that.
| #Racket | Racket | #lang racket
(define file-content
(with-input-from-file "input.txt"
(lambda ()
(let loop ((lst null))
(define new (read-char))
(if (eof-object? new)
(apply string lst)
(loop (append lst (list new))))))))
(with-output-to-file "output.txt"
(lambda ()
(write file-content))) |
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
| #AsciiDots | AsciiDots |
/--#$--\
| |
>-*>{+}/
| \+-/
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
| #C.2B.2B | C++ | #include <iostream>
#include <iomanip>
#include <vector>
#include <algorithm>
#include <iterator>
std::vector<int> GenerateFactors(int n) {
std::vector<int> factors = { 1, n };
for (int i = 2; i * i <= n; ++i) {
if (n % i == 0) {
factors.push_back(i);
if (i * i != n)
factors.push_back(n / i);
}
}
std::sort(factors.begin(), factors.end());
return factors;
}
int main() {
const int SampleNumbers[] = { 3135, 45, 60, 81 };
for (size_t i = 0; i < sizeof(SampleNumbers) / sizeof(int); ++i) {
std::vector<int> factors = GenerateFactors(SampleNumbers[i]);
std::cout << "Factors of ";
std::cout.width(4);
std::cout << SampleNumbers[i] << " are: ";
std::copy(factors.begin(), factors.end(), std::ostream_iterator<int>(std::cout, " "));
std::cout << std::endl;
}
return EXIT_SUCCESS;
} |
http://rosettacode.org/wiki/Fast_Fourier_transform | Fast Fourier transform | Task
Calculate the FFT (Fast Fourier Transform) of an input sequence.
The most general case allows for complex numbers at the input
and results in a sequence of equal length, again of complex numbers.
If you need to restrict yourself to real numbers, the output should
be the magnitude (i.e.: sqrt(re2 + im2)) of the complex result.
The classic version is the recursive Cooley–Tukey FFT. Wikipedia has pseudo-code for that.
Further optimizations are possible but not required.
| #Mathematica_.2F_Wolfram_Language | Mathematica / Wolfram Language |
Fourier[{1,1,1,1,0,0,0,0}, FourierParameters->{1,-1}]
|
http://rosettacode.org/wiki/Factors_of_a_Mersenne_number | Factors of a Mersenne number | A Mersenne number is a number in the form of 2P-1.
If P is prime, the Mersenne number may be a Mersenne prime
(if P is not prime, the Mersenne number is also not prime).
In the search for Mersenne prime numbers it is advantageous to eliminate exponents by finding a small factor before starting a, potentially lengthy, Lucas-Lehmer test.
There are very efficient algorithms for determining if a number divides 2P-1 (or equivalently, if 2P mod (the number) = 1).
Some languages already have built-in implementations of this exponent-and-mod operation (called modPow or similar).
The following is how to implement this modPow yourself:
For example, let's compute 223 mod 47.
Convert the exponent 23 to binary, you get 10111. Starting with square = 1, repeatedly square it.
Remove the top bit of the exponent, and if it's 1 multiply square by the base of the exponentiation (2), then compute square modulo 47.
Use the result of the modulo from the last step as the initial value of square in the next step:
remove optional
square top bit multiply by 2 mod 47
──────────── ─────── ───────────── ──────
1*1 = 1 1 0111 1*2 = 2 2
2*2 = 4 0 111 no 4
4*4 = 16 1 11 16*2 = 32 32
32*32 = 1024 1 1 1024*2 = 2048 27
27*27 = 729 1 729*2 = 1458 1
Since 223 mod 47 = 1, 47 is a factor of 2P-1.
(To see this, subtract 1 from both sides: 223-1 = 0 mod 47.)
Since we've shown that 47 is a factor, 223-1 is not prime.
Further properties of Mersenne numbers allow us to refine the process even more.
Any factor q of 2P-1 must be of the form 2kP+1, k being a positive integer or zero. Furthermore, q must be 1 or 7 mod 8.
Finally any potential factor q must be prime.
As in other trial division algorithms, the algorithm stops when 2kP+1 > sqrt(N).
These primality tests only work on Mersenne numbers where P is prime. For example, M4=15 yields no factors using these techniques, but factors into 3 and 5, neither of which fit 2kP+1.
Task
Using the above method find a factor of 2929-1 (aka M929)
Related tasks
count in factors
prime decomposition
factors of an integer
Sieve of Eratosthenes
primality by trial division
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
See also
Computers in 1948: 2127 - 1
(Note: This video is no longer available because the YouTube account associated with this video has been terminated.)
| #PicoLisp | PicoLisp | (de **Mod (X Y N)
(let M 1
(loop
(when (bit? 1 Y)
(setq M (% (* M X) N)) )
(T (=0 (setq Y (>> 1 Y)))
M )
(setq X (% (* X X) N)) ) ) )
(de prime? (N)
(or
(= N 2)
(and
(> N 1)
(bit? 1 N)
(let S (sqrt N)
(for (D 3 T (+ D 2))
(T (> D S) T)
(T (=0 (% N D)) NIL) ) ) ) ) )
(de mFactor (P)
(let (Lim (sqrt (dec (** 2 P))) K 0 Q)
(loop
(setq Q (inc (* 2 (inc 'K) P)))
(T (>= Q Lim) NIL)
(T
(and
(member (% Q 8) (1 7))
(prime? Q)
(= 1 (**Mod 2 P Q)) )
Q ) ) ) ) |
http://rosettacode.org/wiki/Factors_of_a_Mersenne_number | Factors of a Mersenne number | A Mersenne number is a number in the form of 2P-1.
If P is prime, the Mersenne number may be a Mersenne prime
(if P is not prime, the Mersenne number is also not prime).
In the search for Mersenne prime numbers it is advantageous to eliminate exponents by finding a small factor before starting a, potentially lengthy, Lucas-Lehmer test.
There are very efficient algorithms for determining if a number divides 2P-1 (or equivalently, if 2P mod (the number) = 1).
Some languages already have built-in implementations of this exponent-and-mod operation (called modPow or similar).
The following is how to implement this modPow yourself:
For example, let's compute 223 mod 47.
Convert the exponent 23 to binary, you get 10111. Starting with square = 1, repeatedly square it.
Remove the top bit of the exponent, and if it's 1 multiply square by the base of the exponentiation (2), then compute square modulo 47.
Use the result of the modulo from the last step as the initial value of square in the next step:
remove optional
square top bit multiply by 2 mod 47
──────────── ─────── ───────────── ──────
1*1 = 1 1 0111 1*2 = 2 2
2*2 = 4 0 111 no 4
4*4 = 16 1 11 16*2 = 32 32
32*32 = 1024 1 1 1024*2 = 2048 27
27*27 = 729 1 729*2 = 1458 1
Since 223 mod 47 = 1, 47 is a factor of 2P-1.
(To see this, subtract 1 from both sides: 223-1 = 0 mod 47.)
Since we've shown that 47 is a factor, 223-1 is not prime.
Further properties of Mersenne numbers allow us to refine the process even more.
Any factor q of 2P-1 must be of the form 2kP+1, k being a positive integer or zero. Furthermore, q must be 1 or 7 mod 8.
Finally any potential factor q must be prime.
As in other trial division algorithms, the algorithm stops when 2kP+1 > sqrt(N).
These primality tests only work on Mersenne numbers where P is prime. For example, M4=15 yields no factors using these techniques, but factors into 3 and 5, neither of which fit 2kP+1.
Task
Using the above method find a factor of 2929-1 (aka M929)
Related tasks
count in factors
prime decomposition
factors of an integer
Sieve of Eratosthenes
primality by trial division
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
See also
Computers in 1948: 2127 - 1
(Note: This video is no longer available because the YouTube account associated with this video has been terminated.)
| #Prolog | Prolog |
mersenne_factor(P, F) :-
prime(P),
once((
between(1, 100_000, K), % Fail if we can't find a small factor
Q is 2*K*P + 1,
test_factor(Q, P, F))).
test_factor(Q, P, prime) :- Q*Q > (1 << P - 1), !.
test_factor(Q, P, Q) :-
R is Q /\ 7, member(R, [1, 7]),
prime(Q),
powm(2, P, Q) =:= 1.
wheel235(L) :-
W = [4, 2, 4, 2, 4, 6, 2, 6 | W],
L = [1, 2, 2 | W].
prime(N) :-
N >= 2,
wheel235(W),
prime(N, 2, W).
prime(N, D, _) :- D*D > N, !.
prime(N, D, [A|As]) :-
N mod D =\= 0,
D2 is D + A, prime(N, D2, As).
|
http://rosettacode.org/wiki/Farey_sequence | Farey sequence | The Farey sequence Fn of order n is the sequence of completely reduced fractions between 0 and 1 which, when in lowest terms, have denominators less than or equal to n, arranged in order of increasing size.
The Farey sequence is sometimes incorrectly called a Farey series.
Each Farey sequence:
starts with the value 0 (zero), denoted by the fraction
0
1
{\displaystyle {\frac {0}{1}}}
ends with the value 1 (unity), denoted by the fraction
1
1
{\displaystyle {\frac {1}{1}}}
.
The Farey sequences of orders 1 to 5 are:
F
1
=
0
1
,
1
1
{\displaystyle {\bf {\it {F}}}_{1}={\frac {0}{1}},{\frac {1}{1}}}
F
2
=
0
1
,
1
2
,
1
1
{\displaystyle {\bf {\it {F}}}_{2}={\frac {0}{1}},{\frac {1}{2}},{\frac {1}{1}}}
F
3
=
0
1
,
1
3
,
1
2
,
2
3
,
1
1
{\displaystyle {\bf {\it {F}}}_{3}={\frac {0}{1}},{\frac {1}{3}},{\frac {1}{2}},{\frac {2}{3}},{\frac {1}{1}}}
F
4
=
0
1
,
1
4
,
1
3
,
1
2
,
2
3
,
3
4
,
1
1
{\displaystyle {\bf {\it {F}}}_{4}={\frac {0}{1}},{\frac {1}{4}},{\frac {1}{3}},{\frac {1}{2}},{\frac {2}{3}},{\frac {3}{4}},{\frac {1}{1}}}
F
5
=
0
1
,
1
5
,
1
4
,
1
3
,
2
5
,
1
2
,
3
5
,
2
3
,
3
4
,
4
5
,
1
1
{\displaystyle {\bf {\it {F}}}_{5}={\frac {0}{1}},{\frac {1}{5}},{\frac {1}{4}},{\frac {1}{3}},{\frac {2}{5}},{\frac {1}{2}},{\frac {3}{5}},{\frac {2}{3}},{\frac {3}{4}},{\frac {4}{5}},{\frac {1}{1}}}
Task
Compute and show the Farey sequence for orders 1 through 11 (inclusive).
Compute and display the number of fractions in the Farey sequence for order 100 through 1,000 (inclusive) by hundreds.
Show the fractions as n/d (using the solidus [or slash] to separate the numerator from the denominator).
The length (the number of fractions) of a Farey sequence asymptotically approaches:
3 × n2 ÷
π
{\displaystyle \pi }
2
See also
OEIS sequence A006842 numerators of Farey series of order 1, 2, ···
OEIS sequence A006843 denominators of Farey series of order 1, 2, ···
OEIS sequence A005728 number of fractions in Farey series of order n
MathWorld entry Farey sequence
Wikipedia entry Farey sequence
| #Vala | Vala | struct Fraction {
public uint d;
public uint n;
}
void farey(uint n) {
Fraction f1 = {0, 1};
Fraction f2 = {1, n};
print("0/1 1/%u ", n);
while (f2.n > 1) {
var k = (n + f1.n) / f2.n;
var aux = f1;
f1 = f2;
f2 = {f2.d * k - aux.d, f2.n * k - aux.n};
print("%u/%u ", f2.d, f2.n);
}
print("\n");
}
uint fareyLength(uint n, uint[] cache) {
if (n >= cache.length) {
uint newLen = cache.length;
if (newLen == 0)
newLen = 16;
while (newLen <= n)
newLen *= 2;
cache.resize((int)newLen);
}
else if (cache[n] != 0)
return cache[n];
uint length = n * (n + 3) / 2;
for (uint p = 2, q = 2; p <= n; p = q) {
q = n / (n / p) + 1;
length -= fareyLength(n / p, cache) * (q - p);
}
cache[n] = length;
return length;
}
void main() {
for (uint n = 1; n < 12; n++)
{
print("%8u: ", n);
farey(n);
}
uint[] cache = new uint[0];
for (uint n = 100; n <= 1000; n += 100)
print("%8u: %14u items\n", n, fareyLength(n, cache));
} |
http://rosettacode.org/wiki/Fibonacci_n-step_number_sequences | Fibonacci n-step number sequences | These number series are an expansion of the ordinary Fibonacci sequence where:
For
n
=
2
{\displaystyle n=2}
we have the Fibonacci sequence; with initial values
[
1
,
1
]
{\displaystyle [1,1]}
and
F
k
2
=
F
k
−
1
2
+
F
k
−
2
2
{\displaystyle F_{k}^{2}=F_{k-1}^{2}+F_{k-2}^{2}}
For
n
=
3
{\displaystyle n=3}
we have the tribonacci sequence; with initial values
[
1
,
1
,
2
]
{\displaystyle [1,1,2]}
and
F
k
3
=
F
k
−
1
3
+
F
k
−
2
3
+
F
k
−
3
3
{\displaystyle F_{k}^{3}=F_{k-1}^{3}+F_{k-2}^{3}+F_{k-3}^{3}}
For
n
=
4
{\displaystyle n=4}
we have the tetranacci sequence; with initial values
[
1
,
1
,
2
,
4
]
{\displaystyle [1,1,2,4]}
and
F
k
4
=
F
k
−
1
4
+
F
k
−
2
4
+
F
k
−
3
4
+
F
k
−
4
4
{\displaystyle F_{k}^{4}=F_{k-1}^{4}+F_{k-2}^{4}+F_{k-3}^{4}+F_{k-4}^{4}}
...
For general
n
>
2
{\displaystyle n>2}
we have the Fibonacci
n
{\displaystyle n}
-step sequence -
F
k
n
{\displaystyle F_{k}^{n}}
; with initial values of the first
n
{\displaystyle n}
values of the
(
n
−
1
)
{\displaystyle (n-1)}
'th Fibonacci
n
{\displaystyle n}
-step sequence
F
k
n
−
1
{\displaystyle F_{k}^{n-1}}
; and
k
{\displaystyle k}
'th value of this
n
{\displaystyle n}
'th sequence being
F
k
n
=
∑
i
=
1
(
n
)
F
k
−
i
(
n
)
{\displaystyle F_{k}^{n}=\sum _{i=1}^{(n)}{F_{k-i}^{(n)}}}
For small values of
n
{\displaystyle n}
, Greek numeric prefixes are sometimes used to individually name each series.
Fibonacci
n
{\displaystyle n}
-step sequences
n
{\displaystyle n}
Series name
Values
2
fibonacci
1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 ...
3
tribonacci
1 1 2 4 7 13 24 44 81 149 274 504 927 1705 3136 ...
4
tetranacci
1 1 2 4 8 15 29 56 108 208 401 773 1490 2872 5536 ...
5
pentanacci
1 1 2 4 8 16 31 61 120 236 464 912 1793 3525 6930 ...
6
hexanacci
1 1 2 4 8 16 32 63 125 248 492 976 1936 3840 7617 ...
7
heptanacci
1 1 2 4 8 16 32 64 127 253 504 1004 2000 3984 7936 ...
8
octonacci
1 1 2 4 8 16 32 64 128 255 509 1016 2028 4048 8080 ...
9
nonanacci
1 1 2 4 8 16 32 64 128 256 511 1021 2040 4076 8144 ...
10
decanacci
1 1 2 4 8 16 32 64 128 256 512 1023 2045 4088 8172 ...
Allied sequences can be generated where the initial values are changed:
The Lucas series sums the two preceding values like the fibonacci series for
n
=
2
{\displaystyle n=2}
but uses
[
2
,
1
]
{\displaystyle [2,1]}
as its initial values.
Task
Write a function to generate Fibonacci
n
{\displaystyle n}
-step number sequences given its initial values and assuming the number of initial values determines how many previous values are summed to make the next number of the series.
Use this to print and show here at least the first ten members of the Fibo/tribo/tetra-nacci and Lucas sequences.
Related tasks
Fibonacci sequence
Wolfram Mathworld
Hofstadter Q sequence
Leonardo numbers
Also see
Lucas Numbers - Numberphile (Video)
Tribonacci Numbers (and the Rauzy Fractal) - Numberphile (Video)
Wikipedia, Lucas number
MathWorld, Fibonacci Number
Some identities for r-Fibonacci numbers
OEIS Fibonacci numbers
OEIS Lucas numbers
| #Lua | Lua | function nStepFibs (seq, limit)
local iMax, sum = #seq - 1
while #seq < limit do
sum = 0
for i = 0, iMax do sum = sum + seq[#seq - i] end
table.insert(seq, sum)
end
return seq
end
local fibSeqs = {
{name = "Fibonacci", values = {1, 1} },
{name = "Tribonacci", values = {1, 1, 2} },
{name = "Tetranacci", values = {1, 1, 2, 4}},
{name = "Lucas", values = {2, 1} }
}
for _, sequence in pairs(fibSeqs) do
io.write(sequence.name .. ": ")
print(table.concat(nStepFibs(sequence.values, 10), " "))
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.
| #Julia | Julia | @show filter(iseven, 1:10) |
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
| #KQL | KQL |
range i from 1 to 100 step 1
| project Result =
case(
i % 15 == 0, "FizzBuzz",
i % 3 == 0, "Fizz",
i % 5 == 0, "Buzz",
tostring(i)
)
|
http://rosettacode.org/wiki/File_input/output | File input/output | File input/output is part of Short Circuit's Console Program Basics selection.
Task
Create a file called "output.txt", and place in it the contents of the file "input.txt", via an intermediate variable.
In other words, your program will demonstrate:
how to read from a file into a variable
how to write a variable's contents into a file
Oneliners that skip the intermediate variable are of secondary interest — operating systems have copy commands for that.
| #Raku | Raku | spurt "output.txt", slurp "input.txt"; |
http://rosettacode.org/wiki/File_input/output | File input/output | File input/output is part of Short Circuit's Console Program Basics selection.
Task
Create a file called "output.txt", and place in it the contents of the file "input.txt", via an intermediate variable.
In other words, your program will demonstrate:
how to read from a file into a variable
how to write a variable's contents into a file
Oneliners that skip the intermediate variable are of secondary interest — operating systems have copy commands for that.
| #RapidQ | RapidQ | $INCLUDE "rapidq.inc"
DIM File1 AS QFileStream
DIM File2 AS QFileStream
File1.Open("input.txt", fmOpenRead)
File2.Open("output.txt", fmCreate)
WHILE NOT File1.EOF
data$ = File1.ReadLine
File2.WriteLine(data$)
WEND
File1.Close
File2.Close |
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
| #ATS | ATS |
fun fib_rec(n: int): int =
if n >= 2 then fib_rec(n-1) + fib_rec(n-2) else n
|
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
| #Ceylon | Ceylon | shared void run() {
{Integer*} getFactors(Integer n) =>
(1..n).filter((Integer element) => element.divides(n));
for(Integer i in 1..100) {
print("the factors of ``i`` are ``getFactors(i)``");
}
} |
http://rosettacode.org/wiki/Fast_Fourier_transform | Fast Fourier transform | Task
Calculate the FFT (Fast Fourier Transform) of an input sequence.
The most general case allows for complex numbers at the input
and results in a sequence of equal length, again of complex numbers.
If you need to restrict yourself to real numbers, the output should
be the magnitude (i.e.: sqrt(re2 + im2)) of the complex result.
The classic version is the recursive Cooley–Tukey FFT. Wikipedia has pseudo-code for that.
Further optimizations are possible but not required.
| #MATLAB_.2F_Octave | MATLAB / Octave | fft([1,1,1,1,0,0,0,0]')
|
http://rosettacode.org/wiki/Factors_of_a_Mersenne_number | Factors of a Mersenne number | A Mersenne number is a number in the form of 2P-1.
If P is prime, the Mersenne number may be a Mersenne prime
(if P is not prime, the Mersenne number is also not prime).
In the search for Mersenne prime numbers it is advantageous to eliminate exponents by finding a small factor before starting a, potentially lengthy, Lucas-Lehmer test.
There are very efficient algorithms for determining if a number divides 2P-1 (or equivalently, if 2P mod (the number) = 1).
Some languages already have built-in implementations of this exponent-and-mod operation (called modPow or similar).
The following is how to implement this modPow yourself:
For example, let's compute 223 mod 47.
Convert the exponent 23 to binary, you get 10111. Starting with square = 1, repeatedly square it.
Remove the top bit of the exponent, and if it's 1 multiply square by the base of the exponentiation (2), then compute square modulo 47.
Use the result of the modulo from the last step as the initial value of square in the next step:
remove optional
square top bit multiply by 2 mod 47
──────────── ─────── ───────────── ──────
1*1 = 1 1 0111 1*2 = 2 2
2*2 = 4 0 111 no 4
4*4 = 16 1 11 16*2 = 32 32
32*32 = 1024 1 1 1024*2 = 2048 27
27*27 = 729 1 729*2 = 1458 1
Since 223 mod 47 = 1, 47 is a factor of 2P-1.
(To see this, subtract 1 from both sides: 223-1 = 0 mod 47.)
Since we've shown that 47 is a factor, 223-1 is not prime.
Further properties of Mersenne numbers allow us to refine the process even more.
Any factor q of 2P-1 must be of the form 2kP+1, k being a positive integer or zero. Furthermore, q must be 1 or 7 mod 8.
Finally any potential factor q must be prime.
As in other trial division algorithms, the algorithm stops when 2kP+1 > sqrt(N).
These primality tests only work on Mersenne numbers where P is prime. For example, M4=15 yields no factors using these techniques, but factors into 3 and 5, neither of which fit 2kP+1.
Task
Using the above method find a factor of 2929-1 (aka M929)
Related tasks
count in factors
prime decomposition
factors of an integer
Sieve of Eratosthenes
primality by trial division
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
See also
Computers in 1948: 2127 - 1
(Note: This video is no longer available because the YouTube account associated with this video has been terminated.)
| #Python | Python | def is_prime(number):
return True # code omitted - see Primality by Trial Division
def m_factor(p):
max_k = 16384 / p # arbitrary limit; since Python automatically uses long's, it doesn't overflow
for k in xrange(max_k):
q = 2*p*k + 1
if not is_prime(q):
continue
elif q % 8 != 1 and q % 8 != 7:
continue
elif pow(2, p, q) == 1:
return q
return None
if __name__ == '__main__':
exponent = int(raw_input("Enter exponent of Mersenne number: "))
if not is_prime(exponent):
print "Exponent is not prime: %d" % exponent
else:
factor = m_factor(exponent)
if not factor:
print "No factor found for M%d" % exponent
else:
print "M%d has a factor: %d" % (exponent, factor) |
http://rosettacode.org/wiki/Factors_of_a_Mersenne_number | Factors of a Mersenne number | A Mersenne number is a number in the form of 2P-1.
If P is prime, the Mersenne number may be a Mersenne prime
(if P is not prime, the Mersenne number is also not prime).
In the search for Mersenne prime numbers it is advantageous to eliminate exponents by finding a small factor before starting a, potentially lengthy, Lucas-Lehmer test.
There are very efficient algorithms for determining if a number divides 2P-1 (or equivalently, if 2P mod (the number) = 1).
Some languages already have built-in implementations of this exponent-and-mod operation (called modPow or similar).
The following is how to implement this modPow yourself:
For example, let's compute 223 mod 47.
Convert the exponent 23 to binary, you get 10111. Starting with square = 1, repeatedly square it.
Remove the top bit of the exponent, and if it's 1 multiply square by the base of the exponentiation (2), then compute square modulo 47.
Use the result of the modulo from the last step as the initial value of square in the next step:
remove optional
square top bit multiply by 2 mod 47
──────────── ─────── ───────────── ──────
1*1 = 1 1 0111 1*2 = 2 2
2*2 = 4 0 111 no 4
4*4 = 16 1 11 16*2 = 32 32
32*32 = 1024 1 1 1024*2 = 2048 27
27*27 = 729 1 729*2 = 1458 1
Since 223 mod 47 = 1, 47 is a factor of 2P-1.
(To see this, subtract 1 from both sides: 223-1 = 0 mod 47.)
Since we've shown that 47 is a factor, 223-1 is not prime.
Further properties of Mersenne numbers allow us to refine the process even more.
Any factor q of 2P-1 must be of the form 2kP+1, k being a positive integer or zero. Furthermore, q must be 1 or 7 mod 8.
Finally any potential factor q must be prime.
As in other trial division algorithms, the algorithm stops when 2kP+1 > sqrt(N).
These primality tests only work on Mersenne numbers where P is prime. For example, M4=15 yields no factors using these techniques, but factors into 3 and 5, neither of which fit 2kP+1.
Task
Using the above method find a factor of 2929-1 (aka M929)
Related tasks
count in factors
prime decomposition
factors of an integer
Sieve of Eratosthenes
primality by trial division
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
See also
Computers in 1948: 2127 - 1
(Note: This video is no longer available because the YouTube account associated with this video has been terminated.)
| #Racket | Racket |
#lang racket
(define (number->digits n)
(map (compose1 string->number string)
(string->list (number->string n 2))))
(define (modpow exp base)
(for/fold ([square 1])
([d (number->digits exp)])
(modulo (* (if (= d 1) 2 1) square square) base)))
; Search through all integers from 1 on to find the first divisor.
; Returns #f if 2^p-1 is prime.
(define (mersenne-factor p)
(for/first ([i (in-range 1 (floor (expt 2 (quotient p 2))) (* 2 p))]
#:when (and (member (modulo i 8) '(1 7))
(= 1 (modpow p i))))
i))
(mersenne-factor 929)
|
http://rosettacode.org/wiki/Farey_sequence | Farey sequence | The Farey sequence Fn of order n is the sequence of completely reduced fractions between 0 and 1 which, when in lowest terms, have denominators less than or equal to n, arranged in order of increasing size.
The Farey sequence is sometimes incorrectly called a Farey series.
Each Farey sequence:
starts with the value 0 (zero), denoted by the fraction
0
1
{\displaystyle {\frac {0}{1}}}
ends with the value 1 (unity), denoted by the fraction
1
1
{\displaystyle {\frac {1}{1}}}
.
The Farey sequences of orders 1 to 5 are:
F
1
=
0
1
,
1
1
{\displaystyle {\bf {\it {F}}}_{1}={\frac {0}{1}},{\frac {1}{1}}}
F
2
=
0
1
,
1
2
,
1
1
{\displaystyle {\bf {\it {F}}}_{2}={\frac {0}{1}},{\frac {1}{2}},{\frac {1}{1}}}
F
3
=
0
1
,
1
3
,
1
2
,
2
3
,
1
1
{\displaystyle {\bf {\it {F}}}_{3}={\frac {0}{1}},{\frac {1}{3}},{\frac {1}{2}},{\frac {2}{3}},{\frac {1}{1}}}
F
4
=
0
1
,
1
4
,
1
3
,
1
2
,
2
3
,
3
4
,
1
1
{\displaystyle {\bf {\it {F}}}_{4}={\frac {0}{1}},{\frac {1}{4}},{\frac {1}{3}},{\frac {1}{2}},{\frac {2}{3}},{\frac {3}{4}},{\frac {1}{1}}}
F
5
=
0
1
,
1
5
,
1
4
,
1
3
,
2
5
,
1
2
,
3
5
,
2
3
,
3
4
,
4
5
,
1
1
{\displaystyle {\bf {\it {F}}}_{5}={\frac {0}{1}},{\frac {1}{5}},{\frac {1}{4}},{\frac {1}{3}},{\frac {2}{5}},{\frac {1}{2}},{\frac {3}{5}},{\frac {2}{3}},{\frac {3}{4}},{\frac {4}{5}},{\frac {1}{1}}}
Task
Compute and show the Farey sequence for orders 1 through 11 (inclusive).
Compute and display the number of fractions in the Farey sequence for order 100 through 1,000 (inclusive) by hundreds.
Show the fractions as n/d (using the solidus [or slash] to separate the numerator from the denominator).
The length (the number of fractions) of a Farey sequence asymptotically approaches:
3 × n2 ÷
π
{\displaystyle \pi }
2
See also
OEIS sequence A006842 numerators of Farey series of order 1, 2, ···
OEIS sequence A006843 denominators of Farey series of order 1, 2, ···
OEIS sequence A005728 number of fractions in Farey series of order n
MathWorld entry Farey sequence
Wikipedia entry Farey sequence
| #Vlang | Vlang | struct Frac {
num int
den int
}
fn (f Frac) str() string {
return "$f.num/$f.den"
}
fn f(l Frac, r Frac, n int) {
m := Frac{l.num + r.num, l.den + r.den}
if m.den <= n {
f(l, m, n)
print("$m ")
f(m, r, n)
}
}
fn main() {
// task 1. solution by recursive generation of mediants
for n := 1; n <= 11; n++ {
l := Frac{0, 1}
r := Frac{1, 1}
print("F($n): $l ")
f(l, r, n)
println(r)
}
// task 2. direct solution by summing totient fntion
// 2.1 generate primes to 1000
mut composite := [1001]bool{}
for p in [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31] {
for n := p * 2; n <= 1000; n += p {
composite[n] = true
}
}
// 2.2 generate totients to 1000
mut tot := [1001]int{init: 1}
for n := 2; n <= 1000; n++ {
if !composite[n] {
tot[n] = n - 1
for a := n * 2; a <= 1000; a += n {
mut f := n - 1
for r := a / n; r%n == 0; r /= n {
f *= n
}
tot[a] *= f
}
}
}
// 2.3 sum totients
for n, sum := 1, 1; n <= 1000; n++ {
sum += tot[n]
if n%100 == 0 {
println("|F($n)|: $sum")
}
}
} |
http://rosettacode.org/wiki/Fibonacci_n-step_number_sequences | Fibonacci n-step number sequences | These number series are an expansion of the ordinary Fibonacci sequence where:
For
n
=
2
{\displaystyle n=2}
we have the Fibonacci sequence; with initial values
[
1
,
1
]
{\displaystyle [1,1]}
and
F
k
2
=
F
k
−
1
2
+
F
k
−
2
2
{\displaystyle F_{k}^{2}=F_{k-1}^{2}+F_{k-2}^{2}}
For
n
=
3
{\displaystyle n=3}
we have the tribonacci sequence; with initial values
[
1
,
1
,
2
]
{\displaystyle [1,1,2]}
and
F
k
3
=
F
k
−
1
3
+
F
k
−
2
3
+
F
k
−
3
3
{\displaystyle F_{k}^{3}=F_{k-1}^{3}+F_{k-2}^{3}+F_{k-3}^{3}}
For
n
=
4
{\displaystyle n=4}
we have the tetranacci sequence; with initial values
[
1
,
1
,
2
,
4
]
{\displaystyle [1,1,2,4]}
and
F
k
4
=
F
k
−
1
4
+
F
k
−
2
4
+
F
k
−
3
4
+
F
k
−
4
4
{\displaystyle F_{k}^{4}=F_{k-1}^{4}+F_{k-2}^{4}+F_{k-3}^{4}+F_{k-4}^{4}}
...
For general
n
>
2
{\displaystyle n>2}
we have the Fibonacci
n
{\displaystyle n}
-step sequence -
F
k
n
{\displaystyle F_{k}^{n}}
; with initial values of the first
n
{\displaystyle n}
values of the
(
n
−
1
)
{\displaystyle (n-1)}
'th Fibonacci
n
{\displaystyle n}
-step sequence
F
k
n
−
1
{\displaystyle F_{k}^{n-1}}
; and
k
{\displaystyle k}
'th value of this
n
{\displaystyle n}
'th sequence being
F
k
n
=
∑
i
=
1
(
n
)
F
k
−
i
(
n
)
{\displaystyle F_{k}^{n}=\sum _{i=1}^{(n)}{F_{k-i}^{(n)}}}
For small values of
n
{\displaystyle n}
, Greek numeric prefixes are sometimes used to individually name each series.
Fibonacci
n
{\displaystyle n}
-step sequences
n
{\displaystyle n}
Series name
Values
2
fibonacci
1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 ...
3
tribonacci
1 1 2 4 7 13 24 44 81 149 274 504 927 1705 3136 ...
4
tetranacci
1 1 2 4 8 15 29 56 108 208 401 773 1490 2872 5536 ...
5
pentanacci
1 1 2 4 8 16 31 61 120 236 464 912 1793 3525 6930 ...
6
hexanacci
1 1 2 4 8 16 32 63 125 248 492 976 1936 3840 7617 ...
7
heptanacci
1 1 2 4 8 16 32 64 127 253 504 1004 2000 3984 7936 ...
8
octonacci
1 1 2 4 8 16 32 64 128 255 509 1016 2028 4048 8080 ...
9
nonanacci
1 1 2 4 8 16 32 64 128 256 511 1021 2040 4076 8144 ...
10
decanacci
1 1 2 4 8 16 32 64 128 256 512 1023 2045 4088 8172 ...
Allied sequences can be generated where the initial values are changed:
The Lucas series sums the two preceding values like the fibonacci series for
n
=
2
{\displaystyle n=2}
but uses
[
2
,
1
]
{\displaystyle [2,1]}
as its initial values.
Task
Write a function to generate Fibonacci
n
{\displaystyle n}
-step number sequences given its initial values and assuming the number of initial values determines how many previous values are summed to make the next number of the series.
Use this to print and show here at least the first ten members of the Fibo/tribo/tetra-nacci and Lucas sequences.
Related tasks
Fibonacci sequence
Wolfram Mathworld
Hofstadter Q sequence
Leonardo numbers
Also see
Lucas Numbers - Numberphile (Video)
Tribonacci Numbers (and the Rauzy Fractal) - Numberphile (Video)
Wikipedia, Lucas number
MathWorld, Fibonacci Number
Some identities for r-Fibonacci numbers
OEIS Fibonacci numbers
OEIS Lucas numbers
| #Maple | Maple | numSequence := proc(initValues :: Array)
local n, i, values;
n := numelems(initValues);
values := copy(initValues);
for i from (n+1) to 15 do
values(i) := add(values[i-n..i-1]);
end do;
return values;
end proc:
initValues := Array([1]):
for i from 2 to 10 do
initValues(i) := add(initValues):
printf ("nacci(%d): %a\n", i, convert(numSequence(initValues), list));
end do:
printf ("lucas: %a\n", convert(numSequence(Array([2, 1])), list)); |
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.
| #K | K | / even is a boolean function
even:{0=x!2}
even 1 2 3 4 5
0 1 0 1 0
/ filtering the even numbers
a@&even'a:1+!10
2 4 6 8 10
/ as a function
evens:{x@&even'x}
a:10?100
45 5 79 77 44 15 83 88 33 99
evens a
44 88 |
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
| #KSI | KSI |
`plain
[1 100] `for pos : n ~
out = []
n `mod 3 == 0 ? out.# = 'Fizz' ;
n `mod 5 == 0 ? out.# = 'Buzz' ;
(out `or n) #write_ln #
;
|
http://rosettacode.org/wiki/File_input/output | File input/output | File input/output is part of Short Circuit's Console Program Basics selection.
Task
Create a file called "output.txt", and place in it the contents of the file "input.txt", via an intermediate variable.
In other words, your program will demonstrate:
how to read from a file into a variable
how to write a variable's contents into a file
Oneliners that skip the intermediate variable are of secondary interest — operating systems have copy commands for that.
| #Raven | Raven | 'input.txt' read 'output.txt' write |
http://rosettacode.org/wiki/File_input/output | File input/output | File input/output is part of Short Circuit's Console Program Basics selection.
Task
Create a file called "output.txt", and place in it the contents of the file "input.txt", via an intermediate variable.
In other words, your program will demonstrate:
how to read from a file into a variable
how to write a variable's contents into a file
Oneliners that skip the intermediate variable are of secondary interest — operating systems have copy commands for that.
| #REALbasic | REALbasic |
Sub WriteToFile(input As FolderItem, output As FolderItem)
Dim tis As TextInputStream
Dim tos As TextOutputStream
tis = tis.Open(input)
tos = tos.Create(output)
While Not tis.EOF
tos.WriteLine(tis.ReadLine)
Wend
tis.Close
tos.Close
End Sub
|
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
| #AutoHotkey | AutoHotkey | Loop, 5
MsgBox % fib(A_Index)
Return
fib(n)
{
If (n < 2)
Return n
i := last := this := 1
While (i <= n)
{
new := last + this
last := this
this := new
i++
}
Return this
} |
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
| #Chapel | Chapel | iter factors(n) {
for i in 1..floor(sqrt(n)):int {
if n % i == 0 then {
yield i;
yield n / i;
}
}
} |
http://rosettacode.org/wiki/Fast_Fourier_transform | Fast Fourier transform | Task
Calculate the FFT (Fast Fourier Transform) of an input sequence.
The most general case allows for complex numbers at the input
and results in a sequence of equal length, again of complex numbers.
If you need to restrict yourself to real numbers, the output should
be the magnitude (i.e.: sqrt(re2 + im2)) of the complex result.
The classic version is the recursive Cooley–Tukey FFT. Wikipedia has pseudo-code for that.
Further optimizations are possible but not required.
| #Maxima | Maxima | load(fft)$
fft([1, 2, 3, 4]);
[2.5, -0.5 * %i - 0.5, -0.5, 0.5 * %i - 0.5] |
http://rosettacode.org/wiki/Factors_of_a_Mersenne_number | Factors of a Mersenne number | A Mersenne number is a number in the form of 2P-1.
If P is prime, the Mersenne number may be a Mersenne prime
(if P is not prime, the Mersenne number is also not prime).
In the search for Mersenne prime numbers it is advantageous to eliminate exponents by finding a small factor before starting a, potentially lengthy, Lucas-Lehmer test.
There are very efficient algorithms for determining if a number divides 2P-1 (or equivalently, if 2P mod (the number) = 1).
Some languages already have built-in implementations of this exponent-and-mod operation (called modPow or similar).
The following is how to implement this modPow yourself:
For example, let's compute 223 mod 47.
Convert the exponent 23 to binary, you get 10111. Starting with square = 1, repeatedly square it.
Remove the top bit of the exponent, and if it's 1 multiply square by the base of the exponentiation (2), then compute square modulo 47.
Use the result of the modulo from the last step as the initial value of square in the next step:
remove optional
square top bit multiply by 2 mod 47
──────────── ─────── ───────────── ──────
1*1 = 1 1 0111 1*2 = 2 2
2*2 = 4 0 111 no 4
4*4 = 16 1 11 16*2 = 32 32
32*32 = 1024 1 1 1024*2 = 2048 27
27*27 = 729 1 729*2 = 1458 1
Since 223 mod 47 = 1, 47 is a factor of 2P-1.
(To see this, subtract 1 from both sides: 223-1 = 0 mod 47.)
Since we've shown that 47 is a factor, 223-1 is not prime.
Further properties of Mersenne numbers allow us to refine the process even more.
Any factor q of 2P-1 must be of the form 2kP+1, k being a positive integer or zero. Furthermore, q must be 1 or 7 mod 8.
Finally any potential factor q must be prime.
As in other trial division algorithms, the algorithm stops when 2kP+1 > sqrt(N).
These primality tests only work on Mersenne numbers where P is prime. For example, M4=15 yields no factors using these techniques, but factors into 3 and 5, neither of which fit 2kP+1.
Task
Using the above method find a factor of 2929-1 (aka M929)
Related tasks
count in factors
prime decomposition
factors of an integer
Sieve of Eratosthenes
primality by trial division
trial factoring of a Mersenne number
partition an integer X into N primes
sequence of primes by Trial Division
See also
Computers in 1948: 2127 - 1
(Note: This video is no longer available because the YouTube account associated with this video has been terminated.)
| #Raku | Raku | sub mtest($bits, $p) {
my @bits = $bits.base(2).comb;
loop (my $sq = 1; @bits; $sq %= $p) {
$sq ×= $sq;
$sq += $sq if 1 == @bits.shift;
}
$sq == 1;
}
for flat 2 .. 60, 929 -> $m {
next unless is-prime($m);
my $f = 0;
my $x = 2**$m - 1;
my $q;
for 1..* -> $k {
$q = 2 × $k × $m + 1;
next unless $q % 8 == 1|7 or is-prime($q);
last if $q × $q > $x or $f = mtest($m, $q);
}
say $f ?? "M$m = $x\n\t= $q × { $x div $q }"
!! "M$m = $x is prime";
} |
http://rosettacode.org/wiki/Farey_sequence | Farey sequence | The Farey sequence Fn of order n is the sequence of completely reduced fractions between 0 and 1 which, when in lowest terms, have denominators less than or equal to n, arranged in order of increasing size.
The Farey sequence is sometimes incorrectly called a Farey series.
Each Farey sequence:
starts with the value 0 (zero), denoted by the fraction
0
1
{\displaystyle {\frac {0}{1}}}
ends with the value 1 (unity), denoted by the fraction
1
1
{\displaystyle {\frac {1}{1}}}
.
The Farey sequences of orders 1 to 5 are:
F
1
=
0
1
,
1
1
{\displaystyle {\bf {\it {F}}}_{1}={\frac {0}{1}},{\frac {1}{1}}}
F
2
=
0
1
,
1
2
,
1
1
{\displaystyle {\bf {\it {F}}}_{2}={\frac {0}{1}},{\frac {1}{2}},{\frac {1}{1}}}
F
3
=
0
1
,
1
3
,
1
2
,
2
3
,
1
1
{\displaystyle {\bf {\it {F}}}_{3}={\frac {0}{1}},{\frac {1}{3}},{\frac {1}{2}},{\frac {2}{3}},{\frac {1}{1}}}
F
4
=
0
1
,
1
4
,
1
3
,
1
2
,
2
3
,
3
4
,
1
1
{\displaystyle {\bf {\it {F}}}_{4}={\frac {0}{1}},{\frac {1}{4}},{\frac {1}{3}},{\frac {1}{2}},{\frac {2}{3}},{\frac {3}{4}},{\frac {1}{1}}}
F
5
=
0
1
,
1
5
,
1
4
,
1
3
,
2
5
,
1
2
,
3
5
,
2
3
,
3
4
,
4
5
,
1
1
{\displaystyle {\bf {\it {F}}}_{5}={\frac {0}{1}},{\frac {1}{5}},{\frac {1}{4}},{\frac {1}{3}},{\frac {2}{5}},{\frac {1}{2}},{\frac {3}{5}},{\frac {2}{3}},{\frac {3}{4}},{\frac {4}{5}},{\frac {1}{1}}}
Task
Compute and show the Farey sequence for orders 1 through 11 (inclusive).
Compute and display the number of fractions in the Farey sequence for order 100 through 1,000 (inclusive) by hundreds.
Show the fractions as n/d (using the solidus [or slash] to separate the numerator from the denominator).
The length (the number of fractions) of a Farey sequence asymptotically approaches:
3 × n2 ÷
π
{\displaystyle \pi }
2
See also
OEIS sequence A006842 numerators of Farey series of order 1, 2, ···
OEIS sequence A006843 denominators of Farey series of order 1, 2, ···
OEIS sequence A005728 number of fractions in Farey series of order n
MathWorld entry Farey sequence
Wikipedia entry Farey sequence
| #Wren | Wren | import "/math" for Int
import "/trait" for Stepped
import "/fmt" for Fmt
import "/rat" for Rat
var f //recursive
f = Fn.new { |l, r, n|
var m = Rat.new(l.num + r.num, l.den + r.den)
if (m.den <= n) {
f.call(l, m, n)
System.write("%(m) ")
f.call(m, r, n)
}
}
/* Task 1: solution by recursive generation of mediants. */
for (n in 1..11) {
var l = Rat.zero
var r = Rat.one
System.write("F(%(n)): %(l) ")
f.call(l, r, n)
System.print(r)
}
System.print()
/* Task 2: direct solution by summing totient function. */
// generate primes to 1000
var comp = Int.primeSieve(1001, false)
// generate totients to 1000
var tot = List.filled(1001, 1)
for (n in 2..1000) {
if (!comp[n]) {
tot[n] = n - 1
for (a in Stepped.ascend(n*2..1000, n)) {
var f = n - 1
var r = (a/n).floor
while (r%n == 0) {
f = f * n
r = (r/n).floor
}
tot[a] = tot[a] * f
}
}
}
// sum totients
var sum = 1
for (n in 1..1000) {
sum = sum + tot[n]
if (n%100 == 0) System.print("F(%(Fmt.d(4, n))): %(Fmt.dc(7, sum))")
} |
http://rosettacode.org/wiki/Fibonacci_n-step_number_sequences | Fibonacci n-step number sequences | These number series are an expansion of the ordinary Fibonacci sequence where:
For
n
=
2
{\displaystyle n=2}
we have the Fibonacci sequence; with initial values
[
1
,
1
]
{\displaystyle [1,1]}
and
F
k
2
=
F
k
−
1
2
+
F
k
−
2
2
{\displaystyle F_{k}^{2}=F_{k-1}^{2}+F_{k-2}^{2}}
For
n
=
3
{\displaystyle n=3}
we have the tribonacci sequence; with initial values
[
1
,
1
,
2
]
{\displaystyle [1,1,2]}
and
F
k
3
=
F
k
−
1
3
+
F
k
−
2
3
+
F
k
−
3
3
{\displaystyle F_{k}^{3}=F_{k-1}^{3}+F_{k-2}^{3}+F_{k-3}^{3}}
For
n
=
4
{\displaystyle n=4}
we have the tetranacci sequence; with initial values
[
1
,
1
,
2
,
4
]
{\displaystyle [1,1,2,4]}
and
F
k
4
=
F
k
−
1
4
+
F
k
−
2
4
+
F
k
−
3
4
+
F
k
−
4
4
{\displaystyle F_{k}^{4}=F_{k-1}^{4}+F_{k-2}^{4}+F_{k-3}^{4}+F_{k-4}^{4}}
...
For general
n
>
2
{\displaystyle n>2}
we have the Fibonacci
n
{\displaystyle n}
-step sequence -
F
k
n
{\displaystyle F_{k}^{n}}
; with initial values of the first
n
{\displaystyle n}
values of the
(
n
−
1
)
{\displaystyle (n-1)}
'th Fibonacci
n
{\displaystyle n}
-step sequence
F
k
n
−
1
{\displaystyle F_{k}^{n-1}}
; and
k
{\displaystyle k}
'th value of this
n
{\displaystyle n}
'th sequence being
F
k
n
=
∑
i
=
1
(
n
)
F
k
−
i
(
n
)
{\displaystyle F_{k}^{n}=\sum _{i=1}^{(n)}{F_{k-i}^{(n)}}}
For small values of
n
{\displaystyle n}
, Greek numeric prefixes are sometimes used to individually name each series.
Fibonacci
n
{\displaystyle n}
-step sequences
n
{\displaystyle n}
Series name
Values
2
fibonacci
1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 ...
3
tribonacci
1 1 2 4 7 13 24 44 81 149 274 504 927 1705 3136 ...
4
tetranacci
1 1 2 4 8 15 29 56 108 208 401 773 1490 2872 5536 ...
5
pentanacci
1 1 2 4 8 16 31 61 120 236 464 912 1793 3525 6930 ...
6
hexanacci
1 1 2 4 8 16 32 63 125 248 492 976 1936 3840 7617 ...
7
heptanacci
1 1 2 4 8 16 32 64 127 253 504 1004 2000 3984 7936 ...
8
octonacci
1 1 2 4 8 16 32 64 128 255 509 1016 2028 4048 8080 ...
9
nonanacci
1 1 2 4 8 16 32 64 128 256 511 1021 2040 4076 8144 ...
10
decanacci
1 1 2 4 8 16 32 64 128 256 512 1023 2045 4088 8172 ...
Allied sequences can be generated where the initial values are changed:
The Lucas series sums the two preceding values like the fibonacci series for
n
=
2
{\displaystyle n=2}
but uses
[
2
,
1
]
{\displaystyle [2,1]}
as its initial values.
Task
Write a function to generate Fibonacci
n
{\displaystyle n}
-step number sequences given its initial values and assuming the number of initial values determines how many previous values are summed to make the next number of the series.
Use this to print and show here at least the first ten members of the Fibo/tribo/tetra-nacci and Lucas sequences.
Related tasks
Fibonacci sequence
Wolfram Mathworld
Hofstadter Q sequence
Leonardo numbers
Also see
Lucas Numbers - Numberphile (Video)
Tribonacci Numbers (and the Rauzy Fractal) - Numberphile (Video)
Wikipedia, Lucas number
MathWorld, Fibonacci Number
Some identities for r-Fibonacci numbers
OEIS Fibonacci numbers
OEIS Lucas numbers
| #Mathematica_.2F_Wolfram_Language | Mathematica / Wolfram Language |
f2=Function[{l,k},
Module[{n=Length@l,m},
m=SparseArray[{{i_,j_}/;i==1||i==j+1->1},{n,n}];
NestList[m.#&,l,k]]];
Table[Last/@f2[{1,1}~Join~Table[0,{n-2}],15+n][[-18;;]],{n,2,10}]//TableForm
Table[Last/@f2[{1,2}~Join~Table[0,{n-2}],15+n][[-18;;]],{n,2,10}]//TableForm
|
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.
| #Kotlin | Kotlin | // version 1.0.5-2
fun main(args: Array<String>) {
val array = arrayOf(1, 2, 3, 4, 5, 6, 7, 8, 9)
println(array.joinToString(" "))
val filteredArray = array.filter{ it % 2 == 0 }
println(filteredArray.joinToString(" "))
val mutableList = array.toMutableList()
mutableList.retainAll { it % 2 == 0 }
println(mutableList.joinToString(" "))
} |
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
| #LabVIEW | LabVIEW |
1. direct:
{S.map
{lambda {:i}
{if {= {% :i 15} 0}
then fizzbuzz
else {if {= {% :i 3} 0}
then fizz
else {if {= {% :i 5} 0}
then buzz
else :i}}}}
{S.serie 1 100}}
-> 1 2 fizz 4 buzz fizz 7 8 fizz buzz 11 fizz 13 14 fizzbuzz 16 17 fizz 19 buzz fizz 22 23 fizz buzz 26 fizz 28 29 fizzbuzz 31 32 fizz 34 buzz fizz 37 38 fizz buzz 41 fizz 43 44 fizzbuzz 46 47 fizz 49 buzz fizz 52 53 fizz buzz 56 fizz 58 59 fizzbuzz 61 62 fizz 64 buzz fizz 67 68 fizz buzz 71 fizz 73 74 fizzbuzz 76 77 fizz 79 buzz fizz 82 83 fizz buzz 86 fizz 88 89 fizzbuzz 91 92 fizz 94 buzz fizz 97 98 fizz buzz
2. via a function
{def fizzbuzz
{lambda {:i :n}
{if {> :i :n}
then .
else {if {= {% :i 15} 0}
then fizzbuzz
else {if {= {% :i 3} 0}
then fizz
else {if {= {% :i 5} 0}
then buzz
else :i}}} {fizzbuzz {+ :i 1} :n}
}}}
-> fizzbuzz
{fizzbuzz 1 100}
-> same as above.
|
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