task_url
stringlengths 30
116
| task_name
stringlengths 2
86
| task_description
stringlengths 0
14.4k
| language_url
stringlengths 2
53
| language_name
stringlengths 1
52
| code
stringlengths 0
61.9k
|
---|---|---|---|---|---|
http://rosettacode.org/wiki/Function_definition
|
Function definition
|
A function is a body of code that returns a value.
The value returned may depend on arguments provided to the function.
Task
Write a definition of a function called "multiply" that takes two arguments and returns their product.
(Argument types should be chosen so as not to distract from showing how functions are created and values returned).
Related task
Function prototype
|
#Tcl
|
Tcl
|
proc multiply { arg1 arg2 } {
return [expr {$arg1 * $arg2}]
}
|
http://rosettacode.org/wiki/Forward_difference
|
Forward difference
|
Task
Provide code that produces a list of numbers which is the nth order forward difference, given a non-negative integer (specifying the order) and a list of numbers.
The first-order forward difference of a list of numbers A is a new list B, where Bn = An+1 - An.
List B should have one fewer element as a result.
The second-order forward difference of A will be:
tdefmodule Diff do
def forward(arr,i\\1) do
forward(arr,[],i)
end
def forward([_|[]],diffs,i) do
if i == 1 do
IO.inspect diffs
else
forward(diffs,[],i-1)
end
end
def forward([val1|[val2|vals]],diffs,i) do
forward([val2|vals],diffs++[val2-val1],i)
end
end
The same as the first-order forward difference of B.
That new list will have two fewer elements than A and one less than B.
The goal of this task is to repeat this process up to the desired order.
For a more formal description, see the related Mathworld article.
Algorithmic options
Iterate through all previous forward differences and re-calculate a new array each time.
Use this formula (from Wikipedia):
Δ
n
[
f
]
(
x
)
=
∑
k
=
0
n
(
n
k
)
(
−
1
)
n
−
k
f
(
x
+
k
)
{\displaystyle \Delta ^{n}[f](x)=\sum _{k=0}^{n}{n \choose k}(-1)^{n-k}f(x+k)}
(Pascal's Triangle may be useful for this option.)
|
#Pop11
|
Pop11
|
define forward_difference(l);
lvars res = [], prev, el;
if l = [] then
return([]);
endif;
front(l) -> prev;
for el in back(l) do
cons(el - prev, res) -> res;
el -> prev;
endfor;
rev(res);
enddefine;
define nth_difference(l, n);
lvars res = l, i;
for i from 1 to n do
forward_difference(res) -> res;
endfor;
res;
enddefine;
|
http://rosettacode.org/wiki/Forward_difference
|
Forward difference
|
Task
Provide code that produces a list of numbers which is the nth order forward difference, given a non-negative integer (specifying the order) and a list of numbers.
The first-order forward difference of a list of numbers A is a new list B, where Bn = An+1 - An.
List B should have one fewer element as a result.
The second-order forward difference of A will be:
tdefmodule Diff do
def forward(arr,i\\1) do
forward(arr,[],i)
end
def forward([_|[]],diffs,i) do
if i == 1 do
IO.inspect diffs
else
forward(diffs,[],i-1)
end
end
def forward([val1|[val2|vals]],diffs,i) do
forward([val2|vals],diffs++[val2-val1],i)
end
end
The same as the first-order forward difference of B.
That new list will have two fewer elements than A and one less than B.
The goal of this task is to repeat this process up to the desired order.
For a more formal description, see the related Mathworld article.
Algorithmic options
Iterate through all previous forward differences and re-calculate a new array each time.
Use this formula (from Wikipedia):
Δ
n
[
f
]
(
x
)
=
∑
k
=
0
n
(
n
k
)
(
−
1
)
n
−
k
f
(
x
+
k
)
{\displaystyle \Delta ^{n}[f](x)=\sum _{k=0}^{n}{n \choose k}(-1)^{n-k}f(x+k)}
(Pascal's Triangle may be useful for this option.)
|
#PowerShell
|
PowerShell
|
function Forward-Difference( [UInt64] $n, [Array] $f )
{
$flen = $f.length
if( $flen -gt [Math]::Max( 1, $n ) )
{
0..( $flen - $n - 1 ) | ForEach-Object {
$l=0;
for( $k = 0; $k -le $n; $k++ )
{
$j = 1
for( $i = 1; $i -le $k; $i++ )
{
$j *= ( ( $n - $k + $i ) / $i )
}
$l += $j * ( 1 - 2 * ( ( $n - $k ) % 2 ) ) * $f[ $_ + $k ]
}
$l
}
}
}
Forward-Difference 2 1,2,4,5
|
http://rosettacode.org/wiki/Formatted_numeric_output
|
Formatted numeric output
|
Task
Express a number in decimal as a fixed-length string with leading zeros.
For example, the number 7.125 could be expressed as 00007.125.
|
#Toka
|
Toka
|
needs values
value n
123 to n
2 import printf
" %08d" n printf
|
http://rosettacode.org/wiki/Formatted_numeric_output
|
Formatted numeric output
|
Task
Express a number in decimal as a fixed-length string with leading zeros.
For example, the number 7.125 could be expressed as 00007.125.
|
#Ursala
|
Ursala
|
#import flo
x = 7.125
#show+
t = <printf/'%09.3f' x>
|
http://rosettacode.org/wiki/Four_bit_adder
|
Four bit adder
|
Task
"Simulate" a four-bit adder.
This design can be realized using four 1-bit full adders.
Each of these 1-bit full adders can be built with two half adders and an or gate. ;
Finally a half adder can be made using an xor gate and an and gate.
The xor gate can be made using two nots, two ands and one or.
Not, or and and, the only allowed "gates" for the task, can be "imitated" by using the bitwise operators of your language.
If there is not a bit type in your language, to be sure that the not does not "invert" all the other bits of the basic type (e.g. a byte) we are not interested in, you can use an extra nand (and then not) with the constant 1 on one input.
Instead of optimizing and reducing the number of gates used for the final 4-bit adder, build it in the most straightforward way, connecting the other "constructive blocks", in turn made of "simpler" and "smaller" ones.
Schematics of the "constructive blocks"
(Xor gate with ANDs, ORs and NOTs)
(A half adder)
(A full adder)
(A 4-bit adder)
Solutions should try to be as descriptive as possible, making it as easy as possible to identify "connections" between higher-order "blocks".
It is not mandatory to replicate the syntax of higher-order blocks in the atomic "gate" blocks, i.e. basic "gate" operations can be performed as usual bitwise operations, or they can be "wrapped" in a block in order to expose the same syntax of higher-order blocks, at implementers' choice.
To test the implementation, show the sum of two four-bit numbers (in binary).
|
#Racket
|
Racket
|
#lang racket
(define (adder-and a b)
(if (= 2 (+ a b)) 1 0)) ; Defining the basic and function
(define (adder-not a)
(if (zero? a) 1 0)) ; Defining the basic not function
(define (adder-or a b)
(if (> (+ a b) 0) 1 0)) ; Defining the basic or function
(define (adder-xor a b)
(adder-or
(adder-and
(adder-not a)
b)
(adder-and
a
(adder-not b)))) ; Defines the xor function based on the basic functions
(define (half-adder a b)
(list (adder-xor a b) (adder-and a b))) ; Creates the half adder, returning '(sum carry)
(define (adder a b c0)
(define half-a (half-adder c0 a))
(define half-b (half-adder (car half-a) b))
(list
(car half-b)
(adder-or (cadr half-a) (cadr half-b)))) ; Creates the full adder, returns '(sum carry)
(define (n-bit-adder 4a 4b) ; Creates the n-bit adder, it receives 2 lists of same length
(let-values ; Lists of the form '([01]+)
(((4s v) ; for/fold form will return 2 values, receiving this here
(for/fold ((S null) (c 0)) ;initializes the full sum and carry
((a (in-list (reverse 4a))) (b (in-list (reverse 4b))))
;here it prepares variables for summing each element, starting from the least important bits
(define added
(adder a b c))
(values
(cons (car added) S) ; changes S and c to it's new values in the next iteration
(cadr added)))))
(if (zero? v)
4s
(cons v 4s))))
(n-bit-adder '(1 0 1 0) '(0 1 1 1)) ;-> '(1 0 0 0 1)
|
http://rosettacode.org/wiki/Fivenum
|
Fivenum
|
Many big data or scientific programs use boxplots to show distributions of data. In addition, sometimes saving large arrays for boxplots can be impractical and use extreme amounts of RAM. It can be useful to save large arrays as arrays with five numbers to save memory.
For example, the R programming language implements Tukey's five-number summary as the fivenum function.
Task
Given an array of numbers, compute the five-number summary.
Note
While these five numbers can be used to draw a boxplot, statistical packages will typically need extra data.
Moreover, while there is a consensus about the "box" of the boxplot, there are variations among statistical packages for the whiskers.
|
#MATLAB_.2F_Octave
|
MATLAB / Octave
|
function r = fivenum(x)
r = quantile(x,[0:4]/4);
end;
|
http://rosettacode.org/wiki/Fivenum
|
Fivenum
|
Many big data or scientific programs use boxplots to show distributions of data. In addition, sometimes saving large arrays for boxplots can be impractical and use extreme amounts of RAM. It can be useful to save large arrays as arrays with five numbers to save memory.
For example, the R programming language implements Tukey's five-number summary as the fivenum function.
Task
Given an array of numbers, compute the five-number summary.
Note
While these five numbers can be used to draw a boxplot, statistical packages will typically need extra data.
Moreover, while there is a consensus about the "box" of the boxplot, there are variations among statistical packages for the whiskers.
|
#Mathematica_.2F_Wolfram_Language
|
Mathematica / Wolfram Language
|
ClearAll[FiveNum]
FiveNum[x_List] := Quantile[x, Range[0, 1, 1/4]]
FiveNum[RandomVariate[NormalDistribution[], 10000]]
|
http://rosettacode.org/wiki/Find_the_missing_permutation
|
Find the missing permutation
|
ABCD
CABD
ACDB
DACB
BCDA
ACBD
ADCB
CDAB
DABC
BCAD
CADB
CDBA
CBAD
ABDC
ADBC
BDCA
DCBA
BACD
BADC
BDAC
CBDA
DBCA
DCAB
Listed above are all-but-one of the permutations of the symbols A, B, C, and D, except for one permutation that's not listed.
Task
Find that missing permutation.
Methods
Obvious method:
enumerate all permutations of A, B, C, and D,
and then look for the missing permutation.
alternate method:
Hint: if all permutations were shown above, how many
times would A appear in each position?
What is the parity of this number?
another alternate method:
Hint: if you add up the letter values of each column,
does a missing letter A, B, C, and D from each
column cause the total value for each column to be unique?
Related task
Permutations)
|
#AppleScript
|
AppleScript
|
use framework "Foundation" -- ( sort )
--------------- RAREST LETTER IN EACH COLUMN -------------
on run
concat(map(composeList({¬
head, ¬
minimumBy(comparing(|length|)), ¬
group, ¬
sort}), ¬
transpose(map(chars, ¬
|words|("ABCD CABD ACDB DACB BCDA ACBD " & ¬
"ADCB CDAB DABC BCAD CADB CDBA " & ¬
"CBAD ABDC ADBC BDCA DCBA BACD " & ¬
"BADC BDAC CBDA DBCA DCAB")))))
--> "DBAC"
end run
-------------------- GENERIC FUNCTIONS -------------------
-- chars :: String -> [String]
on chars(s)
characters of s
end chars
-- Ordering :: (-1 | 0 | 1)
-- compare :: a -> a -> Ordering
on compare(a, b)
if a < b then
-1
else if a > b then
1
else
0
end if
end compare
-- comparing :: (a -> b) -> (a -> a -> Ordering)
on comparing(f)
script
on |λ|(a, b)
tell mReturn(f) to compare(|λ|(a), |λ|(b))
end |λ|
end script
end comparing
-- composeList :: [(a -> a)] -> (a -> a)
on composeList(fs)
script
on |λ|(x)
script go
on |λ|(f, a)
mReturn(f)'s |λ|(a)
end |λ|
end script
foldr(go, x, fs)
end |λ|
end script
end composeList
-- concat :: [[a]] -> [a]
-- concat :: [String] -> String
on concat(xs)
set lng to length of xs
if 0 < lng and string is class of (item 1 of xs) then
set acc to ""
else
set acc to {}
end if
repeat with i from 1 to lng
set acc to acc & item i of xs
end repeat
acc
end concat
-- foldl :: (a -> b -> a) -> a -> [b] -> a
on foldl(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from 1 to lng
set v to |λ|(v, item i of xs, i, xs)
end repeat
return v
end tell
end foldl
-- foldr :: (b -> a -> a) -> a -> [b] -> a
on foldr(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from lng to 1 by -1
set v to |λ|(item i of xs, v, i, xs)
end repeat
return v
end tell
end foldr
-- group :: Eq a => [a] -> [[a]]
on group(xs)
script eq
on |λ|(a, b)
a = b
end |λ|
end script
groupBy(eq, xs)
end group
-- groupBy :: (a -> a -> Bool) -> [a] -> [[a]]
on groupBy(f, xs)
set mf to mReturn(f)
script enGroup
on |λ|(a, x)
if length of (active of a) > 0 then
set h to item 1 of active of a
else
set h to missing value
end if
if h is not missing value and mf's |λ|(h, x) then
{active:(active of a) & x, sofar:sofar of a}
else
{active:{x}, sofar:(sofar of a) & {active of a}}
end if
end |λ|
end script
if length of xs > 0 then
set dct to foldl(enGroup, {active:{item 1 of xs}, sofar:{}}, tail(xs))
if length of (active of dct) > 0 then
sofar of dct & {active of dct}
else
sofar of dct
end if
else
{}
end if
end groupBy
-- head :: [a] -> a
on head(xs)
if length of xs > 0 then
item 1 of xs
else
missing value
end if
end head
-- intercalate :: Text -> [Text] -> Text
on intercalate(strText, lstText)
set {dlm, my text item delimiters} to {my text item delimiters, strText}
set strJoined to lstText as text
set my text item delimiters to dlm
return strJoined
end intercalate
-- length :: [a] -> Int
on |length|(xs)
length of xs
end |length|
-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map
-- minimumBy :: (a -> a -> Ordering) -> [a] -> a
on minimumBy(f)
script
on |λ|(xs)
if length of xs < 1 then return missing value
tell mReturn(f)
set v to item 1 of xs
repeat with x in xs
if |λ|(x, v) < 0 then set v to x
end repeat
return v
end tell
end |λ|
end script
end minimumBy
-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: Handler -> Script
on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn
-- sort :: [a] -> [a]
on sort(xs)
((current application's NSArray's arrayWithArray:xs)'s ¬
sortedArrayUsingSelector:"compare:") as list
end sort
-- tail :: [a] -> [a]
on tail(xs)
if length of xs > 1 then
items 2 thru -1 of xs
else
{}
end if
end tail
-- transpose :: [[a]] -> [[a]]
on transpose(xss)
script column
on |λ|(_, iCol)
script row
on |λ|(xs)
item iCol of xs
end |λ|
end script
map(row, xss)
end |λ|
end script
map(column, item 1 of xss)
end transpose
-- words :: String -> [String]
on |words|(s)
words of s
end |words|
|
http://rosettacode.org/wiki/Find_the_last_Sunday_of_each_month
|
Find the last Sunday of each month
|
Write a program or a script that returns the last Sundays of each month of a given year. The year may be given through any simple input method in your language (command line, std in, etc).
Example of an expected output:
./last_sundays 2013
2013-01-27
2013-02-24
2013-03-31
2013-04-28
2013-05-26
2013-06-30
2013-07-28
2013-08-25
2013-09-29
2013-10-27
2013-11-24
2013-12-29
Related tasks
Day of the week
Five weekends
Last Friday of each month
|
#Befunge
|
Befunge
|
":raeY",,,,,&>55+,:::45*:*%\"d"%!*\4%+!3v
v2++6**"I"5\+/*:*54\-/"d"\/4::-1::p53+g5<
>:00p5g4-+7%\:0\v>,"-",5g+:55+/68*+,55+%v
^<<_$$vv*86%+55:<^+*86%+55,+*86/+55:-1:<6
>$$^@$<>+\55+/:#^_$>:#,_$"-",\:04-\-00g^8
^<# #"#"##"#"##!` +76:+1g00,+55,+*<
|
http://rosettacode.org/wiki/Find_the_intersection_of_two_lines
|
Find the intersection of two lines
|
[1]
Task
Find the point of intersection of two lines in 2D.
The 1st line passes though (4,0) and (6,10) .
The 2nd line passes though (0,3) and (10,7) .
|
#C.23
|
C#
|
using System;
using System.Drawing;
public class Program
{
static PointF FindIntersection(PointF s1, PointF e1, PointF s2, PointF e2) {
float a1 = e1.Y - s1.Y;
float b1 = s1.X - e1.X;
float c1 = a1 * s1.X + b1 * s1.Y;
float a2 = e2.Y - s2.Y;
float b2 = s2.X - e2.X;
float c2 = a2 * s2.X + b2 * s2.Y;
float delta = a1 * b2 - a2 * b1;
//If lines are parallel, the result will be (NaN, NaN).
return delta == 0 ? new PointF(float.NaN, float.NaN)
: new PointF((b2 * c1 - b1 * c2) / delta, (a1 * c2 - a2 * c1) / delta);
}
static void Main() {
Func<float, float, PointF> p = (x, y) => new PointF(x, y);
Console.WriteLine(FindIntersection(p(4f, 0f), p(6f, 10f), p(0f, 3f), p(10f, 7f)));
Console.WriteLine(FindIntersection(p(0f, 0f), p(1f, 1f), p(1f, 2f), p(4f, 5f)));
}
}
|
http://rosettacode.org/wiki/Find_the_intersection_of_a_line_with_a_plane
|
Find the intersection of a line with a plane
|
Finding the intersection of an infinite ray with a plane in 3D is an important topic in collision detection.
Task
Find the point of intersection for the infinite ray with direction (0, -1, -1) passing through position (0, 0, 10) with the infinite plane with a normal vector of (0, 0, 1) and which passes through [0, 0, 5].
|
#C.23
|
C#
|
using System;
namespace FindIntersection {
class Vector3D {
private double x, y, z;
public Vector3D(double x, double y, double z) {
this.x = x;
this.y = y;
this.z = z;
}
public static Vector3D operator +(Vector3D lhs, Vector3D rhs) {
return new Vector3D(lhs.x + rhs.x, lhs.y + rhs.y, lhs.z + rhs.z);
}
public static Vector3D operator -(Vector3D lhs, Vector3D rhs) {
return new Vector3D(lhs.x - rhs.x, lhs.y - rhs.y, lhs.z - rhs.z);
}
public static Vector3D operator *(Vector3D lhs, double rhs) {
return new Vector3D(lhs.x * rhs, lhs.y * rhs, lhs.z * rhs);
}
public double Dot(Vector3D rhs) {
return x * rhs.x + y * rhs.y + z * rhs.z;
}
public override string ToString() {
return string.Format("({0:F}, {1:F}, {2:F})", x, y, z);
}
}
class Program {
static Vector3D IntersectPoint(Vector3D rayVector, Vector3D rayPoint, Vector3D planeNormal, Vector3D planePoint) {
var diff = rayPoint - planePoint;
var prod1 = diff.Dot(planeNormal);
var prod2 = rayVector.Dot(planeNormal);
var prod3 = prod1 / prod2;
return rayPoint - rayVector * prod3;
}
static void Main(string[] args) {
var rv = new Vector3D(0.0, -1.0, -1.0);
var rp = new Vector3D(0.0, 0.0, 10.0);
var pn = new Vector3D(0.0, 0.0, 1.0);
var pp = new Vector3D(0.0, 0.0, 5.0);
var ip = IntersectPoint(rv, rp, pn, pp);
Console.WriteLine("The ray intersects the plane at {0}", ip);
}
}
}
|
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
|
#ActionScript
|
ActionScript
|
for (var i:int = 1; i <= 100; i++) {
if (i % 15 == 0)
trace('FizzBuzz');
else if (i % 5 == 0)
trace('Buzz');
else if (i % 3 == 0)
trace('Fizz');
else
trace(i);
}
|
http://rosettacode.org/wiki/Five_weekends
|
Five weekends
|
The month of October in 2010 has five Fridays, five Saturdays, and five Sundays.
Task
Write a program to show all months that have this same characteristic of five full weekends from the year 1900 through 2100 (Gregorian calendar).
Show the number of months with this property (there should be 201).
Show at least the first and last five dates, in order.
Algorithm suggestions
Count the number of Fridays, Saturdays, and Sundays in every month.
Find all of the 31-day months that begin on Friday.
Extra credit
Count and/or show all of the years which do not have at least one five-weekend month (there should be 29).
Related tasks
Day of the week
Last Friday of each month
Find last sunday of each month
|
#Ceylon
|
Ceylon
|
module rosetta.fiveweekends "1.0.0" {
import ceylon.time "1.2.2";
}
|
http://rosettacode.org/wiki/Five_weekends
|
Five weekends
|
The month of October in 2010 has five Fridays, five Saturdays, and five Sundays.
Task
Write a program to show all months that have this same characteristic of five full weekends from the year 1900 through 2100 (Gregorian calendar).
Show the number of months with this property (there should be 201).
Show at least the first and last five dates, in order.
Algorithm suggestions
Count the number of Fridays, Saturdays, and Sundays in every month.
Find all of the 31-day months that begin on Friday.
Extra credit
Count and/or show all of the years which do not have at least one five-weekend month (there should be 29).
Related tasks
Day of the week
Last Friday of each month
Find last sunday of each month
|
#Clojure
|
Clojure
|
(import java.util.GregorianCalendar
java.text.DateFormatSymbols)
(->> (for [year (range 1900 2101)
month [0 2 4 6 7 9 11] ;; 31 day months
:let [cal (GregorianCalendar. year month 1)
day (.get cal GregorianCalendar/DAY_OF_WEEK)]
:when (= day GregorianCalendar/FRIDAY)]
(println month "-" year))
count
(println "Total Months: " ,))
|
http://rosettacode.org/wiki/First_perfect_square_in_base_n_with_n_unique_digits
|
First perfect square in base n with n unique digits
|
Find the first perfect square in a given base N that has at least N digits and
exactly N significant unique digits when expressed in base N.
E.G. In base 10, the first perfect square with at least 10 unique digits is 1026753849 (32043²).
You may use analytical methods to reduce the search space, but the code must do a search. Do not use magic numbers or just feed the code the answer to verify it is correct.
Task
Find and display here, on this page, the first perfect square in base N, with N significant unique digits when expressed in base N, for each of base 2 through 12. Display each number in the base N for which it was calculated.
(optional) Do the same for bases 13 through 16.
(stretch goal) Continue on for bases 17 - ?? (Big Integer math)
See also
OEIS A260182: smallest square that is pandigital in base n.
Related task
Casting out nines
|
#Julia
|
Julia
|
const num = "0123456789abcdef"
hasallin(n, nums, b) = (s = string(n, base=b); all(x -> occursin(x, s), nums))
function squaresearch(base)
basenumerals = [c for c in num[1:base]]
highest = parse(Int, "10" * num[3:base], base=base)
for n in Int(trunc(sqrt(highest))):highest
if hasallin(n * n, basenumerals, base)
return n
end
end
end
println("Base Root N")
for b in 2:16
n = squaresearch(b)
println(lpad(b, 3), lpad(string(n, base=b), 10), " ", string(n * n, base=b))
end
|
http://rosettacode.org/wiki/First_perfect_square_in_base_n_with_n_unique_digits
|
First perfect square in base n with n unique digits
|
Find the first perfect square in a given base N that has at least N digits and
exactly N significant unique digits when expressed in base N.
E.G. In base 10, the first perfect square with at least 10 unique digits is 1026753849 (32043²).
You may use analytical methods to reduce the search space, but the code must do a search. Do not use magic numbers or just feed the code the answer to verify it is correct.
Task
Find and display here, on this page, the first perfect square in base N, with N significant unique digits when expressed in base N, for each of base 2 through 12. Display each number in the base N for which it was calculated.
(optional) Do the same for bases 13 through 16.
(stretch goal) Continue on for bases 17 - ?? (Big Integer math)
See also
OEIS A260182: smallest square that is pandigital in base n.
Related task
Casting out nines
|
#Kotlin
|
Kotlin
|
import java.math.BigInteger
import java.time.Duration
import java.util.ArrayList
import java.util.HashSet
import kotlin.math.sqrt
const val ALPHABET = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz|"
var base: Byte = 0
var bmo: Byte = 0
var blim: Byte = 0
var ic: Byte = 0
var st0: Long = 0
var bllim: BigInteger? = null
var threshold: BigInteger? = null
var hs: MutableSet<Byte> = HashSet()
var o: MutableSet<Byte> = HashSet()
val chars = ALPHABET.toCharArray()
var limits: MutableList<BigInteger?>? = null
var ms: String? = null
fun indexOf(c: Char): Int {
for (i in chars.indices) {
if (chars[i] == c) {
return i
}
}
return -1
}
// convert BigInteger to string using current base
fun toStr(b: BigInteger): String {
var b2 = b
val bigBase = BigInteger.valueOf(base.toLong())
val res = StringBuilder()
while (b2 > BigInteger.ZERO) {
val divRem = b2.divideAndRemainder(bigBase)
res.append(chars[divRem[1].toInt()])
b2 = divRem[0]
}
return res.toString()
}
// check for a portion of digits, bailing if uneven
fun allInQS(b: BigInteger): Boolean {
var b2 = b
val bigBase = BigInteger.valueOf(base.toLong())
var c = ic.toInt()
hs.clear()
hs.addAll(o)
while (b2 > bllim) {
val divRem = b2.divideAndRemainder(bigBase)
hs.add(divRem[1].toByte())
c++
if (c > hs.size) {
return false
}
b2 = divRem[0]
}
return true
}
// check for a portion of digits, all the way to the end
fun allInS(b: BigInteger): Boolean {
var b2 = b
val bigBase = BigInteger.valueOf(base.toLong())
hs.clear()
hs.addAll(o)
while (b2 > bllim) {
val divRem = b2.divideAndRemainder(bigBase)
hs.add(divRem[1].toByte())
b2 = divRem[0]
}
return hs.size == base.toInt()
}
// check for all digits, bailing if uneven
fun allInQ(b: BigInteger): Boolean {
var b2 = b
val bigBase = BigInteger.valueOf(base.toLong())
var c = 0
hs.clear()
while (b2 > BigInteger.ZERO) {
val divRem = b2.divideAndRemainder(bigBase)
hs.add(divRem[1].toByte())
c++
if (c > hs.size) {
return false
}
b2 = divRem[0]
}
return true
}
// check for all digits, all the way to the end
fun allIn(b: BigInteger): Boolean {
var b2 = b
val bigBase = BigInteger.valueOf(base.toLong())
hs.clear()
while (b2 > BigInteger.ZERO) {
val divRem = b2.divideAndRemainder(bigBase)
hs.add(divRem[1].toByte())
b2 = divRem[0]
}
return hs.size == base.toInt()
}
// parse a string into a BigInteger, using current base
fun to10(s: String?): BigInteger {
val bigBase = BigInteger.valueOf(base.toLong())
var res = BigInteger.ZERO
for (element in s!!) {
val idx = indexOf(element)
val bigIdx = BigInteger.valueOf(idx.toLong())
res = res.multiply(bigBase).add(bigIdx)
}
return res
}
// returns the minimum value string, optionally inserting extra digit
fun fixup(n: Int): String {
var res = ALPHABET.substring(0, base.toInt())
if (n > 0) {
val sb = StringBuilder(res)
sb.insert(n, n)
res = sb.toString()
}
return "10" + res.substring(2)
}
// checks the square against the threshold, advances various limits when needed
fun check(sq: BigInteger) {
if (sq > threshold) {
o.remove(indexOf(ms!![blim.toInt()]).toByte())
blim--
ic--
threshold = limits!![bmo - blim - 1]
bllim = to10(ms!!.substring(0, blim + 1))
}
}
// performs all the calculations for the current base
fun doOne() {
limits = ArrayList()
bmo = (base - 1).toByte()
var dr: Byte = 0
if ((base.toInt() and 1) == 1) {
dr = (base.toInt() shr 1).toByte()
}
o.clear()
blim = 0
var id: Byte = 0
var inc = 1
val st = System.nanoTime()
val sdr = ByteArray(bmo.toInt())
var rc: Byte = 0
for (i in 0 until bmo) {
sdr[i] = (i * i % bmo).toByte()
if (sdr[i] == dr) {
rc = (rc + 1).toByte()
}
if (sdr[i] == 0.toByte()) {
sdr[i] = (sdr[i] + bmo).toByte()
}
}
var i: Long = 0
if (dr > 0) {
id = base
i = 1
while (i <= dr) {
if (sdr[i.toInt()] >= dr) {
if (id > sdr[i.toInt()]) {
id = sdr[i.toInt()]
}
}
i++
}
id = (id - dr).toByte()
i = 0
}
ms = fixup(id.toInt())
var sq = to10(ms)
var rt = BigInteger.valueOf((sqrt(sq.toDouble()) + 1).toLong())
sq = rt.multiply(rt)
if (base > 9) {
for (j in 1 until base) {
limits!!.add(to10(ms!!.substring(0, j) + chars[bmo.toInt()].toString().repeat(base - j + if (rc > 0) 0 else 1)))
}
limits!!.reverse()
while (sq < limits!![0]) {
rt = rt.add(BigInteger.ONE)
sq = rt.multiply(rt)
}
}
var dn = rt.shiftLeft(1).add(BigInteger.ONE)
var d = BigInteger.ONE
if (base > 3 && rc > 0) {
while (sq.remainder(BigInteger.valueOf(bmo.toLong())).compareTo(BigInteger.valueOf(dr.toLong())) != 0) {
rt = rt.add(BigInteger.ONE)
sq = sq.add(dn)
dn = dn.add(BigInteger.TWO)
} // aligns sq to dr
inc = bmo / rc
if (inc > 1) {
dn = dn.add(rt.multiply(BigInteger.valueOf(inc - 2.toLong())).subtract(BigInteger.ONE))
d = BigInteger.valueOf(inc * inc.toLong())
}
dn = dn.add(dn).add(d)
}
d = d.shiftLeft(1)
if (base > 9) {
blim = 0
while (sq < limits!![bmo - blim - 1]) {
blim++
}
ic = (blim + 1).toByte()
threshold = limits!![bmo - blim - 1]
if (blim > 0) {
for (j in 0..blim) {
o.add(indexOf(ms!![j]).toByte())
}
}
bllim = if (blim > 0) {
to10(ms!!.substring(0, blim + 1))
} else {
BigInteger.ZERO
}
if (base > 5 && rc > 0) while (!allInQS(sq)) {
sq = sq.add(dn)
dn = dn.add(d)
i += 1
check(sq)
} else {
while (!allInS(sq)) {
sq = sq.add(dn)
dn = dn.add(d)
i += 1
check(sq)
}
}
} else {
if (base > 5 && rc > 0) {
while (!allInQ(sq)) {
sq = sq.add(dn)
dn = dn.add(d)
i += 1
}
} else {
while (!allIn(sq)) {
sq = sq.add(dn)
dn = dn.add(d)
i += 1
}
}
}
rt = rt.add(BigInteger.valueOf(i * inc))
val delta1 = System.nanoTime() - st
val dur1 = Duration.ofNanos(delta1)
val delta2 = System.nanoTime() - st0
val dur2 = Duration.ofNanos(delta2)
System.out.printf(
"%3d %2d %2s %20s -> %-40s %10d %9s %9s\n",
base, inc, if (id > 0) ALPHABET.substring(id.toInt(), id + 1) else " ", toStr(rt), toStr(sq), i, format(dur1), format(dur2)
)
}
private fun format(d: Duration): String {
val minP = d.toMinutesPart()
val secP = d.toSecondsPart()
val milP = d.toMillisPart()
return String.format("%02d:%02d.%03d", minP, secP, milP)
}
fun main() {
println("base inc id root square test count time total")
st0 = System.nanoTime()
base = 2
while (base < 28) {
doOne()
++base
}
}
|
http://rosettacode.org/wiki/First-class_functions
|
First-class functions
|
A language has first-class functions if it can do each of the following without recursively invoking a compiler or interpreter or otherwise metaprogramming:
Create new functions from preexisting functions at run-time
Store functions in collections
Use functions as arguments to other functions
Use functions as return values of other functions
Task
Write a program to create an ordered collection A of functions of a real number. At least one function should be built-in and at least one should be user-defined; try using the sine, cosine, and cubing functions. Fill another collection B with the inverse of each function in A. Implement function composition as in Functional Composition. Finally, demonstrate that the result of applying the composition of each function in A and its inverse in B to a value, is the original value. (Within the limits of computational accuracy).
(A solution need not actually call the collections "A" and "B". These names are only used in the preceding paragraph for clarity.)
Related task
First-class Numbers
|
#D
|
D
|
void main() {
import std.stdio, std.math, std.typetuple, std.functional;
alias dir = TypeTuple!(sin, cos, x => x ^^ 3);
alias inv = TypeTuple!(asin, acos, cbrt);
// foreach (f, g; staticZip!(dir, inv))
foreach (immutable i, f; dir)
writefln("%6.3f", compose!(f, inv[i])(0.5));
}
|
http://rosettacode.org/wiki/First-class_functions
|
First-class functions
|
A language has first-class functions if it can do each of the following without recursively invoking a compiler or interpreter or otherwise metaprogramming:
Create new functions from preexisting functions at run-time
Store functions in collections
Use functions as arguments to other functions
Use functions as return values of other functions
Task
Write a program to create an ordered collection A of functions of a real number. At least one function should be built-in and at least one should be user-defined; try using the sine, cosine, and cubing functions. Fill another collection B with the inverse of each function in A. Implement function composition as in Functional Composition. Finally, demonstrate that the result of applying the composition of each function in A and its inverse in B to a value, is the original value. (Within the limits of computational accuracy).
(A solution need not actually call the collections "A" and "B". These names are only used in the preceding paragraph for clarity.)
Related task
First-class Numbers
|
#Dart
|
Dart
|
import 'dart:math' as Math;
cube(x) => x*x*x;
cuberoot(x) => Math.pow(x, 1/3);
compose(f,g) => ((x)=>f(g(x)));
main(){
var functions = [Math.sin, Math.exp, cube];
var inverses = [Math.asin, Math.log, cuberoot];
for (int i = 0; i < 3; i++){
print(compose(functions[i], inverses[i])(0.5));
}
}
|
http://rosettacode.org/wiki/Forest_fire
|
Forest fire
|
This page uses content from Wikipedia. The original article was at Forest-fire model. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
Task
Implement the Drossel and Schwabl definition of the forest-fire model.
It is basically a 2D cellular automaton where each cell can be in three distinct states (empty, tree and burning) and evolves according to the following rules (as given by Wikipedia)
A burning cell turns into an empty cell
A tree will burn if at least one neighbor is burning
A tree ignites with probability f even if no neighbor is burning
An empty space fills with a tree with probability p
Neighborhood is the Moore neighborhood; boundary conditions are so that on the boundary the cells are always empty ("fixed" boundary condition).
At the beginning, populate the lattice with empty and tree cells according to a specific probability (e.g. a cell has the probability 0.5 to be a tree). Then, let the system evolve.
Task's requirements do not include graphical display or the ability to change parameters (probabilities p and f ) through a graphical or command line interface.
Related tasks
See Conway's Game of Life
See Wireworld.
|
#JAMES_II.2FRule-based_Cellular_Automata
|
JAMES II/Rule-based Cellular Automata
|
@caversion 1;
dimensions 2;
state EMPTY, TREE, BURNING;
// an empty cell grows a tree with a chance of p = 5 %
rule{EMPTY} [0.05] : -> TREE;
// a burning cell turns to a burned cell
rule{BURNING}: -> EMPTY;
// a tree starts burning if there is at least one neighbor burning
rule{TREE} : BURNING{1,} -> BURNING;
// a tree is hit by lightning with a change of f = 0.006 %
rule{TREE} [0.00006] : -> BURNING;
|
http://rosettacode.org/wiki/First_class_environments
|
First class environments
|
According to Wikipedia, "In computing, a first-class object ... is an entity that can be constructed at run-time, passed as a parameter, returned from a subroutine, or assigned into a variable".
Often this term is used in the context of "first class functions". In an analogous way, a programming language may support "first class environments".
The environment is minimally, the set of variables accessible to a statement being executed. Change the environments and the same statement could produce different results when executed.
Often an environment is captured in a closure, which encapsulates a function together with an environment. That environment, however, is not first-class, as it cannot be created, passed etc. independently from the function's code.
Therefore, a first class environment is a set of variable bindings which can be constructed at run-time, passed as a parameter, returned from a subroutine, or assigned into a variable. It is like a closure without code. A statement must be able to be executed within a stored first class environment and act according to the environment variable values stored within.
Task
Build a dozen environments, and a single piece of code to be run repeatedly in each of these environments.
Each environment contains the bindings for two variables:
a value in the Hailstone sequence, and
a count which is incremented until the value drops to 1.
The initial hailstone values are 1 through 12, and the count in each environment is zero.
When the code runs, it calculates the next hailstone step in the current environment (unless the value is already 1) and counts the steps. Then it prints the current value in a tabular form.
When all hailstone values dropped to 1, processing stops, and the total number of hailstone steps for each environment is printed.
|
#Ruby
|
Ruby
|
# Build environments
envs = (1..12).map do |n|
Object.new.instance_eval {@n = n; @cnt = 0; self}
end
# Until all values are 1:
until envs.all? {|e| e.instance_eval{@n} == 1}
envs.each do |e|
e.instance_eval do # Use environment _e_
printf "%4s", @n
if @n > 1
@cnt += 1 # Increment step count
@n = if @n.odd? # Calculate next hailstone value
@n * 3 + 1
else
@n / 2
end
end
end
end
puts
end
puts '=' * 48
envs.each do |e| # For each environment _e_
e.instance_eval do
printf "%4s", @cnt # print the step count
end
end
puts
|
http://rosettacode.org/wiki/First_class_environments
|
First class environments
|
According to Wikipedia, "In computing, a first-class object ... is an entity that can be constructed at run-time, passed as a parameter, returned from a subroutine, or assigned into a variable".
Often this term is used in the context of "first class functions". In an analogous way, a programming language may support "first class environments".
The environment is minimally, the set of variables accessible to a statement being executed. Change the environments and the same statement could produce different results when executed.
Often an environment is captured in a closure, which encapsulates a function together with an environment. That environment, however, is not first-class, as it cannot be created, passed etc. independently from the function's code.
Therefore, a first class environment is a set of variable bindings which can be constructed at run-time, passed as a parameter, returned from a subroutine, or assigned into a variable. It is like a closure without code. A statement must be able to be executed within a stored first class environment and act according to the environment variable values stored within.
Task
Build a dozen environments, and a single piece of code to be run repeatedly in each of these environments.
Each environment contains the bindings for two variables:
a value in the Hailstone sequence, and
a count which is incremented until the value drops to 1.
The initial hailstone values are 1 through 12, and the count in each environment is zero.
When the code runs, it calculates the next hailstone step in the current environment (unless the value is already 1) and counts the steps. Then it prints the current value in a tabular form.
When all hailstone values dropped to 1, processing stops, and the total number of hailstone steps for each environment is printed.
|
#Sidef
|
Sidef
|
func calculator({.is_one} ) { 1 }
func calculator(n {.is_even}) { n / 2 }
func calculator(n ) { 3*n + 1 }
func succ(this {_{:value}.is_one}, _) {
return this
}
func succ(this, get_next) {
this{:value} = get_next(this{:value})
this{:count}++
return this
}
var enviornments = (1..12 -> map {|i| Hash(value => i, count => 0) });
while (!enviornments.map{ _{:value} }.all { .is_one }) {
say enviornments.map {|h| "%4s" % h{:value} }.join;
enviornments.range.each { |i|
enviornments[i] = succ(enviornments[i], calculator);
}
}
say 'Counts';
say enviornments.map{ |h| "%4s" % h{:count} }.join;
|
http://rosettacode.org/wiki/Flatten_a_list
|
Flatten a list
|
Task
Write a function to flatten the nesting in an arbitrary list of values.
Your program should work on the equivalent of this list:
[[1], 2, [[3, 4], 5], [[[]]], [[[6]]], 7, 8, []]
Where the correct result would be the list:
[1, 2, 3, 4, 5, 6, 7, 8]
Related task
Tree traversal
|
#Ela
|
Ela
|
xs = [[1], 2, [[3,4], 5], [[[]]], [[[6]]], 7, 8, []]
flat = flat' []
where flat' n [] = n
flat' n (x::xs)
| x is List = flat' (flat' n xs) x
| else = x :: flat' n xs
flat xs
|
http://rosettacode.org/wiki/Flipping_bits_game
|
Flipping bits game
|
The game
Given an N×N square array of zeroes or ones in an initial configuration, and a target configuration of zeroes and ones.
The game is to transform one to the other in as few moves as possible by inverting whole numbered rows or whole lettered
columns at once (as one move).
In an inversion. any 1 becomes 0, and any 0 becomes 1 for that whole row or column.
Task
Create a program to score for the Flipping bits game.
The game should create an original random target configuration and a starting configuration.
Ensure that the starting position is never the target position.
The target position must be guaranteed as reachable from the starting position. (One possible way to do this is to generate the start position by legal flips from a random target position. The flips will always be reversible back to the target from the given start position).
The number of moves taken so far should be shown.
Show an example of a short game here, on this page, for a 3×3 array of bits.
|
#Raku
|
Raku
|
sub MAIN ($square = 4) {
say "{$square}? Seriously?" and exit if $square < 1 or $square > 26;
my %bits = map { $_ => %( map { $_ => 0 }, ('A' .. *)[^ $square] ) },
(1 .. *)[^ $square];
scramble %bits;
my $target = build %bits;
scramble %bits until build(%bits) ne $target;
display($target, %bits);
my $turns = 0;
while my $flip = prompt "Turn {++$turns}: Flip which row / column? " {
flip $flip.match(/\w/).uc, %bits;
if display($target, %bits) {
say "Hurray! You solved it in $turns turns.";
last;
}
}
}
sub display($goal, %hash) {
shell('clear');
say "Goal\n$goal\nYou";
my $this = build %hash;
say $this;
return ($goal eq $this);
}
sub flip ($a, %hash) {
given $a {
when any(keys %hash) {
%hash{$a}{$_} = %hash{$a}{$_} +^ 1 for %hash{$a}.keys
};
when any(keys %hash{'1'}) {
%hash{$_}{$a} = %hash{$_}{$a} +^ 1 for %hash.keys
};
}
}
sub build (%hash) {
my $string = ' ';
$string ~= sprintf "%2s ", $_ for sort keys %hash{'1'};
$string ~= "\n";
for %hash.keys.sort: +* -> $key {
$string ~= sprintf "%2s ", $key;
$string ~= sprintf "%2s ", %hash{$key}{$_} for sort keys %hash{$key};
$string ~= "\n";
};
$string
}
sub scramble(%hash) {
my @keys = keys %hash;
@keys.push: | keys %hash{'1'};
flip $_, %hash for @keys.pick( @keys/2 );
}
|
http://rosettacode.org/wiki/First_power_of_2_that_has_leading_decimal_digits_of_12
|
First power of 2 that has leading decimal digits of 12
|
(This task is taken from a Project Euler problem.)
(All numbers herein are expressed in base ten.)
27 = 128 and 7 is
the first power of 2 whose leading decimal digits are 12.
The next power of 2 whose leading decimal digits
are 12 is 80,
280 = 1208925819614629174706176.
Define p(L,n) to be the nth-smallest
value of j such that the base ten representation
of 2j begins with the digits of L .
So p(12, 1) = 7 and
p(12, 2) = 80
You are also given that:
p(123, 45) = 12710
Task
find:
p(12, 1)
p(12, 2)
p(123, 45)
p(123, 12345)
p(123, 678910)
display the results here, on this page.
|
#REXX
|
REXX
|
/*REXX program computes powers of two whose leading decimal digits are "12" (in base 10)*/
parse arg L n b . /*obtain optional arguments from the CL*/
if L=='' | L=="," then L= 12 /*Not specified? Then use the default.*/
if n=='' | n=="," then n= 1 /* " " " " " " */
if b=='' | b=="," then b= 2 /* " " " " " " */
LL= length(L) /*obtain the length of L for compares*/
fd= left(L, 1) /*obtain the first dec. digit of L.*/
fr= substr(L, 2) /* " " rest of dec. digits " " */
numeric digits max(20, LL+2) /*use an appropriate value of dec. digs*/
rest= LL - 1 /*the length of the rest of the digits.*/
#= 0 /*the number of occurrences of a result*/
x= 1 /*start with a product of unity (B**0).*/
do j=1 until #==n; x= x * b /*raise B to a whole bunch of powers.*/
parse var x _ 2 /*obtain the first decimal digit of X.*/
if _ \== fd then iterate /*check only the 1st digit at this time*/
if LL>1 then do /*check the rest of the digits, maybe. */
$= format(x, , , , 0) /*express X in exponential format. */
parse var $ '.' +1 f +(rest) /*obtain the rest of the digits. */
if f \== fr then iterate /*verify that X has the rest of digs.*/
end /* [↓] found an occurrence of an answer*/
#= # + 1 /*bump the number of occurrences so far*/
end /*j*/
say 'The ' th(n) ' occurrence of ' b ' raised to a power whose product starts with' ,
' "'L"'" ' is ' commas(j).
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: arg _; do c=length(_)-3 to 1 by -3; _= insert(',', _, c); end; return _
th: arg _; return _ || word('th st nd rd', 1 +_//10 * (_//100 % 10\==1) * (_//10 <4))
|
http://rosettacode.org/wiki/First-class_functions/Use_numbers_analogously
|
First-class functions/Use numbers analogously
|
In First-class functions, a language is showing how its manipulation of functions is similar to its manipulation of other types.
This tasks aim is to compare and contrast a language's implementation of first class functions, with its normal handling of numbers.
Write a program to create an ordered collection of a mixture of literally typed and expressions producing a real number, together with another ordered collection of their multiplicative inverses. Try and use the following pseudo-code to generate the numbers for the ordered collections:
x = 2.0
xi = 0.5
y = 4.0
yi = 0.25
z = x + y
zi = 1.0 / ( x + y )
Create a function multiplier, that given two numbers as arguments returns a function that when called with one argument, returns the result of multiplying the two arguments to the call to multiplier that created it and the argument in the call:
new_function = multiplier(n1,n2)
# where new_function(m) returns the result of n1 * n2 * m
Applying the multiplier of a number and its inverse from the two ordered collections of numbers in pairs, show that the result in each case is one.
Compare and contrast the resultant program with the corresponding entry in First-class functions. They should be close.
To paraphrase the task description: Do what was done before, but with numbers rather than functions
|
#PicoLisp
|
PicoLisp
|
(load "@lib/math.l")
(de multiplier (N1 N2)
(curry (N1 N2) (X)
(*/ N1 N2 X `(* 1.0 1.0)) ) )
(let (X 2.0 Xi 0.5 Y 4.0 Yi 0.25 Z (+ X Y) Zi (*/ 1.0 1.0 Z))
(mapc
'((Num Inv)
(prinl (format ((multiplier Inv Num) 0.5) *Scl)) )
(list X Y Z)
(list Xi Yi Zi) ) )
|
http://rosettacode.org/wiki/First-class_functions/Use_numbers_analogously
|
First-class functions/Use numbers analogously
|
In First-class functions, a language is showing how its manipulation of functions is similar to its manipulation of other types.
This tasks aim is to compare and contrast a language's implementation of first class functions, with its normal handling of numbers.
Write a program to create an ordered collection of a mixture of literally typed and expressions producing a real number, together with another ordered collection of their multiplicative inverses. Try and use the following pseudo-code to generate the numbers for the ordered collections:
x = 2.0
xi = 0.5
y = 4.0
yi = 0.25
z = x + y
zi = 1.0 / ( x + y )
Create a function multiplier, that given two numbers as arguments returns a function that when called with one argument, returns the result of multiplying the two arguments to the call to multiplier that created it and the argument in the call:
new_function = multiplier(n1,n2)
# where new_function(m) returns the result of n1 * n2 * m
Applying the multiplier of a number and its inverse from the two ordered collections of numbers in pairs, show that the result in each case is one.
Compare and contrast the resultant program with the corresponding entry in First-class functions. They should be close.
To paraphrase the task description: Do what was done before, but with numbers rather than functions
|
#Python
|
Python
|
IDLE 2.6.1
>>> # Number literals
>>> x,xi, y,yi = 2.0,0.5, 4.0,0.25
>>> # Numbers from calculation
>>> z = x + y
>>> zi = 1.0 / (x + y)
>>> # The multiplier function is similar to 'compose' but with numbers
>>> multiplier = lambda n1, n2: (lambda m: n1 * n2 * m)
>>> # Numbers as members of collections
>>> numlist = [x, y, z]
>>> numlisti = [xi, yi, zi]
>>> # Apply numbers from list
>>> [multiplier(inversen, n)(.5) for n, inversen in zip(numlist, numlisti)]
[0.5, 0.5, 0.5]
>>>
|
http://rosettacode.org/wiki/Flow-control_structures
|
Flow-control structures
|
Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Task
Document common flow-control structures.
One common example of a flow-control structure is the goto construct.
Note that Conditional Structures and Loop Structures have their own articles/categories.
Related tasks
Conditional Structures
Loop Structures
|
#REBOL
|
REBOL
|
rebol [
Title: "Flow Control"
URL: http://rosettacode.org/wiki/Flow_Control_Structures
]
; return -- Return early from function (normally, functions return
; result of last evaluation).
hatefive: func [
"Prints value unless it's the number 5."
value "Value to print."
][
if value = 5 [return "I hate five!"]
print value
]
print "Function hatefive, with various values:"
hatefive 99
hatefive 13
hatefive 5
hatefive 3
; break -- Break out of current loop.
print [crlf "Loop to 10, but break out at five:"]
repeat i 10 [
if i = 5 [break]
print i
]
; catch/throw -- throw breaks out of a code block to enclosing catch.
print [crlf "Start to print two lines, but throw out after the first:"]
catch [
print "First"
throw "I'm done!"
print "Second"
]
; Using named catch blocks, you can select which catcher you want when throwing.
print [crlf "Throw from inner code block, caught by outer:"]
catch/name [
print "Outer catch block."
catch/name [
print "Inner catch block."
throw/name "I'm done!" 'Johnson
print "We never get here."
] 'Clemens
print "We never get here, either."
] 'Johnson
; try
div: func [
"Divide first number by second."
a b
/local r "Result"
][
if error? try [r: a / b] [r: "Error!"]
r ; Functions return last value evaluated.
]
print [crlf "Report error on bad division:"]
print div 10 4
print div 10 2
print div 10 1
print div 10 0
|
http://rosettacode.org/wiki/Flow-control_structures
|
Flow-control structures
|
Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Task
Document common flow-control structures.
One common example of a flow-control structure is the goto construct.
Note that Conditional Structures and Loop Structures have their own articles/categories.
Related tasks
Conditional Structures
Loop Structures
|
#Relation
|
Relation
|
call routineName /*no arguments passed to routine.*/
call routineName 50 /*one argument (fifty) passed. */
call routineName 50,60 /*two arguments passed. */
call routineName 50, 60 /*(same as above) */
call routineName 50 ,60 /*(same as above) */
call routineName 10*5 , 8**4 - 4 /*(same as above) */
call routineName 50 , , , 70 /*4 args passed, 2nd&3rd omitted.*/
/*omitted args are NOT null. */
call routineName ,,,,,,,,,,,,,,,,800 /*17 args passed, 16 omitted. */
call date /*looks for DATE internally first*/
call 'DATE' /* " " " BIF | externally*/
|
http://rosettacode.org/wiki/Floyd%27s_triangle
|
Floyd's triangle
|
Floyd's triangle lists the natural numbers in a right triangle aligned to the left where
the first row is 1 (unity)
successive rows start towards the left with the next number followed by successive naturals listing one more number than the line above.
The first few lines of a Floyd triangle looks like this:
1
2 3
4 5 6
7 8 9 10
11 12 13 14 15
Task
Write a program to generate and display here the first n lines of a Floyd triangle.
(Use n=5 and n=14 rows).
Ensure that when displayed in a mono-space font, the numbers line up in vertical columns as shown and that only one space separates numbers of the last row.
|
#FreeBASIC
|
FreeBASIC
|
' version 19-09-2015
' compile with: fbc -s console
Sub pascal_triangle(n As UInteger)
Dim As UInteger a = 1, b, i, j, switch = n + 1
Dim As String frmt, frmt_1, frmt_2
' last number of the last line
i = (n * (n + 1)) \ 2
frmt_2 = String(Len(Str(i)) + 1, "#")
' first number of the last line
i = ((n - 1) * n) \ 2 + 1
frmt_1 = String(Len(Str(i)) + 1, "#")
' we have 2 different formats strings
' find the point where we have to make the switch
If frmt_1 <> frmt_2 Then
j = i + 1
While Len(Str(i)) = Len(Str(J))
j = j + 1
Wend
switch = j - i
End If
Print "output for "; Str(n) : Print
For i = 1 To n
frmt = frmt_1
b = (i * (i + 1)) \ 2
For j = a To b
' if we have the switching point change format string
If j - a = switch Then frmt = frmt_2
Print Using frmt; j;
Next j
Print
a = b + 1
Next i
Print
End Sub
' ------=< MAIN >=------
pascal_triangle(5)
pascal_triangle(14)
' empty keyboard buffer
While Inkey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
|
http://rosettacode.org/wiki/Floyd-Warshall_algorithm
|
Floyd-Warshall algorithm
|
The Floyd–Warshall algorithm is an algorithm for finding shortest paths in a weighted graph with positive or negative edge weights.
Task
Find the lengths of the shortest paths between all pairs of vertices of the given directed graph. Your code may assume that the input has already been checked for loops, parallel edges and negative cycles.
Print the pair, the distance and (optionally) the path.
Example
pair dist path
1 -> 2 -1 1 -> 3 -> 4 -> 2
1 -> 3 -2 1 -> 3
1 -> 4 0 1 -> 3 -> 4
2 -> 1 4 2 -> 1
2 -> 3 2 2 -> 1 -> 3
2 -> 4 4 2 -> 1 -> 3 -> 4
3 -> 1 5 3 -> 4 -> 2 -> 1
3 -> 2 1 3 -> 4 -> 2
3 -> 4 2 3 -> 4
4 -> 1 3 4 -> 2 -> 1
4 -> 2 -1 4 -> 2
4 -> 3 1 4 -> 2 -> 1 -> 3
See also
Floyd-Warshall Algorithm - step by step guide (youtube)
|
#Phix
|
Phix
|
constant inf = 1e300*1e300
function Path(integer u, integer v, sequence next)
if next[u,v]=null then
return ""
end if
sequence path = {sprintf("%d",u)}
while u!=v do
u = next[u,v]
path = append(path,sprintf("%d",u))
end while
return join(path,"->")
end function
procedure FloydWarshall(integer V, sequence weights)
sequence dist = repeat(repeat(inf,V),V)
sequence next = repeat(repeat(null,V),V)
for k=1 to length(weights) do
integer {u,v,w} = weights[k]
dist[u,v] := w -- the weight of the edge (u,v)
next[u,v] := v
end for
-- standard Floyd-Warshall implementation
for k=1 to V do
for i=1 to V do
for j=1 to V do
atom d = dist[i,k] + dist[k,j]
if dist[i,j] > d then
dist[i,j] := d
next[i,j] := next[i,k]
end if
end for
end for
end for
printf(1,"pair dist path\n")
for u=1 to V do
for v=1 to V do
if u!=v then
printf(1,"%d->%d %2d %s\n",{u,v,dist[u,v],Path(u,v,next)})
end if
end for
end for
end procedure
constant V = 4
constant weights = {{1, 3, -2}, {2, 1, 4}, {2, 3, 3}, {3, 4, 2}, {4, 2, -1}}
FloydWarshall(V,weights)
|
http://rosettacode.org/wiki/Function_definition
|
Function definition
|
A function is a body of code that returns a value.
The value returned may depend on arguments provided to the function.
Task
Write a definition of a function called "multiply" that takes two arguments and returns their product.
(Argument types should be chosen so as not to distract from showing how functions are created and values returned).
Related task
Function prototype
|
#TI-89_BASIC
|
TI-89 BASIC
|
multiply(a, b)
Func
Return a * b
EndFunc
|
http://rosettacode.org/wiki/Forward_difference
|
Forward difference
|
Task
Provide code that produces a list of numbers which is the nth order forward difference, given a non-negative integer (specifying the order) and a list of numbers.
The first-order forward difference of a list of numbers A is a new list B, where Bn = An+1 - An.
List B should have one fewer element as a result.
The second-order forward difference of A will be:
tdefmodule Diff do
def forward(arr,i\\1) do
forward(arr,[],i)
end
def forward([_|[]],diffs,i) do
if i == 1 do
IO.inspect diffs
else
forward(diffs,[],i-1)
end
end
def forward([val1|[val2|vals]],diffs,i) do
forward([val2|vals],diffs++[val2-val1],i)
end
end
The same as the first-order forward difference of B.
That new list will have two fewer elements than A and one less than B.
The goal of this task is to repeat this process up to the desired order.
For a more formal description, see the related Mathworld article.
Algorithmic options
Iterate through all previous forward differences and re-calculate a new array each time.
Use this formula (from Wikipedia):
Δ
n
[
f
]
(
x
)
=
∑
k
=
0
n
(
n
k
)
(
−
1
)
n
−
k
f
(
x
+
k
)
{\displaystyle \Delta ^{n}[f](x)=\sum _{k=0}^{n}{n \choose k}(-1)^{n-k}f(x+k)}
(Pascal's Triangle may be useful for this option.)
|
#PureBasic
|
PureBasic
|
Procedure forward_difference(List a())
If ListSize(a()) <= 1
ClearList(a()): ProcedureReturn
EndIf
Protected NewList b()
CopyList(a(), b())
LastElement(a()): DeleteElement(a())
SelectElement(b(), 1)
ForEach a()
a() - b(): NextElement(b())
Next
EndProcedure
Procedure nth_difference(List a(), List b(), n)
Protected i
CopyList(a(), b())
For i = 1 To n
forward_difference(b())
Next
EndProcedure
Procedure.s display(List a())
Protected output.s
ForEach a()
output + Str(a()) + ","
Next
ProcedureReturn RTrim(output,",")
EndProcedure
DataSection
;list data
Data.i 10 ;element count
Data.i 90, 47, 58, 29, 22, 32, 55, 5, 55, 73
EndDataSection
;create and fill list
Define i
NewList a()
Read.i i
While i > 0
AddElement(a()): Read.i a(): i - 1
Wend
If OpenConsole()
NewList b()
For i = 1 To 10
nth_difference(a(), b(), i)
PrintN(Str(i) + " [" + display(b()) + "]")
Next
Print(#CRLF$ + #CRLF$ + "Press ENTER to exit"): Input()
CloseConsole()
EndIf
|
http://rosettacode.org/wiki/Formatted_numeric_output
|
Formatted numeric output
|
Task
Express a number in decimal as a fixed-length string with leading zeros.
For example, the number 7.125 could be expressed as 00007.125.
|
#Vala
|
Vala
|
void main() {
double r = 7.125;
print(" %9.3f\n", -r);
print(" %9.3f\n",r);
print(" %-9.3f\n",r);
print(" %09.3f\n",-r);
print(" %09.3f\n",r);
print(" %-09.3f\n",r);
}
|
http://rosettacode.org/wiki/Formatted_numeric_output
|
Formatted numeric output
|
Task
Express a number in decimal as a fixed-length string with leading zeros.
For example, the number 7.125 could be expressed as 00007.125.
|
#VBA
|
VBA
|
Option Explicit
Sub Main()
Debug.Print fFormat(13, 2, 1230.3333)
Debug.Print fFormat(2, 13, 1230.3333)
Debug.Print fFormat(10, 5, 0.3333)
Debug.Print fFormat(13, 2, 1230)
End Sub
Private Function fFormat(NbInt As Integer, NbDec As Integer, Nb As Double) As String
'NbInt : Lenght of integral part
'NbDec : Lenght of decimal part
'Nb : decimal on integer number
Dim u As String, v As String, i As Integer
u = CStr(Nb)
i = InStr(u, Application.DecimalSeparator)
If i > 0 Then
v = Mid(u, i + 1)
u = Left(u, i - 1)
fFormat = Right(String(NbInt, "0") & u, NbInt) & Application.DecimalSeparator & Left(v & String(NbDec, "0"), NbDec)
Else
fFormat = Right(String(NbInt, "0") & u, NbInt) & Application.DecimalSeparator & String(NbDec, "0")
End If
End Function
|
http://rosettacode.org/wiki/Four_bit_adder
|
Four bit adder
|
Task
"Simulate" a four-bit adder.
This design can be realized using four 1-bit full adders.
Each of these 1-bit full adders can be built with two half adders and an or gate. ;
Finally a half adder can be made using an xor gate and an and gate.
The xor gate can be made using two nots, two ands and one or.
Not, or and and, the only allowed "gates" for the task, can be "imitated" by using the bitwise operators of your language.
If there is not a bit type in your language, to be sure that the not does not "invert" all the other bits of the basic type (e.g. a byte) we are not interested in, you can use an extra nand (and then not) with the constant 1 on one input.
Instead of optimizing and reducing the number of gates used for the final 4-bit adder, build it in the most straightforward way, connecting the other "constructive blocks", in turn made of "simpler" and "smaller" ones.
Schematics of the "constructive blocks"
(Xor gate with ANDs, ORs and NOTs)
(A half adder)
(A full adder)
(A 4-bit adder)
Solutions should try to be as descriptive as possible, making it as easy as possible to identify "connections" between higher-order "blocks".
It is not mandatory to replicate the syntax of higher-order blocks in the atomic "gate" blocks, i.e. basic "gate" operations can be performed as usual bitwise operations, or they can be "wrapped" in a block in order to expose the same syntax of higher-order blocks, at implementers' choice.
To test the implementation, show the sum of two four-bit numbers (in binary).
|
#Raku
|
Raku
|
sub xor ($a, $b) { (($a and not $b) or (not $a and $b)) ?? 1 !! 0 }
sub half-adder ($a, $b) {
return xor($a, $b), ($a and $b);
}
sub full-adder ($a, $b, $c0) {
my ($ha0_s, $ha0_c) = half-adder($c0, $a);
my ($ha1_s, $ha1_c) = half-adder($ha0_s, $b);
return $ha1_s, ($ha0_c or $ha1_c);
}
sub four-bit-adder ($a0, $a1, $a2, $a3, $b0, $b1, $b2, $b3) {
my ($fa0_s, $fa0_c) = full-adder($a0, $b0, 0);
my ($fa1_s, $fa1_c) = full-adder($a1, $b1, $fa0_c);
my ($fa2_s, $fa2_c) = full-adder($a2, $b2, $fa1_c);
my ($fa3_s, $fa3_c) = full-adder($a3, $b3, $fa2_c);
return $fa0_s, $fa1_s, $fa2_s, $fa3_s, $fa3_c;
}
{
use Test;
is four-bit-adder(1, 0, 0, 0, 1, 0, 0, 0), (0, 1, 0, 0, 0), '1 + 1 == 2';
is four-bit-adder(1, 0, 1, 0, 1, 0, 1, 0), (0, 1, 0, 1, 0), '5 + 5 == 10';
is four-bit-adder(1, 0, 0, 1, 1, 1, 1, 0)[4], 1, '7 + 9 == overflow';
}
|
http://rosettacode.org/wiki/Fivenum
|
Fivenum
|
Many big data or scientific programs use boxplots to show distributions of data. In addition, sometimes saving large arrays for boxplots can be impractical and use extreme amounts of RAM. It can be useful to save large arrays as arrays with five numbers to save memory.
For example, the R programming language implements Tukey's five-number summary as the fivenum function.
Task
Given an array of numbers, compute the five-number summary.
Note
While these five numbers can be used to draw a boxplot, statistical packages will typically need extra data.
Moreover, while there is a consensus about the "box" of the boxplot, there are variations among statistical packages for the whiskers.
|
#Modula-2
|
Modula-2
|
MODULE Fivenum;
FROM FormatString IMPORT FormatString;
FROM LongStr IMPORT RealToStr;
FROM Terminal IMPORT WriteString,WriteLn,ReadChar;
PROCEDURE WriteLongReal(v : LONGREAL);
VAR buf : ARRAY[0..63] OF CHAR;
BEGIN
RealToStr(v, buf);
WriteString(buf)
END WriteLongReal;
PROCEDURE WriteArray(arr : ARRAY OF LONGREAL);
VAR i : CARDINAL;
BEGIN
WriteString("[");
FOR i:=0 TO HIGH(arr) DO
WriteLongReal(arr[i]);
WriteString(", ")
END;
WriteString("]")
END WriteArray;
(* Assumes that the input is sorted *)
PROCEDURE Median(x : ARRAY OF LONGREAL; beg,end : CARDINAL) : LONGREAL;
VAR m,cnt : CARDINAL;
BEGIN
cnt := end - beg + 1;
m := cnt / 2;
IF cnt MOD 2 = 1 THEN
RETURN x[beg + m]
END;
RETURN (x[beg + m - 1] + x[beg + m]) / 2.0
END Median;
TYPE Summary = ARRAY[0..4] OF LONGREAL;
PROCEDURE Fivenum(input : ARRAY OF LONGREAL) : Summary;
PROCEDURE Sort();
VAR
i,j : CARDINAL;
t : LONGREAL;
BEGIN
FOR i:=0 TO HIGH(input) DO
FOR j:=0 TO HIGH(input) DO
IF (i#j) AND (input[i] < input[j]) THEN
t := input[i];
input[i] := input[j];
input[j] := t
END
END
END
END Sort;
VAR
result : Summary;
size,m,low : CARDINAL;
BEGIN
size := HIGH(input);
Sort();
result[0] := input[0];
result[2] := Median(input,0,size);
result[4] := input[size];
m := size / 2;
IF (size MOD 2 = 1) THEN
low := m
ELSE
low := m - 1
END;
result[1] := Median(input, 0, m);
result[3] := Median(input, m+1, size);
RETURN result;
END Fivenum;
TYPE
A6 = ARRAY[0..5] OF LONGREAL;
A11 = ARRAY[0..10] OF LONGREAL;
A20 = ARRAY[0..19] OF LONGREAL;
VAR
a6 : A6;
a11 : A11;
a20 : A20;
BEGIN
a11 := A11{15.0, 6.0, 42.0, 41.0, 7.0, 36.0, 49.0, 40.0, 39.0, 47.0, 43.0};
WriteArray(Fivenum(a11));
WriteLn;
WriteLn;
a6 := A6{36.0, 40.0, 7.0, 39.0, 41.0, 15.0};
WriteArray(Fivenum(a6));
WriteLn;
WriteLn;
a20 := A20{
0.14082834, 0.09748790, 1.73131507, 0.87636009, -1.95059594, 0.73438555,
-0.03035726, 1.46675970, -0.74621349, -0.72588772, 0.63905160, 0.61501527,
-0.98983780, -1.00447874, -0.62759469, 0.66206163, 1.04312009, -0.10305385,
0.75775634, 0.32566578
};
WriteArray(Fivenum(a20));
WriteLn;
ReadChar
END Fivenum.
|
http://rosettacode.org/wiki/Find_the_missing_permutation
|
Find the missing permutation
|
ABCD
CABD
ACDB
DACB
BCDA
ACBD
ADCB
CDAB
DABC
BCAD
CADB
CDBA
CBAD
ABDC
ADBC
BDCA
DCBA
BACD
BADC
BDAC
CBDA
DBCA
DCAB
Listed above are all-but-one of the permutations of the symbols A, B, C, and D, except for one permutation that's not listed.
Task
Find that missing permutation.
Methods
Obvious method:
enumerate all permutations of A, B, C, and D,
and then look for the missing permutation.
alternate method:
Hint: if all permutations were shown above, how many
times would A appear in each position?
What is the parity of this number?
another alternate method:
Hint: if you add up the letter values of each column,
does a missing letter A, B, C, and D from each
column cause the total value for each column to be unique?
Related task
Permutations)
|
#Arturo
|
Arturo
|
perms: [
"ABCD" "CABD" "ACDB" "DACB" "BCDA" "ACBD" "ADCB" "CDAB" "DABC"
"BCAD" "CADB" "CDBA" "CBAD" "ABDC" "ADBC" "BDCA" "DCBA" "BACD"
"BADC" "BDAC" "CBDA" "DBCA" "DCAB"
]
allPerms: map permutate split "ABCD" => join
print first difference allPerms perms
|
http://rosettacode.org/wiki/Find_the_last_Sunday_of_each_month
|
Find the last Sunday of each month
|
Write a program or a script that returns the last Sundays of each month of a given year. The year may be given through any simple input method in your language (command line, std in, etc).
Example of an expected output:
./last_sundays 2013
2013-01-27
2013-02-24
2013-03-31
2013-04-28
2013-05-26
2013-06-30
2013-07-28
2013-08-25
2013-09-29
2013-10-27
2013-11-24
2013-12-29
Related tasks
Day of the week
Five weekends
Last Friday of each month
|
#C
|
C
|
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv[])
{
int days[] = {31,29,31,30,31,30,31,31,30,31,30,31};
int m, y, w;
if (argc < 2 || (y = atoi(argv[1])) <= 1752) return 1;
days[1] -= (y % 4) || (!(y % 100) && (y % 400));
w = y * 365 + 97 * (y - 1) / 400 + 4;
for(m = 0; m < 12; m++) {
w = (w + days[m]) % 7;
printf("%d-%02d-%d\n", y, m + 1,days[m] - w);
}
return 0;
}
|
http://rosettacode.org/wiki/Find_the_intersection_of_two_lines
|
Find the intersection of two lines
|
[1]
Task
Find the point of intersection of two lines in 2D.
The 1st line passes though (4,0) and (6,10) .
The 2nd line passes though (0,3) and (10,7) .
|
#C.2B.2B
|
C++
|
#include <iostream>
#include <cmath>
#include <cassert>
using namespace std;
/** Calculate determinant of matrix:
[a b]
[c d]
*/
inline double Det(double a, double b, double c, double d)
{
return a*d - b*c;
}
/// Calculate intersection of two lines.
///\return true if found, false if not found or error
bool LineLineIntersect(double x1, double y1, // Line 1 start
double x2, double y2, // Line 1 end
double x3, double y3, // Line 2 start
double x4, double y4, // Line 2 end
double &ixOut, double &iyOut) // Output
{
double detL1 = Det(x1, y1, x2, y2);
double detL2 = Det(x3, y3, x4, y4);
double x1mx2 = x1 - x2;
double x3mx4 = x3 - x4;
double y1my2 = y1 - y2;
double y3my4 = y3 - y4;
double denom = Det(x1mx2, y1my2, x3mx4, y3my4);
if(denom == 0.0) // Lines don't seem to cross
{
ixOut = NAN;
iyOut = NAN;
return false;
}
double xnom = Det(detL1, x1mx2, detL2, x3mx4);
double ynom = Det(detL1, y1my2, detL2, y3my4);
ixOut = xnom / denom;
iyOut = ynom / denom;
if(!isfinite(ixOut) || !isfinite(iyOut)) // Probably a numerical issue
return false;
return true; //All OK
}
int main()
{
// **Simple crossing diagonal lines**
// Line 1
double x1=4.0, y1=0.0;
double x2=6.0, y2=10.0;
// Line 2
double x3=0.0, y3=3.0;
double x4=10.0, y4=7.0;
double ix = -1.0, iy = -1.0;
bool result = LineLineIntersect(x1, y1, x2, y2, x3, y3, x4, y4, ix, iy);
cout << "result " << result << "," << ix << "," << iy << endl;
double eps = 1e-6;
assert(result == true);
assert(fabs(ix - 5.0) < eps);
assert(fabs(iy - 5.0) < eps);
return 0;
}
|
http://rosettacode.org/wiki/Find_the_intersection_of_a_line_with_a_plane
|
Find the intersection of a line with a plane
|
Finding the intersection of an infinite ray with a plane in 3D is an important topic in collision detection.
Task
Find the point of intersection for the infinite ray with direction (0, -1, -1) passing through position (0, 0, 10) with the infinite plane with a normal vector of (0, 0, 1) and which passes through [0, 0, 5].
|
#C.2B.2B
|
C++
|
#include <iostream>
#include <sstream>
class Vector3D {
public:
Vector3D(double x, double y, double z) {
this->x = x;
this->y = y;
this->z = z;
}
double dot(const Vector3D& rhs) const {
return x * rhs.x + y * rhs.y + z * rhs.z;
}
Vector3D operator-(const Vector3D& rhs) const {
return Vector3D(x - rhs.x, y - rhs.y, z - rhs.z);
}
Vector3D operator*(double rhs) const {
return Vector3D(rhs*x, rhs*y, rhs*z);
}
friend std::ostream& operator<<(std::ostream&, const Vector3D&);
private:
double x, y, z;
};
std::ostream & operator<<(std::ostream & os, const Vector3D &f) {
std::stringstream ss;
ss << "(" << f.x << ", " << f.y << ", " << f.z << ")";
return os << ss.str();
}
Vector3D intersectPoint(Vector3D rayVector, Vector3D rayPoint, Vector3D planeNormal, Vector3D planePoint) {
Vector3D diff = rayPoint - planePoint;
double prod1 = diff.dot(planeNormal);
double prod2 = rayVector.dot(planeNormal);
double prod3 = prod1 / prod2;
return rayPoint - rayVector * prod3;
}
int main() {
Vector3D rv = Vector3D(0.0, -1.0, -1.0);
Vector3D rp = Vector3D(0.0, 0.0, 10.0);
Vector3D pn = Vector3D(0.0, 0.0, 1.0);
Vector3D pp = Vector3D(0.0, 0.0, 5.0);
Vector3D ip = intersectPoint(rv, rp, pn, pp);
std::cout << "The ray intersects the plane at " << ip << std::endl;
return 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
|
#Ada
|
Ada
|
with Ada.Text_IO; use Ada.Text_IO;
procedure Fizzbuzz is
begin
for I in 1..100 loop
if I mod 15 = 0 then
Put_Line("FizzBuzz");
elsif I mod 5 = 0 then
Put_Line("Buzz");
elsif I mod 3 = 0 then
Put_Line("Fizz");
else
Put_Line(Integer'Image(I));
end if;
end loop;
end Fizzbuzz;
|
http://rosettacode.org/wiki/Five_weekends
|
Five weekends
|
The month of October in 2010 has five Fridays, five Saturdays, and five Sundays.
Task
Write a program to show all months that have this same characteristic of five full weekends from the year 1900 through 2100 (Gregorian calendar).
Show the number of months with this property (there should be 201).
Show at least the first and last five dates, in order.
Algorithm suggestions
Count the number of Fridays, Saturdays, and Sundays in every month.
Find all of the 31-day months that begin on Friday.
Extra credit
Count and/or show all of the years which do not have at least one five-weekend month (there should be 29).
Related tasks
Day of the week
Last Friday of each month
Find last sunday of each month
|
#COBOL
|
COBOL
|
program-id. five-we.
data division.
working-storage section.
1 wk binary.
2 int-date pic 9(8).
2 dow pic 9(4).
2 friday pic 9(4) value 5.
2 mo-sub pic 9(4).
2 months-with-5 pic 9(4) value 0.
2 years-no-5 pic 9(4) value 0.
2 5-we-flag pic 9(4) value 0.
88 5-we-true value 1 when false 0.
1 31-day-mos pic 9(14) value 01030507081012.
1 31-day-table redefines 31-day-mos.
2 mo-no occurs 7 pic 99.
1 cal-date.
2 yr pic 9(4).
2 mo pic 9(2).
2 da pic 9(2) value 1.
procedure division.
perform varying yr from 1900 by 1
until yr > 2100
set 5-we-true to false
perform varying mo-sub from 1 by 1
until mo-sub > 7
move mo-no (mo-sub) to mo
compute int-date = function
integer-of-date (function numval (cal-date))
compute dow = function mod
((int-date - 1) 7) + 1
if dow = friday
perform output-date
add 1 to months-with-5
set 5-we-true to true
end-if
end-perform
if not 5-we-true
add 1 to years-no-5
end-if
end-perform
perform output-counts
stop run
.
output-counts.
display "Months with 5 weekends: " months-with-5
display "Years without 5 weekends: " years-no-5
.
output-date.
display yr "-" mo
.
end program five-we.
|
http://rosettacode.org/wiki/First_perfect_square_in_base_n_with_n_unique_digits
|
First perfect square in base n with n unique digits
|
Find the first perfect square in a given base N that has at least N digits and
exactly N significant unique digits when expressed in base N.
E.G. In base 10, the first perfect square with at least 10 unique digits is 1026753849 (32043²).
You may use analytical methods to reduce the search space, but the code must do a search. Do not use magic numbers or just feed the code the answer to verify it is correct.
Task
Find and display here, on this page, the first perfect square in base N, with N significant unique digits when expressed in base N, for each of base 2 through 12. Display each number in the base N for which it was calculated.
(optional) Do the same for bases 13 through 16.
(stretch goal) Continue on for bases 17 - ?? (Big Integer math)
See also
OEIS A260182: smallest square that is pandigital in base n.
Related task
Casting out nines
|
#Mathematica_.2F_Wolfram_Language
|
Mathematica / Wolfram Language
|
ClearAll[FirstSquare]
FirstSquare[b_Integer] := Module[{n, alldigits, digs, start},
digs = Range[0, b - 1];
digs[[{2, 1}]] //= Reverse;
start = Floor[Sqrt[FromDigits[digs, b]]];
n = start;
alldigits = Range[0, b - 1];
While[! ContainsAll[IntegerDigits[n^2, b], alldigits], n++];
{b, n, start, BaseForm[n, b]}
]
Scan[Print@*FirstSquare, Range[2, 16]]
|
http://rosettacode.org/wiki/First_perfect_square_in_base_n_with_n_unique_digits
|
First perfect square in base n with n unique digits
|
Find the first perfect square in a given base N that has at least N digits and
exactly N significant unique digits when expressed in base N.
E.G. In base 10, the first perfect square with at least 10 unique digits is 1026753849 (32043²).
You may use analytical methods to reduce the search space, but the code must do a search. Do not use magic numbers or just feed the code the answer to verify it is correct.
Task
Find and display here, on this page, the first perfect square in base N, with N significant unique digits when expressed in base N, for each of base 2 through 12. Display each number in the base N for which it was calculated.
(optional) Do the same for bases 13 through 16.
(stretch goal) Continue on for bases 17 - ?? (Big Integer math)
See also
OEIS A260182: smallest square that is pandigital in base n.
Related task
Casting out nines
|
#Nim
|
Nim
|
import algorithm, math, strformat
const Alphabet = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"
func toBaseN(num, base: Natural): string =
doAssert(base in 2..Alphabet.len, &"base must be in 2..{Alphabet.len}")
var num = num
while true:
result.add(Alphabet[num mod base])
num = num div base
if num == 0: break
result.reverse()
func countUnique(str: string): int =
var charset: set['0'..'Z']
for ch in str: charset.incl(ch)
result = charset.card
proc find(base: Natural) =
var n = pow(base.toFloat, (base - 1) / 2).int
while true:
let sq = n * n
let sqstr = sq.toBaseN(base)
if sqstr.len >= base and countUnique(sqstr) == base:
let nstr = n.toBaseN(base)
echo &"Base {base:2d}: {nstr:>8s}² = {sqstr:<16s}"
break
inc n
when isMainModule:
for base in 2..16:
base.find()
|
http://rosettacode.org/wiki/First-class_functions
|
First-class functions
|
A language has first-class functions if it can do each of the following without recursively invoking a compiler or interpreter or otherwise metaprogramming:
Create new functions from preexisting functions at run-time
Store functions in collections
Use functions as arguments to other functions
Use functions as return values of other functions
Task
Write a program to create an ordered collection A of functions of a real number. At least one function should be built-in and at least one should be user-defined; try using the sine, cosine, and cubing functions. Fill another collection B with the inverse of each function in A. Implement function composition as in Functional Composition. Finally, demonstrate that the result of applying the composition of each function in A and its inverse in B to a value, is the original value. (Within the limits of computational accuracy).
(A solution need not actually call the collections "A" and "B". These names are only used in the preceding paragraph for clarity.)
Related task
First-class Numbers
|
#Delphi
|
Delphi
|
program First_class_functions;
{$APPTYPE CONSOLE}
uses
System.SysUtils,
System.Math;
type
TFunctionTuple = record
forward, backward: TFunc<Double, Double>;
procedure Assign(forward, backward: TFunc<Double, Double>);
end;
TFunctionTuples = array of TFunctionTuple;
var
cube, croot, fsin, fcos, faSin, faCos: TFunc<Double, Double>;
FunctionTuples: TFunctionTuples;
ft: TFunctionTuple;
{ TFunctionTuple }
procedure TFunctionTuple.Assign(forward, backward: TFunc<Double, Double>);
begin
self.forward := forward;
self.backward := backward;
end;
begin
cube :=
function(x: Double): Double
begin
result := x * x * x;
end;
croot :=
function(x: Double): Double
begin
result := Power(x, 1 / 3.0);
end;
fsin :=
function(x: Double): Double
begin
result := Sin(x);
end;
fcos :=
function(x: Double): Double
begin
result := Cos(x);
end;
faSin :=
function(x: Double): Double
begin
result := ArcSin(x);
end;
faCos :=
function(x: Double): Double
begin
result := ArcCos(x);
end;
SetLength(FunctionTuples, 3);
FunctionTuples[0].Assign(fsin, faSin);
FunctionTuples[1].Assign(fcos, faCos);
FunctionTuples[2].Assign(cube, croot);
for ft in FunctionTuples do
Writeln(ft.backward(ft.forward(0.5)):2:2);
readln;
end.
|
http://rosettacode.org/wiki/Forest_fire
|
Forest fire
|
This page uses content from Wikipedia. The original article was at Forest-fire model. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
Task
Implement the Drossel and Schwabl definition of the forest-fire model.
It is basically a 2D cellular automaton where each cell can be in three distinct states (empty, tree and burning) and evolves according to the following rules (as given by Wikipedia)
A burning cell turns into an empty cell
A tree will burn if at least one neighbor is burning
A tree ignites with probability f even if no neighbor is burning
An empty space fills with a tree with probability p
Neighborhood is the Moore neighborhood; boundary conditions are so that on the boundary the cells are always empty ("fixed" boundary condition).
At the beginning, populate the lattice with empty and tree cells according to a specific probability (e.g. a cell has the probability 0.5 to be a tree). Then, let the system evolve.
Task's requirements do not include graphical display or the ability to change parameters (probabilities p and f ) through a graphical or command line interface.
Related tasks
See Conway's Game of Life
See Wireworld.
|
#Java
|
Java
|
import java.util.Arrays;
import java.util.LinkedList;
import java.util.List;
public class Fire {
private static final char BURNING = 'w'; //w looks like fire, right?
private static final char TREE = 'T';
private static final char EMPTY = '.';
private static final double F = 0.2;
private static final double P = 0.4;
private static final double TREE_PROB = 0.5;
private static List<String> process(List<String> land){
List<String> newLand = new LinkedList<String>();
for(int i = 0; i < land.size(); i++){
String rowAbove, thisRow = land.get(i), rowBelow;
if(i == 0){//first row
rowAbove = null;
rowBelow = land.get(i + 1);
}else if(i == land.size() - 1){//last row
rowBelow = null;
rowAbove = land.get(i - 1);
}else{//middle
rowBelow = land.get(i + 1);
rowAbove = land.get(i - 1);
}
newLand.add(processRows(rowAbove, thisRow, rowBelow));
}
return newLand;
}
private static String processRows(String rowAbove, String thisRow,
String rowBelow){
String newRow = "";
for(int i = 0; i < thisRow.length();i++){
switch(thisRow.charAt(i)){
case BURNING:
newRow+= EMPTY;
break;
case EMPTY:
newRow+= Math.random() < P ? TREE : EMPTY;
break;
case TREE:
String neighbors = "";
if(i == 0){//first char
neighbors+= rowAbove == null ? "" : rowAbove.substring(i, i + 2);
neighbors+= thisRow.charAt(i + 1);
neighbors+= rowBelow == null ? "" : rowBelow.substring(i, i + 2);
if(neighbors.contains(Character.toString(BURNING))){
newRow+= BURNING;
break;
}
}else if(i == thisRow.length() - 1){//last char
neighbors+= rowAbove == null ? "" : rowAbove.substring(i - 1, i + 1);
neighbors+= thisRow.charAt(i - 1);
neighbors+= rowBelow == null ? "" : rowBelow.substring(i - 1, i + 1);
if(neighbors.contains(Character.toString(BURNING))){
newRow+= BURNING;
break;
}
}else{//middle
neighbors+= rowAbove == null ? "" : rowAbove.substring(i - 1, i + 2);
neighbors+= thisRow.charAt(i + 1);
neighbors+= thisRow.charAt(i - 1);
neighbors+= rowBelow == null ? "" : rowBelow.substring(i - 1, i + 2);
if(neighbors.contains(Character.toString(BURNING))){
newRow+= BURNING;
break;
}
}
newRow+= Math.random() < F ? BURNING : TREE;
}
}
return newRow;
}
public static List<String> populate(int width, int height){
List<String> land = new LinkedList<String>();
for(;height > 0; height--){//height is just a copy anyway
StringBuilder line = new StringBuilder(width);
for(int i = width; i > 0; i--){
line.append((Math.random() < TREE_PROB) ? TREE : EMPTY);
}
land.add(line.toString());
}
return land;
}
//process the land n times
public static void processN(List<String> land, int n){
for(int i = 0;i < n; i++){
land = process(land);
}
}
//process the land n times and print each step along the way
public static void processNPrint(List<String> land, int n){
for(int i = 0;i < n; i++){
land = process(land);
print(land);
}
}
//print the land
public static void print(List<String> land){
for(String row: land){
System.out.println(row);
}
System.out.println();
}
public static void main(String[] args){
List<String> land = Arrays.asList(".TTT.T.T.TTTT.T",
"T.T.T.TT..T.T..",
"TT.TTTT...T.TT.",
"TTT..TTTTT.T..T",
".T.TTT....TT.TT",
"...T..TTT.TT.T.",
".TT.TT...TT..TT",
".TT.T.T..T.T.T.",
"..TTT.TT.T..T..",
".T....T.....TTT",
"T..TTT..T..T...",
"TTT....TTTTTT.T",
"......TwTTT...T",
"..T....TTTTTTTT",
".T.T.T....TT...");
print(land);
processNPrint(land, 10);
System.out.println("Random land test:");
land = populate(10, 10);
print(land);
processNPrint(land, 10);
}
}
|
http://rosettacode.org/wiki/First_class_environments
|
First class environments
|
According to Wikipedia, "In computing, a first-class object ... is an entity that can be constructed at run-time, passed as a parameter, returned from a subroutine, or assigned into a variable".
Often this term is used in the context of "first class functions". In an analogous way, a programming language may support "first class environments".
The environment is minimally, the set of variables accessible to a statement being executed. Change the environments and the same statement could produce different results when executed.
Often an environment is captured in a closure, which encapsulates a function together with an environment. That environment, however, is not first-class, as it cannot be created, passed etc. independently from the function's code.
Therefore, a first class environment is a set of variable bindings which can be constructed at run-time, passed as a parameter, returned from a subroutine, or assigned into a variable. It is like a closure without code. A statement must be able to be executed within a stored first class environment and act according to the environment variable values stored within.
Task
Build a dozen environments, and a single piece of code to be run repeatedly in each of these environments.
Each environment contains the bindings for two variables:
a value in the Hailstone sequence, and
a count which is incremented until the value drops to 1.
The initial hailstone values are 1 through 12, and the count in each environment is zero.
When the code runs, it calculates the next hailstone step in the current environment (unless the value is already 1) and counts the steps. Then it prints the current value in a tabular form.
When all hailstone values dropped to 1, processing stops, and the total number of hailstone steps for each environment is printed.
|
#Tcl
|
Tcl
|
package require Tcl 8.5
for {set i 1} {$i <= 12} {incr i} {
dict set hailenv hail$i [dict create num $i steps 0]
}
while 1 {
set loopagain false
foreach k [dict keys $hailenv] {
dict with hailenv $k {
puts -nonewline [format %4d $num]
if {$num == 1} {
continue
} elseif {$num & 1} {
set num [expr {3*$num + 1}]
} else {
set num [expr {$num / 2}]
}
set loopagain true
incr steps
}
}
puts ""
if {!$loopagain} break
}
puts "Counts..."
foreach k [dict keys $hailenv] {
dict with hailenv $k {
puts -nonewline [format %4d $steps]
}
}
puts ""
|
http://rosettacode.org/wiki/First_class_environments
|
First class environments
|
According to Wikipedia, "In computing, a first-class object ... is an entity that can be constructed at run-time, passed as a parameter, returned from a subroutine, or assigned into a variable".
Often this term is used in the context of "first class functions". In an analogous way, a programming language may support "first class environments".
The environment is minimally, the set of variables accessible to a statement being executed. Change the environments and the same statement could produce different results when executed.
Often an environment is captured in a closure, which encapsulates a function together with an environment. That environment, however, is not first-class, as it cannot be created, passed etc. independently from the function's code.
Therefore, a first class environment is a set of variable bindings which can be constructed at run-time, passed as a parameter, returned from a subroutine, or assigned into a variable. It is like a closure without code. A statement must be able to be executed within a stored first class environment and act according to the environment variable values stored within.
Task
Build a dozen environments, and a single piece of code to be run repeatedly in each of these environments.
Each environment contains the bindings for two variables:
a value in the Hailstone sequence, and
a count which is incremented until the value drops to 1.
The initial hailstone values are 1 through 12, and the count in each environment is zero.
When the code runs, it calculates the next hailstone step in the current environment (unless the value is already 1) and counts the steps. Then it prints the current value in a tabular form.
When all hailstone values dropped to 1, processing stops, and the total number of hailstone steps for each environment is printed.
|
#Wren
|
Wren
|
import "/fmt" for Fmt
var environment = Fn.new {
class E {
construct new(value, count) {
_value = value
_count = count
}
value { _value }
count { _count }
hailstone() {
Fmt.write("$4d", _value)
if (_value == 1) return
_count = _count + 1
_value = (_value%2 == 0) ? _value/2 : 3*_value + 1
}
}
return E
}
// create and initialize the environments
var jobs = 12
var envs = List.filled(jobs, null)
for (i in 0...jobs) envs[i] = environment.call().new(i+1, 0)
System.print("Sequences:")
var done = false
while (!done) {
for (env in envs) env.hailstone()
System.print()
done = true
for (env in envs) {
if (env.value != 1) {
done = false
break
}
}
}
System.print("Counts:")
for (env in envs) Fmt.write("$4d", env.count)
System.print()
|
http://rosettacode.org/wiki/Flatten_a_list
|
Flatten a list
|
Task
Write a function to flatten the nesting in an arbitrary list of values.
Your program should work on the equivalent of this list:
[[1], 2, [[3, 4], 5], [[[]]], [[[6]]], 7, 8, []]
Where the correct result would be the list:
[1, 2, 3, 4, 5, 6, 7, 8]
Related task
Tree traversal
|
#Elixir
|
Elixir
|
defmodule RC do
def flatten([]), do: []
def flatten([h|t]), do: flatten(h) ++ flatten(t)
def flatten(h), do: [h]
end
list = [[1], 2, [[3,4], 5], [[[]]], [[[6]]], 7, 8, []]
# Our own implementation
IO.inspect RC.flatten(list)
# Library function
IO.inspect List.flatten(list)
|
http://rosettacode.org/wiki/Flipping_bits_game
|
Flipping bits game
|
The game
Given an N×N square array of zeroes or ones in an initial configuration, and a target configuration of zeroes and ones.
The game is to transform one to the other in as few moves as possible by inverting whole numbered rows or whole lettered
columns at once (as one move).
In an inversion. any 1 becomes 0, and any 0 becomes 1 for that whole row or column.
Task
Create a program to score for the Flipping bits game.
The game should create an original random target configuration and a starting configuration.
Ensure that the starting position is never the target position.
The target position must be guaranteed as reachable from the starting position. (One possible way to do this is to generate the start position by legal flips from a random target position. The flips will always be reversible back to the target from the given start position).
The number of moves taken so far should be shown.
Show an example of a short game here, on this page, for a 3×3 array of bits.
|
#Red
|
Red
|
Red []
random/seed now/time/precise ;; start random generator
kRows: kCols: 3 ;; define board size, 3x3 upto 9x9 possible
;; create series of 3 empty blocks:
loop kRows [ append/only board: [] copy [] ] ;; ( this is actually a bit tricky, normally you'd have to use "copy" [] inside a loop )
kValid: "1A" ;; generate string for input validation "321ABC"
loop (kRows - 1 ) [insert kValid (first kValid) + 1 ]
loop (kCols - 1 ) [append kValid (last kValid) + 1 ]
repeat row kRows [ loop kCols [ append board/:row -1 + random 2 ] ] ;; fill board with random 0 / 1
;;--------------------------------------
xorme: func ['val][ set val 1 xor get val ] ;; function: flip the given board position
;;--------------------------------------
flip: func [ what [string!] ] [ ;; flip complete row or column of board
row: -48 + to-integer first what ;; convert string to integer row/column index
if row <= kRows [ repeat col kCols [ xorme board/:row/:col] return 0 ]
repeat row2 kRows [ xorme board/:row2/(row - 16)]
]
;;--------------------------------------
showboard: func [title [string!] b] [ ;; function: show board name + board or target
prin [title newline newline" " letter: #"A" ] ;; ( prin doesn't print newline at end )
loop ( kCols - 1) [ prin ["" letter: letter + 1] ] print "" ;; print column letters
repeat row kRows [ ;; print one row
prin row ;; first print row number
repeat col kCols [ prin ["" b/:row/:col ]]
print ""
]
]
showboard "Target" target: copy/deep board ;; create target as copy from board and show
random kvalid
repeat pos 3 [flip copy/part skip kvalid pos 1] ;; now flip board 3 times at random row/column
run: -1
forever [
showboard "Board" board
if board = target [ Print ["You solved it in" run + 1 "move(s) !" ] halt ] ;; count last move
print [newline "moves:" run: run + 1 ] ;; show moves taken so far
until [ find kvalid inp: uppercase ask "Enter Row No or Column Letter to flip ?" ] ;; read valid input character
flip inp
] ;; 42 lines :- )
|
http://rosettacode.org/wiki/First_power_of_2_that_has_leading_decimal_digits_of_12
|
First power of 2 that has leading decimal digits of 12
|
(This task is taken from a Project Euler problem.)
(All numbers herein are expressed in base ten.)
27 = 128 and 7 is
the first power of 2 whose leading decimal digits are 12.
The next power of 2 whose leading decimal digits
are 12 is 80,
280 = 1208925819614629174706176.
Define p(L,n) to be the nth-smallest
value of j such that the base ten representation
of 2j begins with the digits of L .
So p(12, 1) = 7 and
p(12, 2) = 80
You are also given that:
p(123, 45) = 12710
Task
find:
p(12, 1)
p(12, 2)
p(123, 45)
p(123, 12345)
p(123, 678910)
display the results here, on this page.
|
#Ruby
|
Ruby
|
def p(l, n)
test = 0
logv = Math.log(2.0) / Math.log(10.0)
factor = 1
loopv = l
while loopv > 10 do
factor = factor * 10
loopv = loopv / 10
end
while n > 0 do
test = test + 1
val = (factor * (10.0 ** ((test * logv).modulo(1.0)))).floor
if val == l then
n = n - 1
end
end
return test
end
def runTest(l, n)
print "P(%d, %d) = %d\n" % [l, n, p(l, n)]
end
runTest(12, 1)
runTest(12, 2)
runTest(123, 45)
runTest(123, 12345)
runTest(123, 678910)
|
http://rosettacode.org/wiki/First-class_functions/Use_numbers_analogously
|
First-class functions/Use numbers analogously
|
In First-class functions, a language is showing how its manipulation of functions is similar to its manipulation of other types.
This tasks aim is to compare and contrast a language's implementation of first class functions, with its normal handling of numbers.
Write a program to create an ordered collection of a mixture of literally typed and expressions producing a real number, together with another ordered collection of their multiplicative inverses. Try and use the following pseudo-code to generate the numbers for the ordered collections:
x = 2.0
xi = 0.5
y = 4.0
yi = 0.25
z = x + y
zi = 1.0 / ( x + y )
Create a function multiplier, that given two numbers as arguments returns a function that when called with one argument, returns the result of multiplying the two arguments to the call to multiplier that created it and the argument in the call:
new_function = multiplier(n1,n2)
# where new_function(m) returns the result of n1 * n2 * m
Applying the multiplier of a number and its inverse from the two ordered collections of numbers in pairs, show that the result in each case is one.
Compare and contrast the resultant program with the corresponding entry in First-class functions. They should be close.
To paraphrase the task description: Do what was done before, but with numbers rather than functions
|
#R
|
R
|
multiplier <- function(n1,n2) { (function(m){n1*n2*m}) }
x = 2.0
xi = 0.5
y = 4.0
yi = 0.25
z = x + y
zi = 1.0 / ( x + y )
num = c(x,y,z)
inv = c(xi,yi,zi)
multiplier(num,inv)(0.5)
Output
[1] 0.5 0.5 0.5
|
http://rosettacode.org/wiki/First-class_functions/Use_numbers_analogously
|
First-class functions/Use numbers analogously
|
In First-class functions, a language is showing how its manipulation of functions is similar to its manipulation of other types.
This tasks aim is to compare and contrast a language's implementation of first class functions, with its normal handling of numbers.
Write a program to create an ordered collection of a mixture of literally typed and expressions producing a real number, together with another ordered collection of their multiplicative inverses. Try and use the following pseudo-code to generate the numbers for the ordered collections:
x = 2.0
xi = 0.5
y = 4.0
yi = 0.25
z = x + y
zi = 1.0 / ( x + y )
Create a function multiplier, that given two numbers as arguments returns a function that when called with one argument, returns the result of multiplying the two arguments to the call to multiplier that created it and the argument in the call:
new_function = multiplier(n1,n2)
# where new_function(m) returns the result of n1 * n2 * m
Applying the multiplier of a number and its inverse from the two ordered collections of numbers in pairs, show that the result in each case is one.
Compare and contrast the resultant program with the corresponding entry in First-class functions. They should be close.
To paraphrase the task description: Do what was done before, but with numbers rather than functions
|
#Racket
|
Racket
|
#lang racket
(define x 2.0)
(define xi 0.5)
(define y 4.0)
(define yi 0.25)
(define z (+ x y))
(define zi (/ 1.0 (+ x y)))
(define ((multiplier x y) z) (* x y z))
(define numbers (list x y z))
(define inverses (list xi yi zi))
(for/list ([n numbers] [i inverses])
((multiplier n i) 0.5))
;; -> '(0.5 0.5 0.5)
|
http://rosettacode.org/wiki/Flow-control_structures
|
Flow-control structures
|
Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Task
Document common flow-control structures.
One common example of a flow-control structure is the goto construct.
Note that Conditional Structures and Loop Structures have their own articles/categories.
Related tasks
Conditional Structures
Loop Structures
|
#REXX
|
REXX
|
call routineName /*no arguments passed to routine.*/
call routineName 50 /*one argument (fifty) passed. */
call routineName 50,60 /*two arguments passed. */
call routineName 50, 60 /*(same as above) */
call routineName 50 ,60 /*(same as above) */
call routineName 10*5 , 8**4 - 4 /*(same as above) */
call routineName 50 , , , 70 /*4 args passed, 2nd&3rd omitted.*/
/*omitted args are NOT null. */
call routineName ,,,,,,,,,,,,,,,,800 /*17 args passed, 16 omitted. */
call date /*looks for DATE internally first*/
call 'DATE' /* " " " BIF | externally*/
|
http://rosettacode.org/wiki/Flow-control_structures
|
Flow-control structures
|
Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Task
Document common flow-control structures.
One common example of a flow-control structure is the goto construct.
Note that Conditional Structures and Loop Structures have their own articles/categories.
Related tasks
Conditional Structures
Loop Structures
|
#Ring
|
Ring
|
i = 1
while true
see i + nl
if i = 10 see "Break!" exit ok
i = i + 1
end
|
http://rosettacode.org/wiki/Floyd%27s_triangle
|
Floyd's triangle
|
Floyd's triangle lists the natural numbers in a right triangle aligned to the left where
the first row is 1 (unity)
successive rows start towards the left with the next number followed by successive naturals listing one more number than the line above.
The first few lines of a Floyd triangle looks like this:
1
2 3
4 5 6
7 8 9 10
11 12 13 14 15
Task
Write a program to generate and display here the first n lines of a Floyd triangle.
(Use n=5 and n=14 rows).
Ensure that when displayed in a mono-space font, the numbers line up in vertical columns as shown and that only one space separates numbers of the last row.
|
#Gambas
|
Gambas
|
Public Sub Main()
Dim siCount, siNo, siCounter As Short
Dim siLine As Short = 1
Dim siInput As Short[] = [5, 14]
For siCount = 0 To siInput.Max
Print "Floyd's triangle to " & siInput[siCount] & " lines"
Do
Inc siNo
Inc siCounter
Print Format(siNo, "####");
If siLine = siCounter Then
Print
Inc siLine
siCounter = 0
End If
If siLine - 1 = siInput[siCount] Then Break
Loop
siLine = 1
siCounter = 0
siNo = 0
Print
Next
End
|
http://rosettacode.org/wiki/Floyd-Warshall_algorithm
|
Floyd-Warshall algorithm
|
The Floyd–Warshall algorithm is an algorithm for finding shortest paths in a weighted graph with positive or negative edge weights.
Task
Find the lengths of the shortest paths between all pairs of vertices of the given directed graph. Your code may assume that the input has already been checked for loops, parallel edges and negative cycles.
Print the pair, the distance and (optionally) the path.
Example
pair dist path
1 -> 2 -1 1 -> 3 -> 4 -> 2
1 -> 3 -2 1 -> 3
1 -> 4 0 1 -> 3 -> 4
2 -> 1 4 2 -> 1
2 -> 3 2 2 -> 1 -> 3
2 -> 4 4 2 -> 1 -> 3 -> 4
3 -> 1 5 3 -> 4 -> 2 -> 1
3 -> 2 1 3 -> 4 -> 2
3 -> 4 2 3 -> 4
4 -> 1 3 4 -> 2 -> 1
4 -> 2 -1 4 -> 2
4 -> 3 1 4 -> 2 -> 1 -> 3
See also
Floyd-Warshall Algorithm - step by step guide (youtube)
|
#PHP
|
PHP
|
<?php
$graph = array();
for ($i = 0; $i < 10; ++$i) {
$graph[] = array();
for ($j = 0; $j < 10; ++$j)
$graph[$i][] = $i == $j ? 0 : 9999999;
}
for ($i = 1; $i < 10; ++$i) {
$graph[0][$i] = $graph[$i][0] = rand(1, 9);
}
for ($k = 0; $k < 10; ++$k) {
for ($i = 0; $i < 10; ++$i) {
for ($j = 0; $j < 10; ++$j) {
if ($graph[$i][$j] > $graph[$i][$k] + $graph[$k][$j])
$graph[$i][$j] = $graph[$i][$k] + $graph[$k][$j];
}
}
}
print_r($graph);
?>
|
http://rosettacode.org/wiki/Function_definition
|
Function definition
|
A function is a body of code that returns a value.
The value returned may depend on arguments provided to the function.
Task
Write a definition of a function called "multiply" that takes two arguments and returns their product.
(Argument types should be chosen so as not to distract from showing how functions are created and values returned).
Related task
Function prototype
|
#Toka
|
Toka
|
[ ( ab-c ) * ] is multiply
|
http://rosettacode.org/wiki/Function_definition
|
Function definition
|
A function is a body of code that returns a value.
The value returned may depend on arguments provided to the function.
Task
Write a definition of a function called "multiply" that takes two arguments and returns their product.
(Argument types should be chosen so as not to distract from showing how functions are created and values returned).
Related task
Function prototype
|
#Transd
|
Transd
|
multiply: (lambda a Double() b Double() (* a b))
|
http://rosettacode.org/wiki/Forward_difference
|
Forward difference
|
Task
Provide code that produces a list of numbers which is the nth order forward difference, given a non-negative integer (specifying the order) and a list of numbers.
The first-order forward difference of a list of numbers A is a new list B, where Bn = An+1 - An.
List B should have one fewer element as a result.
The second-order forward difference of A will be:
tdefmodule Diff do
def forward(arr,i\\1) do
forward(arr,[],i)
end
def forward([_|[]],diffs,i) do
if i == 1 do
IO.inspect diffs
else
forward(diffs,[],i-1)
end
end
def forward([val1|[val2|vals]],diffs,i) do
forward([val2|vals],diffs++[val2-val1],i)
end
end
The same as the first-order forward difference of B.
That new list will have two fewer elements than A and one less than B.
The goal of this task is to repeat this process up to the desired order.
For a more formal description, see the related Mathworld article.
Algorithmic options
Iterate through all previous forward differences and re-calculate a new array each time.
Use this formula (from Wikipedia):
Δ
n
[
f
]
(
x
)
=
∑
k
=
0
n
(
n
k
)
(
−
1
)
n
−
k
f
(
x
+
k
)
{\displaystyle \Delta ^{n}[f](x)=\sum _{k=0}^{n}{n \choose k}(-1)^{n-k}f(x+k)}
(Pascal's Triangle may be useful for this option.)
|
#Python
|
Python
|
>>> dif = lambda s: [x-s[i] for i,x in enumerate(s[1:])]
>>> # or, dif = lambda s: [x-y for x,y in zip(s[1:],s)]
>>> difn = lambda s, n: difn(dif(s), n-1) if n else s
>>> s = [90, 47, 58, 29, 22, 32, 55, 5, 55, 73]
>>> difn(s, 0)
[90, 47, 58, 29, 22, 32, 55, 5, 55, 73]
>>> difn(s, 1)
[-43, 11, -29, -7, 10, 23, -50, 50, 18]
>>> difn(s, 2)
[54, -40, 22, 17, 13, -73, 100, -32]
>>> from pprint import pprint
>>> pprint( [difn(s, i) for i in xrange(10)] )
[[90, 47, 58, 29, 22, 32, 55, 5, 55, 73],
[-43, 11, -29, -7, 10, 23, -50, 50, 18],
[54, -40, 22, 17, 13, -73, 100, -32],
[-94, 62, -5, -4, -86, 173, -132],
[156, -67, 1, -82, 259, -305],
[-223, 68, -83, 341, -564],
[291, -151, 424, -905],
[-442, 575, -1329],
[1017, -1904],
[-2921]]
|
http://rosettacode.org/wiki/Formatted_numeric_output
|
Formatted numeric output
|
Task
Express a number in decimal as a fixed-length string with leading zeros.
For example, the number 7.125 could be expressed as 00007.125.
|
#VBScript
|
VBScript
|
a = 1234.5678
' Round to three decimal places. Groups by default. Output = "1,234.568".
WScript.Echo FormatNumber(a, 3)
' Truncate to three decimal places. Output = "1234.567".
WScript.Echo Left(a, InStr(a, ".") + 3)
' Round to a whole number. Grouping disabled. Output = "1235".
WScript.Echo FormatNumber(a, 0, , , False)
' Use integer portion only and pad with zeroes to fill 8 chars. Output = "00001234".
WScript.Echo Right("00000000" & Int(a), 8)
|
http://rosettacode.org/wiki/Formatted_numeric_output
|
Formatted numeric output
|
Task
Express a number in decimal as a fixed-length string with leading zeros.
For example, the number 7.125 could be expressed as 00007.125.
|
#Vedit_macro_language
|
Vedit macro language
|
#1 = 7125
Num_Ins(#1, FILL+COUNT, 9) Char(-3) Ins_Char('.')
|
http://rosettacode.org/wiki/Four_bit_adder
|
Four bit adder
|
Task
"Simulate" a four-bit adder.
This design can be realized using four 1-bit full adders.
Each of these 1-bit full adders can be built with two half adders and an or gate. ;
Finally a half adder can be made using an xor gate and an and gate.
The xor gate can be made using two nots, two ands and one or.
Not, or and and, the only allowed "gates" for the task, can be "imitated" by using the bitwise operators of your language.
If there is not a bit type in your language, to be sure that the not does not "invert" all the other bits of the basic type (e.g. a byte) we are not interested in, you can use an extra nand (and then not) with the constant 1 on one input.
Instead of optimizing and reducing the number of gates used for the final 4-bit adder, build it in the most straightforward way, connecting the other "constructive blocks", in turn made of "simpler" and "smaller" ones.
Schematics of the "constructive blocks"
(Xor gate with ANDs, ORs and NOTs)
(A half adder)
(A full adder)
(A 4-bit adder)
Solutions should try to be as descriptive as possible, making it as easy as possible to identify "connections" between higher-order "blocks".
It is not mandatory to replicate the syntax of higher-order blocks in the atomic "gate" blocks, i.e. basic "gate" operations can be performed as usual bitwise operations, or they can be "wrapped" in a block in order to expose the same syntax of higher-order blocks, at implementers' choice.
To test the implementation, show the sum of two four-bit numbers (in binary).
|
#REXX
|
REXX
|
/*REXX program displays (all) the sums of a full 4─bit adder (with carry). */
call hdr1; call hdr2 /*note the order of headers & trailers.*/
/* [↓] traipse thru all possibilities.*/
do j=0 for 16
do m=0 for 4; a.m= bit(j, m)
end /*m*/
do k=0 for 16
do m=0 for 4; b.m= bit(k, m)
end /*m*/
sc= 4bitAdder(a., b.)
z= a.3 a.2 a.1 a.0 '~+~' b.3 b.2 b.1 b.0 "~=~" sc ',' s.3 s.2 s.1 s.0
say translate( space(z, 0), , '~') /*translate tildes (~) to blanks in Z. */
end /*k*/
end /*j*/
call hdr2; call hdr1 /*display two trailers (note the order)*/
exit 0 /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
bit: procedure; parse arg x,y; return substr( reverse( x2b( d2x(x) ) ), y+1, 1)
halfAdder: procedure expose c; parse arg x,y; c= x & y; return x && y
hdr1: say 'aaaa + bbbb = c, sum [c=carry]'; return
hdr2: say '════ ════ ══════' ; return
/*──────────────────────────────────────────────────────────────────────────────────────*/
fullAdder: procedure expose c; parse arg x,y,fc
#1= halfAdder(fc, x); c1= c
#2= halfAdder(#1, y); c= c | c1; return #2
/*──────────────────────────────────────────────────────────────────────────────────────*/
4bitAdder: procedure expose s. a. b.; carry.= 0
do j=0 for 4; n= j - 1
s.j= fullAdder(a.j, b.j, carry.n); carry.j= c
end /*j*/; return c
|
http://rosettacode.org/wiki/Fivenum
|
Fivenum
|
Many big data or scientific programs use boxplots to show distributions of data. In addition, sometimes saving large arrays for boxplots can be impractical and use extreme amounts of RAM. It can be useful to save large arrays as arrays with five numbers to save memory.
For example, the R programming language implements Tukey's five-number summary as the fivenum function.
Task
Given an array of numbers, compute the five-number summary.
Note
While these five numbers can be used to draw a boxplot, statistical packages will typically need extra data.
Moreover, while there is a consensus about the "box" of the boxplot, there are variations among statistical packages for the whiskers.
|
#Nim
|
Nim
|
import algorithm
type FiveNum = array[5, float]
template isOdd(n: SomeInteger): bool = (n and 1) != 0
func median(x: openArray[float]; startIndex, endIndex: Natural): float =
let size = endIndex - startIndex + 1
assert(size > 0, "array slice cannot be empty")
let m = startIndex + size div 2
result = if size.isOdd: x[m] else: (x[m-1] + x[m]) / 2
func fivenum(x: openArray[float]): FiveNum =
let x = sorted(x)
let m = x.len div 2
let lowerEnd = if x.len.isOdd: m else: m - 1
result[0] = x[0]
result[1] = median(x, 0, lowerEnd)
result[2] = median(x, 0, x.high)
result[3] = median(x, m, x.high)
result[4] = x[^1]
const Lists = [@[15.0, 6.0, 42.0, 41.0, 7.0, 36.0, 49.0, 40.0, 39.0, 47.0, 43.0],
@[36.0, 40.0, 7.0, 39.0, 41.0, 15.0],
@[0.14082834, 0.09748790, 1.73131507, 0.87636009, -1.95059594,
0.73438555, -0.03035726, 1.46675970, -0.74621349, -0.72588772,
0.63905160, 0.61501527, -0.98983780, -1.00447874, -0.62759469,
0.66206163, 1.04312009, -0.10305385, 0.75775634, 0.32566578]]
for list in Lists:
echo ""
echo list
echo " → ", list.fivenum
|
http://rosettacode.org/wiki/Fivenum
|
Fivenum
|
Many big data or scientific programs use boxplots to show distributions of data. In addition, sometimes saving large arrays for boxplots can be impractical and use extreme amounts of RAM. It can be useful to save large arrays as arrays with five numbers to save memory.
For example, the R programming language implements Tukey's five-number summary as the fivenum function.
Task
Given an array of numbers, compute the five-number summary.
Note
While these five numbers can be used to draw a boxplot, statistical packages will typically need extra data.
Moreover, while there is a consensus about the "box" of the boxplot, there are variations among statistical packages for the whiskers.
|
#Perl
|
Perl
|
use POSIX qw(ceil floor);
sub fivenum {
my(@array) = @_;
my $n = scalar @array;
die "No values were entered into fivenum!" if $n == 0;
my @x = sort {$a <=> $b} @array;
my $n4 = floor(($n+3)/2)/2;
my @d = (1, $n4, ($n +1)/2, $n+1-$n4, $n);
my @sum_array;
for my $e (0..4) {
my $floor = floor($d[$e]-1);
my $ceil = ceil($d[$e]-1);
push @sum_array, (0.5 * ($x[$floor] + $x[$ceil]));
}
return @sum_array;
}
my @x = (15, 6, 42, 41, 7, 36, 49, 40, 39, 47, 43);
my @tukey = fivenum(\@x);
say join (',', @tukey);
#----------
@x = (36, 40, 7, 39, 41, 15),
@tukey = fivenum(\@x);
say join (',', @tukey);
#----------
@x = (0.14082834, 0.09748790, 1.73131507, 0.87636009, -1.95059594,
0.73438555, -0.03035726, 1.46675970, -0.74621349, -0.72588772,
0.63905160, 0.61501527, -0.98983780, -1.00447874, -0.62759469,
0.66206163, 1.04312009, -0.10305385, 0.75775634, 0.32566578);
@tukey = fivenum(\@x);
say join (',', @tukey);
|
http://rosettacode.org/wiki/Find_the_missing_permutation
|
Find the missing permutation
|
ABCD
CABD
ACDB
DACB
BCDA
ACBD
ADCB
CDAB
DABC
BCAD
CADB
CDBA
CBAD
ABDC
ADBC
BDCA
DCBA
BACD
BADC
BDAC
CBDA
DBCA
DCAB
Listed above are all-but-one of the permutations of the symbols A, B, C, and D, except for one permutation that's not listed.
Task
Find that missing permutation.
Methods
Obvious method:
enumerate all permutations of A, B, C, and D,
and then look for the missing permutation.
alternate method:
Hint: if all permutations were shown above, how many
times would A appear in each position?
What is the parity of this number?
another alternate method:
Hint: if you add up the letter values of each column,
does a missing letter A, B, C, and D from each
column cause the total value for each column to be unique?
Related task
Permutations)
|
#AutoHotkey
|
AutoHotkey
|
IncompleteList := "ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC BCAD CADB CDBA CBAD ABDC ADBC BDCA DCBA BACD BADC BDAC CBDA DBCA DCAB"
CompleteList := Perm( "ABCD" )
Missing := ""
Loop, Parse, CompleteList, `n, `r
If !InStr( IncompleteList , A_LoopField )
Missing .= "`n" A_LoopField
MsgBox Missing Permutation(s):%Missing%
;-------------------------------------------------
; Shortened version of [VxE]'s permutation function
; http://www.autohotkey.com/forum/post-322251.html#322251
Perm( s , dL="" , t="" , p="") {
StringSplit, m, s, % d := SubStr(dL,1,1) , %t%
IfEqual, m0, 1, return m1 d p
Loop %m0%
{
r := m1
Loop % m0-2
x := A_Index + 1, r .= d m%x%
L .= Perm(r, d, t, m%m0% d p)"`n" , mx := m1
Loop % m0-1
x := A_Index + 1, m%A_Index% := m%x%
m%m0% := mx
}
return substr(L, 1, -1)
}
|
http://rosettacode.org/wiki/Find_the_last_Sunday_of_each_month
|
Find the last Sunday of each month
|
Write a program or a script that returns the last Sundays of each month of a given year. The year may be given through any simple input method in your language (command line, std in, etc).
Example of an expected output:
./last_sundays 2013
2013-01-27
2013-02-24
2013-03-31
2013-04-28
2013-05-26
2013-06-30
2013-07-28
2013-08-25
2013-09-29
2013-10-27
2013-11-24
2013-12-29
Related tasks
Day of the week
Five weekends
Last Friday of each month
|
#C.23
|
C#
|
using System;
namespace LastSundayOfEachMonth
{
class Program
{
static void Main()
{
Console.Write("Year to calculate: ");
string strYear = Console.ReadLine();
int year = Convert.ToInt32(strYear);
DateTime date;
for (int i = 1; i <= 12; i++)
{
date = new DateTime(year, i, DateTime.DaysInMonth(year, i), System.Globalization.CultureInfo.CurrentCulture.Calendar);
/* Modification by Albert Zakhia on 2021-16-02
The below code is very slow due to the loop, we will go twice as fast
while (date.DayOfWeek != DayOfWeek.Sunday)
{
date = date.AddDays(-1);
}
*/
// The updated code
int daysOffset = date.DayOfWeek - dayOfWeek; // take the offset to subtract directly instead of looping
if (daysOffset < 0) daysOffset += 7; // if the code is negative, we need to normalize them
date = date.AddDays(-daysOffset ); // now just add the days offset
Console.WriteLine(date.ToString("yyyy-MM-dd"));
}
}
}
}
|
http://rosettacode.org/wiki/Find_the_intersection_of_two_lines
|
Find the intersection of two lines
|
[1]
Task
Find the point of intersection of two lines in 2D.
The 1st line passes though (4,0) and (6,10) .
The 2nd line passes though (0,3) and (10,7) .
|
#Clojure
|
Clojure
|
;; Point is [x y] tuple
(defn compute-line [pt1 pt2]
(let [[x1 y1] pt1
[x2 y2] pt2
m (/ (- y2 y1) (- x2 x1))]
{:slope m
:offset (- y1 (* m x1))}))
(defn intercept [line1 line2]
(let [x (/ (- (:offset line1) (:offset line2))
(- (:slope line2) (:slope line1)))]
{:x x
:y (+ (* (:slope line1) x)
(:offset line1))}))
|
http://rosettacode.org/wiki/Find_the_intersection_of_two_lines
|
Find the intersection of two lines
|
[1]
Task
Find the point of intersection of two lines in 2D.
The 1st line passes though (4,0) and (6,10) .
The 2nd line passes though (0,3) and (10,7) .
|
#Common_Lisp
|
Common Lisp
|
;; Point is [x y] tuple
(defun point-of-intersection (x1 y1 x2 y2 x3 y3 x4 y4)
"Find the point of intersection of the lines defined by the points (x1 y1) (x2 y2) and (x3 y3) (x4 y4)"
(let* ((dx1 (- x2 x1))
(dx2 (- x4 x3))
(dy1 (- y2 y1))
(dy2 (- y4 y3))
(den (- (* dy1 dx2) (* dy2 dx1))) )
(unless (zerop den)
(list (/ (+ (* (- y3 y1) dx1 dx2) (* x1 dy1 dx2) (* -1 x3 dy2 dx1)) den)
(/ (+ (* (+ x3 x1) dy1 dy2) (* -1 y1 dx1 dy2) (* y3 dx2 dy1)) den) ))))
|
http://rosettacode.org/wiki/Find_the_intersection_of_a_line_with_a_plane
|
Find the intersection of a line with a plane
|
Finding the intersection of an infinite ray with a plane in 3D is an important topic in collision detection.
Task
Find the point of intersection for the infinite ray with direction (0, -1, -1) passing through position (0, 0, 10) with the infinite plane with a normal vector of (0, 0, 1) and which passes through [0, 0, 5].
|
#D
|
D
|
import std.stdio;
struct Vector3D {
private real x;
private real y;
private real z;
this(real x, real y, real z) {
this.x = x;
this.y = y;
this.z = z;
}
auto opBinary(string op)(Vector3D rhs) const {
static if (op == "+" || op == "-") {
mixin("return Vector3D(x" ~ op ~ "rhs.x, y" ~ op ~ "rhs.y, z" ~ op ~ "rhs.z);");
}
}
auto opBinary(string op : "*")(real s) const {
return Vector3D(s*x, s*y, s*z);
}
auto dot(Vector3D rhs) const {
return x*rhs.x + y*rhs.y + z*rhs.z;
}
void toString(scope void delegate(const(char)[]) sink) const {
import std.format;
sink("(");
formattedWrite!"%f"(sink, x);
sink(",");
formattedWrite!"%f"(sink, y);
sink(",");
formattedWrite!"%f"(sink, z);
sink(")");
}
}
auto intersectPoint(Vector3D rayVector, Vector3D rayPoint, Vector3D planeNormal, Vector3D planePoint) {
auto diff = rayPoint - planePoint;
auto prod1 = diff.dot(planeNormal);
auto prod2 = rayVector.dot(planeNormal);
auto prod3 = prod1 / prod2;
return rayPoint - rayVector * prod3;
}
void main() {
auto rv = Vector3D(0.0, -1.0, -1.0);
auto rp = Vector3D(0.0, 0.0, 10.0);
auto pn = Vector3D(0.0, 0.0, 1.0);
auto pp = Vector3D(0.0, 0.0, 5.0);
auto ip = intersectPoint(rv, rp, pn, pp);
writeln("The ray intersects the plane at ", ip);
}
|
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
|
#ALGOL_68
|
ALGOL 68
|
main:(
FOR i TO 100 DO
printf(($gl$,
IF i %* 15 = 0 THEN
"FizzBuzz"
ELIF i %* 3 = 0 THEN
"Fizz"
ELIF i %* 5 = 0 THEN
"Buzz"
ELSE
i
FI
))
OD
)
|
http://rosettacode.org/wiki/Five_weekends
|
Five weekends
|
The month of October in 2010 has five Fridays, five Saturdays, and five Sundays.
Task
Write a program to show all months that have this same characteristic of five full weekends from the year 1900 through 2100 (Gregorian calendar).
Show the number of months with this property (there should be 201).
Show at least the first and last five dates, in order.
Algorithm suggestions
Count the number of Fridays, Saturdays, and Sundays in every month.
Find all of the 31-day months that begin on Friday.
Extra credit
Count and/or show all of the years which do not have at least one five-weekend month (there should be 29).
Related tasks
Day of the week
Last Friday of each month
Find last sunday of each month
|
#CoffeeScript
|
CoffeeScript
|
startsOnFriday = (month, year) ->
# 0 is Sunday, 1 is Monday, ... 5 is Friday, 6 is Saturday
new Date(year, month, 1).getDay() == 5
has31Days = (month, year) ->
new Date(year, month, 31).getDate() == 31
checkMonths = (year) ->
month = undefined
count = 0
month = 0
while month < 12
if startsOnFriday(month, year) and has31Days(month, year)
count += 1
console.log year + ' ' + month + ''
month += 1
count
fiveWeekends = ->
startYear = 1900
endYear = 2100
year = undefined
monthTotal = 0
yearsWithoutFiveWeekends = []
total = 0
year = startYear
while year <= endYear
monthTotal = checkMonths(year)
total += monthTotal
# extra credit
if monthTotal == 0
yearsWithoutFiveWeekends.push year
year += 1
console.log 'Total number of months: ' + total + ''
console.log ''
console.log yearsWithoutFiveWeekends + ''
console.log 'Years with no five-weekend months: ' + yearsWithoutFiveWeekends.length + ''
return
fiveWeekends()
|
http://rosettacode.org/wiki/First_perfect_square_in_base_n_with_n_unique_digits
|
First perfect square in base n with n unique digits
|
Find the first perfect square in a given base N that has at least N digits and
exactly N significant unique digits when expressed in base N.
E.G. In base 10, the first perfect square with at least 10 unique digits is 1026753849 (32043²).
You may use analytical methods to reduce the search space, but the code must do a search. Do not use magic numbers or just feed the code the answer to verify it is correct.
Task
Find and display here, on this page, the first perfect square in base N, with N significant unique digits when expressed in base N, for each of base 2 through 12. Display each number in the base N for which it was calculated.
(optional) Do the same for bases 13 through 16.
(stretch goal) Continue on for bases 17 - ?? (Big Integer math)
See also
OEIS A260182: smallest square that is pandigital in base n.
Related task
Casting out nines
|
#Pascal
|
Pascal
|
program project1;
//Find the smallest number n to base b, so that n*n includes all
//digits of base b
{$IFDEF FPC}{$MODE DELPHI}{$ENDIF}
uses
sysutils;
const
charSet : array[0..36] of char ='0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ';
type
tNumtoBase = record
ntb_dgt : array[0..31-4] of byte;
ntb_cnt,
ntb_bas : Word;
end;
var
Num,
sqr2B,
deltaNum : tNumtoBase;
function Minimal_n(base:NativeUint):Uint64;
//' 1023456789ABCDEFGHIJ...'
var
i : NativeUint;
Begin
result := base; // aka '10'
IF base > 2 then
For i := 2 to base-1 do
result := result*base+i;
result := trunc(sqrt(result)+0.99999);
end;
procedure Conv2num(var num:tNumtoBase;n:Uint64;base:NativeUint);
var
quot :UInt64;
i :NativeUint;
Begin
i := 0;
repeat
quot := n div base;
Num.ntb_dgt[i] := n-quot*base;
n := quot;
inc(i);
until n = 0;
Num.ntb_cnt := i;
Num.ntb_bas := base;
//clear upper digits
For i := i to high(tNumtoBase.ntb_dgt) do
Num.ntb_dgt[i] := 0;
end;
procedure OutNum(const num:tNumtoBase);
var
i : NativeInt;
Begin
with num do
Begin
For i := 17-ntb_cnt-1 downto 0 do
write(' ');
For i := ntb_cnt-1 downto 0 do
write(charSet[ntb_dgt[i]]);
end;
end;
procedure IncNumBig(var add1:tNumtoBase;n:NativeUInt);
//prerequisites
//bases are the same,delta : NativeUint
var
i,s,b,carry : NativeInt;
Begin
b := add1.ntb_bas;
i := 0;
carry := 0;
while n > 0 do
Begin
s := add1.ntb_dgt[i]+carry+ n MOD b;
carry := Ord(s>=b);
s := s- (-carry AND b);
add1.ntb_dgt[i] := s;
n := n div b;
inc(i);
end;
while carry <> 0 do
Begin
s := add1.ntb_dgt[i]+carry;
carry := Ord(s>=b);
s := s- (-carry AND b);
add1.ntb_dgt[i] := s;
inc(i);
end;
IF add1.ntb_cnt < i then
add1.ntb_cnt := i;
end;
procedure IncNum(var add1:tNumtoBase;carry:NativeInt);
//prerequisites: bases are the same, carry==delta < base
var
i,s,b : NativeInt;
Begin
b := add1.ntb_bas;
i := 0;
while carry <> 0 do
Begin
s := add1.ntb_dgt[i]+carry;
carry := Ord(s>=b);
s := s- (-carry AND b);
add1.ntb_dgt[i] := s;
inc(i);
end;
IF add1.ntb_cnt < i then
add1.ntb_cnt := i;
end;
procedure AddNum(var add1,add2:tNumtoBase);
//prerequisites
//bases are the same,add1>add2, add1 <= add1+add2;
var
i,carry,s,b : NativeInt;
Begin
b := add1.ntb_bas;
carry := 0;
For i := 0 to add2.ntb_cnt-1 do
begin
s := add1.ntb_dgt[i]+add2.ntb_dgt[i]+carry;
carry := Ord(s>=b);
s := s- (-carry AND b);
add1.ntb_dgt[i] := s;
end;
i := add2.ntb_cnt;
while carry = 1 do
Begin
s := add1.ntb_dgt[i]+carry;
carry := Ord(s>=b);
// remove of if s>b then by bit-twiddling
s := s- (-carry AND b);
add1.ntb_dgt[i] := s;
inc(i);
end;
IF add1.ntb_cnt < i then
add1.ntb_cnt := i;
end;
procedure Test(base:NativeInt);
var
n : Uint64;
i,j,TestSet : NativeInt;
Begin
write(base:5);
n := Minimal_n(base);
Conv2num(sqr2B,n*n,base);
Conv2num(Num,n,base);
deltaNum := num;
AddNum(deltaNum,deltaNum);
IncNum(deltaNum,1);
i := 0;
repeat
//count used digits
TestSet := 0;
For j := sqr2B.ntb_cnt-1 downto 0 do
TestSet := TestSet OR (1 shl sqr2B.ntb_dgt[j]);
inc(TestSet);
IF (1 shl base)=TestSet then
BREAK;
//next square number
AddNum(sqr2B,deltaNum);
IncNum(deltaNum,2);
inc(i);
until false;
IncNumBig(num,i);
OutNum(Num);
OutNum(sqr2B);
Writeln(i:14);
end;
var
T0: TDateTime;
base :nativeInt;
begin
T0 := now;
writeln('base n square(n) Testcnt');
For base := 2 to 16 do
Test(base);
writeln((now-T0)*86400:10:3);
{$IFDEF WINDOWS}readln;{$ENDIF}
end.
|
http://rosettacode.org/wiki/First-class_functions
|
First-class functions
|
A language has first-class functions if it can do each of the following without recursively invoking a compiler or interpreter or otherwise metaprogramming:
Create new functions from preexisting functions at run-time
Store functions in collections
Use functions as arguments to other functions
Use functions as return values of other functions
Task
Write a program to create an ordered collection A of functions of a real number. At least one function should be built-in and at least one should be user-defined; try using the sine, cosine, and cubing functions. Fill another collection B with the inverse of each function in A. Implement function composition as in Functional Composition. Finally, demonstrate that the result of applying the composition of each function in A and its inverse in B to a value, is the original value. (Within the limits of computational accuracy).
(A solution need not actually call the collections "A" and "B". These names are only used in the preceding paragraph for clarity.)
Related task
First-class Numbers
|
#Dyalect
|
Dyalect
|
func apply(fun, x) { y => fun(x, y) }
func sum(x, y) { x + y }
let sum2 = apply(sum, 2)
|
http://rosettacode.org/wiki/First-class_functions
|
First-class functions
|
A language has first-class functions if it can do each of the following without recursively invoking a compiler or interpreter or otherwise metaprogramming:
Create new functions from preexisting functions at run-time
Store functions in collections
Use functions as arguments to other functions
Use functions as return values of other functions
Task
Write a program to create an ordered collection A of functions of a real number. At least one function should be built-in and at least one should be user-defined; try using the sine, cosine, and cubing functions. Fill another collection B with the inverse of each function in A. Implement function composition as in Functional Composition. Finally, demonstrate that the result of applying the composition of each function in A and its inverse in B to a value, is the original value. (Within the limits of computational accuracy).
(A solution need not actually call the collections "A" and "B". These names are only used in the preceding paragraph for clarity.)
Related task
First-class Numbers
|
#D.C3.A9j.C3.A0_Vu
|
Déjà Vu
|
negate:
- 0
set :A [ @++ $ @negate @-- ]
set :B [ @-- $ @++ @negate ]
test n:
for i range 0 -- len A:
if /= n call compose @B! i @A! i n:
return false
true
test to-num !prompt "Enter a number: "
if:
!print "f^-1(f(x)) = x"
else:
!print "Something went wrong."
|
http://rosettacode.org/wiki/Forest_fire
|
Forest fire
|
This page uses content from Wikipedia. The original article was at Forest-fire model. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
Task
Implement the Drossel and Schwabl definition of the forest-fire model.
It is basically a 2D cellular automaton where each cell can be in three distinct states (empty, tree and burning) and evolves according to the following rules (as given by Wikipedia)
A burning cell turns into an empty cell
A tree will burn if at least one neighbor is burning
A tree ignites with probability f even if no neighbor is burning
An empty space fills with a tree with probability p
Neighborhood is the Moore neighborhood; boundary conditions are so that on the boundary the cells are always empty ("fixed" boundary condition).
At the beginning, populate the lattice with empty and tree cells according to a specific probability (e.g. a cell has the probability 0.5 to be a tree). Then, let the system evolve.
Task's requirements do not include graphical display or the ability to change parameters (probabilities p and f ) through a graphical or command line interface.
Related tasks
See Conway's Game of Life
See Wireworld.
|
#JavaScript
|
JavaScript
|
"use strict"
const _ = require('lodash');
const WIDTH_ARGUMENT_POSITION = 2;
const HEIGHT_ARGUMENT_POSITION = 3;
const TREE_PROBABILITY = 0.5;
const NEW_TREE_PROBABILITY = 0.01;
const BURN_PROBABILITY = 0.0001;
const CONSOLE_RED = '\x1b[31m';
const CONSOLE_GREEN = '\x1b[32m';
const CONSOLE_COLOR_CLOSE = '\x1b[91m';
const CONSOLE_CLEAR = '\u001B[2J\u001B[0;0f';
const NEIGHBOURS = [
[-1, -1],
[-1, 0],
[-1, 1],
[ 0, -1],
[ 0, 1],
[ 1, -1],
[ 1, 0],
[ 1, 1]
];
const PRINT_DECODE = {
' ': ' ',
'T': `${CONSOLE_GREEN}T${CONSOLE_COLOR_CLOSE}`,
'B': `${CONSOLE_RED}T${CONSOLE_COLOR_CLOSE}`,
};
const CONDITIONS = {
'T': (forest, y, x) => Math.random() < BURN_PROBABILITY || burningNeighbour(forest, y, x) ? 'B' : 'T',
' ': () => Math.random() < NEW_TREE_PROBABILITY ? 'T' : ' ',
'B': () => ' '
};
const WIDTH = process.argv[WIDTH_ARGUMENT_POSITION] || 20;
const HEIGHT = process.argv[HEIGHT_ARGUMENT_POSITION] || 10;
const update = forest => {
return _.map(forest, (c, ci) => {
return _.map(c, (r, ri) => {
return CONDITIONS[r](forest, ci, ri);
});
});
}
const printForest = forest => {
process.stdout.write(CONSOLE_CLEAR);
_.each(forest, c => {
_.each(c, r => {
process.stdout.write(PRINT_DECODE[r]);
});
process.stdout.write('\n');
})
}
const burningNeighbour = (forest, y, x) => {
return _(NEIGHBOURS)
.map(n => _.isUndefined(forest[y + n[0]]) ? null : forest[y + n[0]][x + n[1]])
.any(_.partial(_.isEqual, 'B'));
};
let forest = _.times(HEIGHT, () => _.times(WIDTH, () => Math.random() < TREE_PROBABILITY ? 'T' : ' '));
setInterval(() => {
forest = update(forest);
printForest(forest)
}, 20);
|
http://rosettacode.org/wiki/First_class_environments
|
First class environments
|
According to Wikipedia, "In computing, a first-class object ... is an entity that can be constructed at run-time, passed as a parameter, returned from a subroutine, or assigned into a variable".
Often this term is used in the context of "first class functions". In an analogous way, a programming language may support "first class environments".
The environment is minimally, the set of variables accessible to a statement being executed. Change the environments and the same statement could produce different results when executed.
Often an environment is captured in a closure, which encapsulates a function together with an environment. That environment, however, is not first-class, as it cannot be created, passed etc. independently from the function's code.
Therefore, a first class environment is a set of variable bindings which can be constructed at run-time, passed as a parameter, returned from a subroutine, or assigned into a variable. It is like a closure without code. A statement must be able to be executed within a stored first class environment and act according to the environment variable values stored within.
Task
Build a dozen environments, and a single piece of code to be run repeatedly in each of these environments.
Each environment contains the bindings for two variables:
a value in the Hailstone sequence, and
a count which is incremented until the value drops to 1.
The initial hailstone values are 1 through 12, and the count in each environment is zero.
When the code runs, it calculates the next hailstone step in the current environment (unless the value is already 1) and counts the steps. Then it prints the current value in a tabular form.
When all hailstone values dropped to 1, processing stops, and the total number of hailstone steps for each environment is printed.
|
#zkl
|
zkl
|
class Env{
var n,cnt=0;
fcn init(_n){n=_n; returnClass(self.f)}
fcn f{
if(n!=1){
cnt += 1;
if(n.isEven) n=n/2; else n=n*3+1;
}
n
}
}
|
http://rosettacode.org/wiki/Flatten_a_list
|
Flatten a list
|
Task
Write a function to flatten the nesting in an arbitrary list of values.
Your program should work on the equivalent of this list:
[[1], 2, [[3, 4], 5], [[[]]], [[[6]]], 7, 8, []]
Where the correct result would be the list:
[1, 2, 3, 4, 5, 6, 7, 8]
Related task
Tree traversal
|
#Elm
|
Elm
|
import Graphics.Element exposing (show)
type Tree a
= Leaf a
| Node (List (Tree a))
flatten : Tree a -> List a
flatten tree =
case tree of
Leaf a -> [a]
Node list -> List.concatMap flatten list
-- [[1], 2, [[3, 4], 5], [[[]]], [[[6]]], 7, 8, []]
tree : Tree Int
tree = Node
[ Node [Leaf 1]
, Leaf 2
, Node [Node [Leaf 3, Leaf 4], Leaf 5]
, Node [Node [Node []]]
, Node [Node [Node [Leaf 6]]]
, Leaf 7
, Leaf 8
, Node []
]
main =
show (flatten tree)
|
http://rosettacode.org/wiki/Flipping_bits_game
|
Flipping bits game
|
The game
Given an N×N square array of zeroes or ones in an initial configuration, and a target configuration of zeroes and ones.
The game is to transform one to the other in as few moves as possible by inverting whole numbered rows or whole lettered
columns at once (as one move).
In an inversion. any 1 becomes 0, and any 0 becomes 1 for that whole row or column.
Task
Create a program to score for the Flipping bits game.
The game should create an original random target configuration and a starting configuration.
Ensure that the starting position is never the target position.
The target position must be guaranteed as reachable from the starting position. (One possible way to do this is to generate the start position by legal flips from a random target position. The flips will always be reversible back to the target from the given start position).
The number of moves taken so far should be shown.
Show an example of a short game here, on this page, for a 3×3 array of bits.
|
#REXX
|
REXX
|
/*REXX program presents a "flipping bit" puzzle. The user can solve via it via C.L. */
parse arg N u seed . /*get optional arguments from the C.L. */
if N=='' | N=="," then N=3 /*Size given? Then use default of 3.*/
if u=='' | u=="," then u=N /*the number of bits initialized to ON.*/
if datatype(seed, 'W') then call random ,,seed /*is there a seed (for repeatability?) */
col@= 'a b c d e f g h i j k l m n o p q r s t u v w x y z' /*literal for column id.*/
cols=space(col@, 0); upper cols /*letters to be used for the columns. */
@.=0; !.=0 /*set both arrays to "off" characters.*/
tries=0 /*number of player's attempts (so far).*/
do while show(0) < u /* [↓] turn "on" U number of bits.*/
r=random(1, N); c=random(1, N) /*get a random row and column. */
@.r.c=1 ; !.r.c=1 /*set (both) row and column to ON. */
end /*while*/ /* [↑] keep going 'til U bits set.*/
oz=z /*save the original array string. */
call show 1, ' ◄═══target═══╣', , 1 /*display the target for user to attain*/
do random(1,2); call flip 'R',random(1,N) /*flip a row of bits. */
call flip 'C',random(1,N) /* " " column " " */
end /*random*/ /* [↑] just perform 1 or 2 times. */
if z==oz then call flip 'R', random(1, N) /*ensure it's not target we're flipping*/
do until z==oz; call prompt /*prompt until they get it right. */
call flip left(?, 1), substr(?, 2) /*flip a user selected row or column. */
call show 0 /*get image (Z) of the updated array. */
end /*until*/
call show 1, ' ◄───your array' /*display the array to the terminal. */
say '─────────Congrats! You did it in' tries "tries."
exit tries /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
halt: say 'program was halted by user.'; exit /*the REXX program was halted by user. */
hdr: aaa=arg(1); if oo==1 then aaa=translate(aaa, "╔═║", '┌─│'); say aaa; return
isInt: return datatype( arg(1), 'W') /*returns 1 if arg is an integer.*/
isLet: return datatype( arg(1), 'M') /*returns 1 if arg is a letter. */
terr: if ok then say '───────── ***error***: illegal' arg(1); ok=0; return
/*──────────────────────────────────────────────────────────────────────────────────────*/
flip: arg x,#; do c=1 for N while x=='R'; @.#.c = \@.#.c; end /*c*/
do r=1 for N while x=='C'; @.r.# = \@.r.#; end /*r*/; return
/*──────────────────────────────────────────────────────────────────────────────────────*/
prompt: if tries\==0 then say '─────────bit array after play: ' tries
signal on halt /*another method for the player to quit*/
!='─────────Please enter a row number or a column letter, or Quit:'
call show 1, ' ◄───your array' /*display the array to the terminal. */
do forever until ok; ok=1; say; say !; pull ? _ . 1 aa
if abbrev('QUIT', ?, 1) then do; say '─────────quitting···'; exit 0; end
if ?=='' then do; call show 1," ◄═══target═══╣",.,1; ok=0
call show 1," ◄───your array"
end /* [↑] reshow targ*/
if _ \== '' then call terr 'too many args entered:' aa
if \isInt(?) & \isLet(?) then call terr 'row/column: ' ?
if isLet(?) then a=pos(?, cols)
if isLet(?) & (a<1 | a>N | length(?)>1) then call terr 'column: ' ?
if isLet(?) then ?='C'pos(?, cols)
if isInt(?) & (?<1 | ?>N) then call terr 'row: ' ?
if isInt(?) then ?='R' || (?/1) /*normalize number*/
end /*forever*/ /*end of da checks*/
tries= tries + 1 /*bump da counter.*/
return ? /*return response.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
show: $=0; _=; parse arg tell,tx,o,oo /*$≡num of ON bits*/
if tell then do; say; say ' ' subword(col@, 1, N) " column letter"
call hdr 'row ┌'copies('─', N+N+1) /*prepend col hdrs*/
end /* [↑] grid hdrs.*/
z= /* [↓] build grid.*/
do r=1 for N /*show grid rows.*/
do c=1 for N; if o==. then do; z=z || !.r.c; _=_ !.r.c; $=$ + !.r.c; end
else do; z=z || @.r.c; _=_ @.r.c; $=$ + @.r.c; end
end /*c*/ /*··· and sum ONs.*/
if tx\=='' then tar.r=_ tx /*build da target?*/
if tell then call hdr right(r, 2) ' │'_ tx; _= /*show the grid? */
end /*r*/ /*show a grid row.*/
if tell then say; return $ /*show blank line?*/
|
http://rosettacode.org/wiki/First_power_of_2_that_has_leading_decimal_digits_of_12
|
First power of 2 that has leading decimal digits of 12
|
(This task is taken from a Project Euler problem.)
(All numbers herein are expressed in base ten.)
27 = 128 and 7 is
the first power of 2 whose leading decimal digits are 12.
The next power of 2 whose leading decimal digits
are 12 is 80,
280 = 1208925819614629174706176.
Define p(L,n) to be the nth-smallest
value of j such that the base ten representation
of 2j begins with the digits of L .
So p(12, 1) = 7 and
p(12, 2) = 80
You are also given that:
p(123, 45) = 12710
Task
find:
p(12, 1)
p(12, 2)
p(123, 45)
p(123, 12345)
p(123, 678910)
display the results here, on this page.
|
#Rust
|
Rust
|
fn power_of_two(l: isize, n: isize) -> isize {
let mut test: isize = 0;
let log: f64 = 2.0_f64.ln() / 10.0_f64.ln();
let mut factor: isize = 1;
let mut looop = l;
let mut nn = n;
while looop > 10 {
factor *= 10;
looop /= 10;
}
while nn > 0 {
test = test + 1;
let val: isize = (factor as f64 * 10.0_f64.powf(test as f64 * log % 1.0)) as isize;
if val == l {
nn = nn - 1;
}
}
test
}
fn run_test(l: isize, n: isize) {
println!("p({}, {}) = {}", l, n, power_of_two(l, n));
}
fn main() {
run_test(12, 1);
run_test(12, 2);
run_test(123, 45);
run_test(123, 12345);
run_test(123, 678910);
}
|
http://rosettacode.org/wiki/First_power_of_2_that_has_leading_decimal_digits_of_12
|
First power of 2 that has leading decimal digits of 12
|
(This task is taken from a Project Euler problem.)
(All numbers herein are expressed in base ten.)
27 = 128 and 7 is
the first power of 2 whose leading decimal digits are 12.
The next power of 2 whose leading decimal digits
are 12 is 80,
280 = 1208925819614629174706176.
Define p(L,n) to be the nth-smallest
value of j such that the base ten representation
of 2j begins with the digits of L .
So p(12, 1) = 7 and
p(12, 2) = 80
You are also given that:
p(123, 45) = 12710
Task
find:
p(12, 1)
p(12, 2)
p(123, 45)
p(123, 12345)
p(123, 678910)
display the results here, on this page.
|
#Scala
|
Scala
|
object FirstPowerOfTwo {
def p(l: Int, n: Int): Int = {
var n2 = n
var test = 0
val log = math.log(2) / math.log(10)
var factor = 1
var loop = l
while (loop > 10) {
factor *= 10
loop /= 10
}
while (n2 > 0) {
test += 1
val value = (factor * math.pow(10, test * log % 1)).asInstanceOf[Int]
if (value == l) {
n2 -= 1
}
}
test
}
def runTest(l: Int, n: Int): Unit = {
printf("p(%d, %d) = %,d%n", l, n, p(l, n))
}
def main(args: Array[String]): Unit = {
runTest(12, 1)
runTest(12, 2)
runTest(123, 45)
runTest(123, 12345)
runTest(123, 678910)
}
}
|
http://rosettacode.org/wiki/First-class_functions/Use_numbers_analogously
|
First-class functions/Use numbers analogously
|
In First-class functions, a language is showing how its manipulation of functions is similar to its manipulation of other types.
This tasks aim is to compare and contrast a language's implementation of first class functions, with its normal handling of numbers.
Write a program to create an ordered collection of a mixture of literally typed and expressions producing a real number, together with another ordered collection of their multiplicative inverses. Try and use the following pseudo-code to generate the numbers for the ordered collections:
x = 2.0
xi = 0.5
y = 4.0
yi = 0.25
z = x + y
zi = 1.0 / ( x + y )
Create a function multiplier, that given two numbers as arguments returns a function that when called with one argument, returns the result of multiplying the two arguments to the call to multiplier that created it and the argument in the call:
new_function = multiplier(n1,n2)
# where new_function(m) returns the result of n1 * n2 * m
Applying the multiplier of a number and its inverse from the two ordered collections of numbers in pairs, show that the result in each case is one.
Compare and contrast the resultant program with the corresponding entry in First-class functions. They should be close.
To paraphrase the task description: Do what was done before, but with numbers rather than functions
|
#Raku
|
Raku
|
sub multiplied ($g, $f) { return { $g * $f * $^x } }
my $x = 2.0;
my $xi = 0.5;
my $y = 4.0;
my $yi = 0.25;
my $z = $x + $y;
my $zi = 1.0 / ( $x + $y );
my @numbers = $x, $y, $z;
my @inverses = $xi, $yi, $zi;
for flat @numbers Z @inverses { say multiplied($^g, $^f)(.5) }
|
http://rosettacode.org/wiki/First-class_functions/Use_numbers_analogously
|
First-class functions/Use numbers analogously
|
In First-class functions, a language is showing how its manipulation of functions is similar to its manipulation of other types.
This tasks aim is to compare and contrast a language's implementation of first class functions, with its normal handling of numbers.
Write a program to create an ordered collection of a mixture of literally typed and expressions producing a real number, together with another ordered collection of their multiplicative inverses. Try and use the following pseudo-code to generate the numbers for the ordered collections:
x = 2.0
xi = 0.5
y = 4.0
yi = 0.25
z = x + y
zi = 1.0 / ( x + y )
Create a function multiplier, that given two numbers as arguments returns a function that when called with one argument, returns the result of multiplying the two arguments to the call to multiplier that created it and the argument in the call:
new_function = multiplier(n1,n2)
# where new_function(m) returns the result of n1 * n2 * m
Applying the multiplier of a number and its inverse from the two ordered collections of numbers in pairs, show that the result in each case is one.
Compare and contrast the resultant program with the corresponding entry in First-class functions. They should be close.
To paraphrase the task description: Do what was done before, but with numbers rather than functions
|
#REXX
|
REXX
|
/*REXX program to use a first-class function to use numbers analogously. */
nums= 2.0 4.0 6.0 /*various numbers, can have fractions.*/
invs= 1/2.0 1/4.0 1/6.0 /*inverses of the above (real) numbers.*/
m= 0.5 /*multiplier when invoking new function*/
do j=1 for words(nums); num= word(nums, j); inv= word(invs, j)
nf= multiplier(num, inv); interpret call nf m /*sets the var RESULT.*/
say 'number=' @(num) 'inverse=' @(inv) 'm=' @(m) 'result=' @(result)
end /*j*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
@: return left( arg(1) / 1, 15) /*format the number, left justified. */
multiplier: procedure expose n1n2; parse arg n1,n2; n1n2= n1 * n2; return 'a_new_func'
a_new_func: return n1n2 * arg(1)
|
http://rosettacode.org/wiki/Flow-control_structures
|
Flow-control structures
|
Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Task
Document common flow-control structures.
One common example of a flow-control structure is the goto construct.
Note that Conditional Structures and Loop Structures have their own articles/categories.
Related tasks
Conditional Structures
Loop Structures
|
#Ruby
|
Ruby
|
begin
# some code that may raise an exception
rescue ExceptionClassA => a
# handle code
rescue ExceptionClassB, ExceptionClassC => b_or_c
# handle ...
rescue
# handle all other exceptions
else
# when no exception occurred, execute this code
ensure
# execute this code always
end
|
http://rosettacode.org/wiki/Flow-control_structures
|
Flow-control structures
|
Control Structures
These are examples of control structures. You may also be interested in:
Conditional structures
Exceptions
Flow-control structures
Loops
Task
Document common flow-control structures.
One common example of a flow-control structure is the goto construct.
Note that Conditional Structures and Loop Structures have their own articles/categories.
Related tasks
Conditional Structures
Loop Structures
|
#SAS
|
SAS
|
/* GOTO: as in other languages
STOP: to stop current data step */
data _null_;
n=1;
p=1;
L1:
put n p;
n=n+1;
if n<=p then goto L1;
p=p+1;
n=1;
if p>10 then stop;
goto L1;
run;
/* LINK: equivalent of GOSUB in BASIC
RETURN: after a LINK, or to return to the beginning of data step */
data _null_;
input a b;
link gcd;
put a b gcd;
return;
gcd:
_a=a;
_b=b;
do while(_b>0);
_r=mod(_a,_b);
_a=_b;
_b=_r;
end;
gcd=_a;
return;
cards;
2 15
533 221
8 44
;
run;
|
http://rosettacode.org/wiki/Floyd%27s_triangle
|
Floyd's triangle
|
Floyd's triangle lists the natural numbers in a right triangle aligned to the left where
the first row is 1 (unity)
successive rows start towards the left with the next number followed by successive naturals listing one more number than the line above.
The first few lines of a Floyd triangle looks like this:
1
2 3
4 5 6
7 8 9 10
11 12 13 14 15
Task
Write a program to generate and display here the first n lines of a Floyd triangle.
(Use n=5 and n=14 rows).
Ensure that when displayed in a mono-space font, the numbers line up in vertical columns as shown and that only one space separates numbers of the last row.
|
#Go
|
Go
|
package main
import "fmt"
func main() {
floyd(5)
floyd(14)
}
func floyd(n int) {
fmt.Printf("Floyd %d:\n", n)
lowerLeftCorner := n*(n-1)/2 + 1
lastInColumn := lowerLeftCorner
lastInRow := 1
for i, row := 1, 1; row <= n; i++ {
w := len(fmt.Sprint(lastInColumn))
if i < lastInRow {
fmt.Printf("%*d ", w, i)
lastInColumn++
} else {
fmt.Printf("%*d\n", w, i)
row++
lastInRow += row
lastInColumn = lowerLeftCorner
}
}
}
|
http://rosettacode.org/wiki/Floyd-Warshall_algorithm
|
Floyd-Warshall algorithm
|
The Floyd–Warshall algorithm is an algorithm for finding shortest paths in a weighted graph with positive or negative edge weights.
Task
Find the lengths of the shortest paths between all pairs of vertices of the given directed graph. Your code may assume that the input has already been checked for loops, parallel edges and negative cycles.
Print the pair, the distance and (optionally) the path.
Example
pair dist path
1 -> 2 -1 1 -> 3 -> 4 -> 2
1 -> 3 -2 1 -> 3
1 -> 4 0 1 -> 3 -> 4
2 -> 1 4 2 -> 1
2 -> 3 2 2 -> 1 -> 3
2 -> 4 4 2 -> 1 -> 3 -> 4
3 -> 1 5 3 -> 4 -> 2 -> 1
3 -> 2 1 3 -> 4 -> 2
3 -> 4 2 3 -> 4
4 -> 1 3 4 -> 2 -> 1
4 -> 2 -1 4 -> 2
4 -> 3 1 4 -> 2 -> 1 -> 3
See also
Floyd-Warshall Algorithm - step by step guide (youtube)
|
#Prolog
|
Prolog
|
:- use_module(library(clpfd)).
path(List, To, From, [From], W) :-
select([To,From,W],List,_).
path(List, To, From, [Link|R], W) :-
select([To,Link,W1],List,Rest),
W #= W1 + W2,
path(Rest, Link, From, R, W2).
find_path(Din, From, To, [From|Pout], Wout) :-
between(1, 4, From),
between(1, 4, To),
dif(From, To),
findall([W,P], (
path(Din, From, To, P, W),
all_distinct(P)
), Paths),
sort(Paths, [[Wout,Pout]|_]).
print_all_paths :-
D = [[1, 3, -2], [2, 3, 3], [2, 1, 4], [3, 4, 2], [4, 2, -1]],
format('Pair\t Dist\tPath~n'),
forall(
find_path(D, From, To, Path, Weight),(
atomic_list_concat(Path, ' -> ', PPath),
format('~p -> ~p\t ~p\t~w~n', [From, To, Weight, PPath]))).
|
http://rosettacode.org/wiki/Function_definition
|
Function definition
|
A function is a body of code that returns a value.
The value returned may depend on arguments provided to the function.
Task
Write a definition of a function called "multiply" that takes two arguments and returns their product.
(Argument types should be chosen so as not to distract from showing how functions are created and values returned).
Related task
Function prototype
|
#TXR
|
TXR
|
@(define multiply (a b out))
@(bind out @(* a b))
@(end)
@(multiply 3 4 result)
|
http://rosettacode.org/wiki/Function_definition
|
Function definition
|
A function is a body of code that returns a value.
The value returned may depend on arguments provided to the function.
Task
Write a definition of a function called "multiply" that takes two arguments and returns their product.
(Argument types should be chosen so as not to distract from showing how functions are created and values returned).
Related task
Function prototype
|
#uBasic.2F4tH_2
|
uBasic/4tH
|
PRINT FUNC (_Multiply (2,3))
END
_Multiply PARAM (2)
RETURN (a@ * b@)
|
http://rosettacode.org/wiki/Forward_difference
|
Forward difference
|
Task
Provide code that produces a list of numbers which is the nth order forward difference, given a non-negative integer (specifying the order) and a list of numbers.
The first-order forward difference of a list of numbers A is a new list B, where Bn = An+1 - An.
List B should have one fewer element as a result.
The second-order forward difference of A will be:
tdefmodule Diff do
def forward(arr,i\\1) do
forward(arr,[],i)
end
def forward([_|[]],diffs,i) do
if i == 1 do
IO.inspect diffs
else
forward(diffs,[],i-1)
end
end
def forward([val1|[val2|vals]],diffs,i) do
forward([val2|vals],diffs++[val2-val1],i)
end
end
The same as the first-order forward difference of B.
That new list will have two fewer elements than A and one less than B.
The goal of this task is to repeat this process up to the desired order.
For a more formal description, see the related Mathworld article.
Algorithmic options
Iterate through all previous forward differences and re-calculate a new array each time.
Use this formula (from Wikipedia):
Δ
n
[
f
]
(
x
)
=
∑
k
=
0
n
(
n
k
)
(
−
1
)
n
−
k
f
(
x
+
k
)
{\displaystyle \Delta ^{n}[f](x)=\sum _{k=0}^{n}{n \choose k}(-1)^{n-k}f(x+k)}
(Pascal's Triangle may be useful for this option.)
|
#Quackery
|
Quackery
|
[ times
[ [] swap behead
swap witheach
[ tuck dip [ - join ] ]
drop ] ] is f-diff ( [ n --> [ )
' [ 90 47 58 29 22 32 55 5 55 73 ]
dup size times
[ dup i^
dup echo say ": "
f-diff echo cr ]
drop
|
http://rosettacode.org/wiki/Forward_difference
|
Forward difference
|
Task
Provide code that produces a list of numbers which is the nth order forward difference, given a non-negative integer (specifying the order) and a list of numbers.
The first-order forward difference of a list of numbers A is a new list B, where Bn = An+1 - An.
List B should have one fewer element as a result.
The second-order forward difference of A will be:
tdefmodule Diff do
def forward(arr,i\\1) do
forward(arr,[],i)
end
def forward([_|[]],diffs,i) do
if i == 1 do
IO.inspect diffs
else
forward(diffs,[],i-1)
end
end
def forward([val1|[val2|vals]],diffs,i) do
forward([val2|vals],diffs++[val2-val1],i)
end
end
The same as the first-order forward difference of B.
That new list will have two fewer elements than A and one less than B.
The goal of this task is to repeat this process up to the desired order.
For a more formal description, see the related Mathworld article.
Algorithmic options
Iterate through all previous forward differences and re-calculate a new array each time.
Use this formula (from Wikipedia):
Δ
n
[
f
]
(
x
)
=
∑
k
=
0
n
(
n
k
)
(
−
1
)
n
−
k
f
(
x
+
k
)
{\displaystyle \Delta ^{n}[f](x)=\sum _{k=0}^{n}{n \choose k}(-1)^{n-k}f(x+k)}
(Pascal's Triangle may be useful for this option.)
|
#R
|
R
|
forwarddif <- function(a, n) {
if ( n == 1 )
a[2:length(a)] - a[1:length(a)-1]
else {
r <- forwarddif(a, 1)
forwarddif(r, n-1)
}
}
fdiff <- function(a, n) {
r <- a
for(i in 1:n) {
r <- r[2:length(r)] - r[1:length(r)-1]
}
r
}
v <- c(90, 47, 58, 29, 22, 32, 55, 5, 55, 73)
print(forwarddif(v, 9))
print(fdiff(v, 9))
|
http://rosettacode.org/wiki/Formatted_numeric_output
|
Formatted numeric output
|
Task
Express a number in decimal as a fixed-length string with leading zeros.
For example, the number 7.125 could be expressed as 00007.125.
|
#Visual_Basic
|
Visual Basic
|
Debug.Print Format$(7.125, "00000.000")
|
http://rosettacode.org/wiki/Formatted_numeric_output
|
Formatted numeric output
|
Task
Express a number in decimal as a fixed-length string with leading zeros.
For example, the number 7.125 could be expressed as 00007.125.
|
#Wren
|
Wren
|
import "/fmt" for Fmt
var n = 7.125
System.print(Fmt.rjust(9, n, "0"))
|
http://rosettacode.org/wiki/Four_bit_adder
|
Four bit adder
|
Task
"Simulate" a four-bit adder.
This design can be realized using four 1-bit full adders.
Each of these 1-bit full adders can be built with two half adders and an or gate. ;
Finally a half adder can be made using an xor gate and an and gate.
The xor gate can be made using two nots, two ands and one or.
Not, or and and, the only allowed "gates" for the task, can be "imitated" by using the bitwise operators of your language.
If there is not a bit type in your language, to be sure that the not does not "invert" all the other bits of the basic type (e.g. a byte) we are not interested in, you can use an extra nand (and then not) with the constant 1 on one input.
Instead of optimizing and reducing the number of gates used for the final 4-bit adder, build it in the most straightforward way, connecting the other "constructive blocks", in turn made of "simpler" and "smaller" ones.
Schematics of the "constructive blocks"
(Xor gate with ANDs, ORs and NOTs)
(A half adder)
(A full adder)
(A 4-bit adder)
Solutions should try to be as descriptive as possible, making it as easy as possible to identify "connections" between higher-order "blocks".
It is not mandatory to replicate the syntax of higher-order blocks in the atomic "gate" blocks, i.e. basic "gate" operations can be performed as usual bitwise operations, or they can be "wrapped" in a block in order to expose the same syntax of higher-order blocks, at implementers' choice.
To test the implementation, show the sum of two four-bit numbers (in binary).
|
#Ring
|
Ring
|
###---------------------------
# Program: 4 Bit Adder - Ring
# Author: Bert Mariani
# Date: 2018-02-28
#
# Bit Adder: Input A B Cin
# Output S Cout
#
# A ^ B => axb XOR gate
# axb ^ C => Sout XOR gate
# axb & C => d AND gate
#
# A & B => anb AND gate
# anb | d => Cout OR gate
#
# Call Adder for number of bit in input fields
###-------------------------------------------
### 4 Bits
Cout = "0"
OutputS = "0000"
InputA = "0101"
InputB = "1101"
See "InputA:.. "+ InputA +nl
See "InputB:.. "+ InputB +nl
BitsAdd(InputA, InputB)
See "Sum...: "+ Cout +" "+ OutputS +nl+nl
###-------------------------------------------
### 32 Bits
Cout = "0"
OutputS = "00000000000000000000000000000000"
InputA = "01010101010101010101010101010101"
InputB = "11011101110111011101110111011101"
See "InputA:.. "+ InputA +nl
See "InputB:.. "+ InputB +nl
BitsAdd(InputA, InputB)
See "Sum...: "+ Cout +" "+ OutputS +nl+nl
###-------------------------------
Func BitsAdd(InputA, InputB)
nbrBits = len(InputA)
for i = nbrBits to 1 step -1
A = InputA[i]
B = InputB[i]
C = Cout
S = Adder(A,B,C)
OutputS[i] = "" + S
next
return
###------------------------
Func Adder(A,B,C)
axb = A ^ B
Sout = axb ^ C
d = axb & C
anb = A & B
Cout = anb | d ### Cout is global
return(Sout)
###------------------------
|
http://rosettacode.org/wiki/Fivenum
|
Fivenum
|
Many big data or scientific programs use boxplots to show distributions of data. In addition, sometimes saving large arrays for boxplots can be impractical and use extreme amounts of RAM. It can be useful to save large arrays as arrays with five numbers to save memory.
For example, the R programming language implements Tukey's five-number summary as the fivenum function.
Task
Given an array of numbers, compute the five-number summary.
Note
While these five numbers can be used to draw a boxplot, statistical packages will typically need extra data.
Moreover, while there is a consensus about the "box" of the boxplot, there are variations among statistical packages for the whiskers.
|
#Phix
|
Phix
|
with javascript_semantics
function median(sequence tbl, integer lo, hi)
integer l = hi-lo+1
integer m = lo+floor(l/2)
if remainder(l,2)=1 then
return tbl[m]
end if
return (tbl[m-1]+tbl[m])/2
end function
function fivenum(sequence tbl)
tbl = sort(deep_copy(tbl))
integer l = length(tbl),
m = floor(l/2)+remainder(l,2)
atom r1 = tbl[1],
r2 = median(tbl,1,m),
r3 = median(tbl,1,l),
r4 = median(tbl,m+1,l),
r5 = tbl[l]
return {r1, r2, r3, r4, r5}
end function
constant x1 = {15, 6, 42, 41, 7, 36, 49, 40, 39, 47, 43},
x2 = {36, 40, 7, 39, 41, 15},
x3 = {0.14082834, 0.09748790, 1.73131507, 0.87636009, -1.95059594,
0.73438555, -0.03035726, 1.46675970, -0.74621349, -0.72588772,
0.63905160, 0.61501527, -0.98983780, -1.00447874, -0.62759469,
0.66206163, 1.04312009, -0.10305385, 0.75775634, 0.32566578}
pp(fivenum(x1))
pp(fivenum(x2))
pp(fivenum(x3))
|
http://rosettacode.org/wiki/Find_the_missing_permutation
|
Find the missing permutation
|
ABCD
CABD
ACDB
DACB
BCDA
ACBD
ADCB
CDAB
DABC
BCAD
CADB
CDBA
CBAD
ABDC
ADBC
BDCA
DCBA
BACD
BADC
BDAC
CBDA
DBCA
DCAB
Listed above are all-but-one of the permutations of the symbols A, B, C, and D, except for one permutation that's not listed.
Task
Find that missing permutation.
Methods
Obvious method:
enumerate all permutations of A, B, C, and D,
and then look for the missing permutation.
alternate method:
Hint: if all permutations were shown above, how many
times would A appear in each position?
What is the parity of this number?
another alternate method:
Hint: if you add up the letter values of each column,
does a missing letter A, B, C, and D from each
column cause the total value for each column to be unique?
Related task
Permutations)
|
#AWK
|
AWK
|
{
split($1,a,"");
for (i=1;i<=4;++i) {
t[i,a[i]]++;
}
}
END {
for (k in t) {
split(k,a,SUBSEP)
for (l in t) {
split(l, b, SUBSEP)
if (a[1] == b[1] && t[k] < t[l]) {
s[a[1]] = a[2]
break
}
}
}
print s[1]s[2]s[3]s[4]
}
|
http://rosettacode.org/wiki/Find_the_last_Sunday_of_each_month
|
Find the last Sunday of each month
|
Write a program or a script that returns the last Sundays of each month of a given year. The year may be given through any simple input method in your language (command line, std in, etc).
Example of an expected output:
./last_sundays 2013
2013-01-27
2013-02-24
2013-03-31
2013-04-28
2013-05-26
2013-06-30
2013-07-28
2013-08-25
2013-09-29
2013-10-27
2013-11-24
2013-12-29
Related tasks
Day of the week
Five weekends
Last Friday of each month
|
#C.2B.2B
|
C++
|
#include <windows.h>
#include <iostream>
#include <string>
//--------------------------------------------------------------------------------------------------
using namespace std;
//--------------------------------------------------------------------------------------------------
class lastSunday
{
public:
lastSunday()
{
m[0] = "JANUARY: "; m[1] = "FEBRUARY: "; m[2] = "MARCH: "; m[3] = "APRIL: ";
m[4] = "MAY: "; m[5] = "JUNE: "; m[6] = "JULY: "; m[7] = "AUGUST: ";
m[8] = "SEPTEMBER: "; m[9] = "OCTOBER: "; m[10] = "NOVEMBER: "; m[11] = "DECEMBER: ";
}
void findLastSunday( int y )
{
year = y;
isleapyear();
int days[] = { 31, isleap ? 29 : 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 },
d;
for( int i = 0; i < 12; i++ )
{
d = days[i];
while( true )
{
if( !getWeekDay( i, d ) ) break;
d--;
}
lastDay[i] = d;
}
display();
}
private:
void isleapyear()
{
isleap = false;
if( !( year % 4 ) )
{
if( year % 100 ) isleap = true;
else if( !( year % 400 ) ) isleap = true;
}
}
void display()
{
system( "cls" );
cout << " YEAR " << year << endl << "=============" << endl;
for( int x = 0; x < 12; x++ )
cout << m[x] << lastDay[x] << endl;
cout << endl << endl;
}
int getWeekDay( int m, int d )
{
int y = year;
int f = y + d + 3 * m - 1;
m++;
if( m < 3 ) y--;
else f -= int( .4 * m + 2.3 );
f += int( y / 4 ) - int( ( y / 100 + 1 ) * 0.75 );
f %= 7;
return f;
}
int lastDay[12], year;
string m[12];
bool isleap;
};
//--------------------------------------------------------------------------------------------------
int main( int argc, char* argv[] )
{
int y;
lastSunday ls;
while( true )
{
system( "cls" );
cout << "Enter the year( yyyy ) --- ( 0 to quit ): ";
cin >> y;
if( !y ) return 0;
ls.findLastSunday( y );
system( "pause" );
}
return 0;
}
//--------------------------------------------------------------------------------------------------
|
http://rosettacode.org/wiki/Find_the_intersection_of_two_lines
|
Find the intersection of two lines
|
[1]
Task
Find the point of intersection of two lines in 2D.
The 1st line passes though (4,0) and (6,10) .
The 2nd line passes though (0,3) and (10,7) .
|
#D
|
D
|
import std.stdio;
struct Point {
real x, y;
void toString(scope void delegate(const(char)[]) sink) const {
import std.format;
sink("{");
sink.formattedWrite!"%f"(x);
sink(", ");
sink.formattedWrite!"%f"(y);
sink("}");
}
}
struct Line {
Point s, e;
}
Point findIntersection(Line l1, Line l2) {
auto a1 = l1.e.y - l1.s.y;
auto b1 = l1.s.x - l1.e.x;
auto c1 = a1 * l1.s.x + b1 * l1.s.y;
auto a2 = l2.e.y - l2.s.y;
auto b2 = l2.s.x - l2.e.x;
auto c2 = a2 * l2.s.x + b2 * l2.s.y;
auto delta = a1 * b2 - a2 * b1;
// If lines are parallel, intersection point will contain infinite values
return Point((b2 * c1 - b1 * c2) / delta, (a1 * c2 - a2 * c1) / delta);
}
void main() {
auto l1 = Line(Point(4.0, 0.0), Point(6.0, 10.0));
auto l2 = Line(Point(0f, 3f), Point(10f, 7f));
writeln(findIntersection(l1, l2));
l1 = Line(Point(0.0, 0.0), Point(1.0, 1.0));
l2 = Line(Point(1.0, 2.0), Point(4.0, 5.0));
writeln(findIntersection(l1, l2));
}
|
http://rosettacode.org/wiki/Find_the_intersection_of_a_line_with_a_plane
|
Find the intersection of a line with a plane
|
Finding the intersection of an infinite ray with a plane in 3D is an important topic in collision detection.
Task
Find the point of intersection for the infinite ray with direction (0, -1, -1) passing through position (0, 0, 10) with the infinite plane with a normal vector of (0, 0, 1) and which passes through [0, 0, 5].
|
#F.23
|
F#
|
open System
type Vector(x : double, y : double, z : double) =
member this.x = x
member this.y = y
member this.z = z
static member (-) (lhs : Vector, rhs : Vector) =
Vector(lhs.x - rhs.x, lhs.y - rhs.y, lhs.z - rhs.z)
static member (*) (lhs : Vector, rhs : double) =
Vector(lhs.x * rhs, lhs.y * rhs, lhs.z * rhs)
override this.ToString() =
String.Format("({0:F}, {1:F}, {2:F})", x, y, z)
let Dot (lhs:Vector) (rhs:Vector) =
lhs.x * rhs.x + lhs.y * rhs.y + lhs.z * rhs.z
let IntersectPoint rayVector rayPoint planeNormal planePoint =
let diff = rayPoint - planePoint
let prod1 = Dot diff planeNormal
let prod2 = Dot rayVector planeNormal
let prod3 = prod1 / prod2
rayPoint - rayVector * prod3
[<EntryPoint>]
let main _ =
let rv = Vector(0.0, -1.0, -1.0)
let rp = Vector(0.0, 0.0, 10.0)
let pn = Vector(0.0, 0.0, 1.0)
let pp = Vector(0.0, 0.0, 5.0)
let ip = IntersectPoint rv rp pn pp
Console.WriteLine("The ray intersects the plane at {0}", ip)
0 // return an integer exit code
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.