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/Yellowstone_sequence | Yellowstone sequence | The Yellowstone sequence, also called the Yellowstone permutation, is defined as:
For n <= 3,
a(n) = n
For n >= 4,
a(n) = the smallest number not already in sequence such that a(n) is relatively prime to a(n-1) and
is not relatively prime to a(n-2).
The sequence is a permutation of the natural numbers, and gets its name from what its authors felt was a spiking, geyser like appearance of a plot of the sequence.
Example
a(4) is 4 because 4 is the smallest number following 1, 2, 3 in the sequence that is relatively prime to the entry before it (3), and is not relatively prime to the number two entries before it (2).
Task
Find and show as output the first 30 Yellowstone numbers.
Extra
Demonstrate how to plot, with x = n and y coordinate a(n), the first 100 Yellowstone numbers.
Related tasks
Greatest common divisor.
Plot coordinate pairs.
See also
The OEIS entry: A098550 The Yellowstone permutation.
Applegate et al, 2015: The Yellowstone Permutation [1].
| #XPL0 | XPL0 | func GCD(N, D); \Return the greatest common divisor of N and D
int N, D, R; \numerator, denominator, remainder
[if D > N then
[R:=D; D:=N; N:=R]; \swap D and N
while D > 0 do
[R:= rem(N/D);
N:= D;
D:= R;
];
return N;
];
int I, A(30+1), N, T;
[for I:= 1 to 3 do A(I):= I; \givens
N:= 4;
repeat T:= 4;
loop [if GCD(T, A(N-1)) = 1 and \relatively prime
GCD(T, A(N-2)) # 1 then \not relatively prime
[loop [for I:= 1 to N-1 do \test if in sequence
if T = A(I) then quit;
quit;
];
if I = N then \T is not in sequence so
[A(N):= T; \ add it in
N:= N+1;
quit;
];
];
T:= T+1; \next trial
];
until N > 30;
for N:= 1 to 30 do
[IntOut(0, A(N)); ChOut(0, ^ )];
\\for N:= 1 to 100 do Point(N, A(N)); \plot demonstration
] |
http://rosettacode.org/wiki/Yellowstone_sequence | Yellowstone sequence | The Yellowstone sequence, also called the Yellowstone permutation, is defined as:
For n <= 3,
a(n) = n
For n >= 4,
a(n) = the smallest number not already in sequence such that a(n) is relatively prime to a(n-1) and
is not relatively prime to a(n-2).
The sequence is a permutation of the natural numbers, and gets its name from what its authors felt was a spiking, geyser like appearance of a plot of the sequence.
Example
a(4) is 4 because 4 is the smallest number following 1, 2, 3 in the sequence that is relatively prime to the entry before it (3), and is not relatively prime to the number two entries before it (2).
Task
Find and show as output the first 30 Yellowstone numbers.
Extra
Demonstrate how to plot, with x = n and y coordinate a(n), the first 100 Yellowstone numbers.
Related tasks
Greatest common divisor.
Plot coordinate pairs.
See also
The OEIS entry: A098550 The Yellowstone permutation.
Applegate et al, 2015: The Yellowstone Permutation [1].
| #zkl | zkl | fcn yellowstoneW{ // --> iterator
Walker.zero().tweak(fcn(a,b){
foreach i in ([1..]){
if(not b.holds(i) and i.gcd(a[-1])==1 and i.gcd(a[-2]) >1){
a.del(0).append(i); // only keep last two terms
b[i]=True;
return(i);
}
}
}.fp(List(2,3), Dictionary(1,True, 2,True, 3,True))).push(1,2,3);
} |
http://rosettacode.org/wiki/Arbitrary-precision_integers_(included) | Arbitrary-precision integers (included) | Using the in-built capabilities of your language, calculate the integer value of:
5
4
3
2
{\displaystyle 5^{4^{3^{2}}}}
Confirm that the first and last twenty digits of the answer are:
62060698786608744707...92256259918212890625
Find and show the number of decimal digits in the answer.
Note: Do not submit an implementation of arbitrary precision arithmetic. The intention is to show the capabilities of the language as supplied. If a language has a single, overwhelming, library of varied modules that is endorsed by its home site – such as CPAN for Perl or Boost for C++ – then that may be used instead.
Strictly speaking, this should not be solved by fixed-precision numeric libraries where the precision has to be manually set to a large value; although if this is the only recourse then it may be used with a note explaining that the precision must be set manually to a large enough value.
Related tasks
Long multiplication
Exponentiation order
exponentiation operator
Exponentiation with infix operators in (or operating on) the base
| #Ol | Ol |
(define x (expt 5 (expt 4 (expt 3 2))))
(print
(div x (expt 10 (- (log 10 x) 20)))
"..."
(mod x (expt 10 20)))
(print "totally digits: " (log 10 x))
|
http://rosettacode.org/wiki/Arbitrary-precision_integers_(included) | Arbitrary-precision integers (included) | Using the in-built capabilities of your language, calculate the integer value of:
5
4
3
2
{\displaystyle 5^{4^{3^{2}}}}
Confirm that the first and last twenty digits of the answer are:
62060698786608744707...92256259918212890625
Find and show the number of decimal digits in the answer.
Note: Do not submit an implementation of arbitrary precision arithmetic. The intention is to show the capabilities of the language as supplied. If a language has a single, overwhelming, library of varied modules that is endorsed by its home site – such as CPAN for Perl or Boost for C++ – then that may be used instead.
Strictly speaking, this should not be solved by fixed-precision numeric libraries where the precision has to be manually set to a large value; although if this is the only recourse then it may be used with a note explaining that the precision must be set manually to a large enough value.
Related tasks
Long multiplication
Exponentiation order
exponentiation operator
Exponentiation with infix operators in (or operating on) the base
| #ooRexx | ooRexx |
--REXX program to show arbitrary precision integers.
numeric digits 200000
check = '62060698786608744707...92256259918212890625'
start = .datetime~new
n = 5 ** (4 ** (3**2))
time = .datetime~new - start
say 'elapsed time for the calculation:' time
say
sampl = left(n, 20)"..."right(n, 20)
say ' check:' check
say 'Sample:' sampl
say 'digits:' length(n)
say
if check=sampl then say 'passed!'
else say 'failed!'
|
http://rosettacode.org/wiki/Zeckendorf_number_representation | Zeckendorf number representation | Just as numbers can be represented in a positional notation as sums of multiples of the powers of ten (decimal) or two (binary); all the positive integers can be represented as the sum of one or zero times the distinct members of the Fibonacci series.
Recall that the first six distinct Fibonacci numbers are: 1, 2, 3, 5, 8, 13.
The decimal number eleven can be written as 0*13 + 1*8 + 0*5 + 1*3 + 0*2 + 0*1 or 010100 in positional notation where the columns represent multiplication by a particular member of the sequence. Leading zeroes are dropped so that 11 decimal becomes 10100.
10100 is not the only way to make 11 from the Fibonacci numbers however; 0*13 + 1*8 + 0*5 + 0*3 + 1*2 + 1*1 or 010011 would also represent decimal 11. For a true Zeckendorf number there is the added restriction that no two consecutive Fibonacci numbers can be used which leads to the former unique solution.
Task
Generate and show here a table of the Zeckendorf number representations of the decimal numbers zero to twenty, in order.
The intention in this task to find the Zeckendorf form of an arbitrary integer. The Zeckendorf form can be iterated by some bit twiddling rather than calculating each value separately but leave that to another separate task.
Also see
OEIS A014417 for the the sequence of required results.
Brown's Criterion - Numberphile
Related task
Fibonacci sequence
| #Plain_TeX | Plain TeX | \def\genfibolist#1{% #creates the fibo list which sum>=#1
\let\fibolist\empty\def\targetsum{#1}\def\fibosum{0}%
\genfibolistaux1,1\relax
}
\def\genfibolistaux#1,#2\relax{%
\ifnum\fibosum<\targetsum\relax
\edef\fibosum{\number\numexpr\fibosum+#2}%
\edef\fibolist{#2,\fibolist}%
\edef\tempfibo{\noexpand\genfibolistaux#2,\number\numexpr#1+#2\relax\relax}%
\expandafter\tempfibo
\fi
}
\def\zeckendorf#1{\expandafter\zeckendorfaux\fibolist,\relax#1\relax\relax0}
\def\zeckendorfaux#1,#2\relax#3\relax#4\relax#5{%
\ifx\relax#2\relax
#4%
\else
\ifnum#3<#1
\edef\temp{#2\relax#3\relax#4\ifnum#5=1 0\fi\relax#5}%
\else
\edef\temp{#2\relax\number\numexpr#3-#1\relax\relax#41\relax1}%
\fi
\expandafter\expandafter\expandafter\zeckendorfaux\expandafter\temp
\fi
}
\newcount\ii
\def\listzeckendorf#1{%
\genfibolist{#1}%
\ii=0
\loop
\ifnum\ii<#1
\advance\ii1
\number\ii: \zeckendorf\ii\endgraf
\repeat
}
\listzeckendorf{20}% any integer accepted
\bye |
http://rosettacode.org/wiki/Zeckendorf_number_representation | Zeckendorf number representation | Just as numbers can be represented in a positional notation as sums of multiples of the powers of ten (decimal) or two (binary); all the positive integers can be represented as the sum of one or zero times the distinct members of the Fibonacci series.
Recall that the first six distinct Fibonacci numbers are: 1, 2, 3, 5, 8, 13.
The decimal number eleven can be written as 0*13 + 1*8 + 0*5 + 1*3 + 0*2 + 0*1 or 010100 in positional notation where the columns represent multiplication by a particular member of the sequence. Leading zeroes are dropped so that 11 decimal becomes 10100.
10100 is not the only way to make 11 from the Fibonacci numbers however; 0*13 + 1*8 + 0*5 + 0*3 + 1*2 + 1*1 or 010011 would also represent decimal 11. For a true Zeckendorf number there is the added restriction that no two consecutive Fibonacci numbers can be used which leads to the former unique solution.
Task
Generate and show here a table of the Zeckendorf number representations of the decimal numbers zero to twenty, in order.
The intention in this task to find the Zeckendorf form of an arbitrary integer. The Zeckendorf form can be iterated by some bit twiddling rather than calculating each value separately but leave that to another separate task.
Also see
OEIS A014417 for the the sequence of required results.
Brown's Criterion - Numberphile
Related task
Fibonacci sequence
| #PowerShell | PowerShell |
function Get-ZeckendorfNumber ( $N )
{
# Calculate relevant portation of Fibonacci series
$Fib = @( 1, 1 )
While ( $Fib[-1] -lt $N ) { $Fib += $Fib[-1] + $Fib[-2] }
# Start with 0
$ZeckendorfNumber = 0
# For each number in the relevant portion of Fibonacci series
For ( $i = $Fib.Count - 1; $i -gt 0; $i-- )
{
# If Fibonacci number is less than or equal to remainder of N
If ( $Fib[$i] -le $N )
{
# Double Z number and add 1 (equivalent to adding a '1' to the end of a binary number)
$ZeckendorfNumber = $ZeckendorfNumber * 2 + 1
# Reduce N by Fibonacci number, skip next Fibonacci number
$N -= $Fib[$i--]
}
# If were aren't finished yet, double Z number
# (equivalent to adding a '0' to the end of a binary number)
If ( $i ) { $ZeckendorfNumber *= 2 }
}
return $ZeckendorfNumber
}
|
http://rosettacode.org/wiki/100_doors | 100 doors | There are 100 doors in a row that are all initially closed.
You make 100 passes by the doors.
The first time through, visit every door and toggle the door (if the door is closed, open it; if it is open, close it).
The second time, only visit every 2nd door (door #2, #4, #6, ...), and toggle it.
The third time, visit every 3rd door (door #3, #6, #9, ...), etc, until you only visit the 100th door.
Task
Answer the question: what state are the doors in after the last pass? Which are open, which are closed?
Alternate:
As noted in this page's discussion page, the only doors that remain open are those whose numbers are perfect squares.
Opening only those doors is an optimization that may also be expressed;
however, as should be obvious, this defeats the intent of comparing implementations across programming languages.
| #D | D | import std.stdio;
const N = 101; // #doors + 1
void main() {
bool[N] doors = false;
for(auto door=1; door<N; door++ ) {
for(auto i=door; i<N; i+=door ) doors[i] = !doors[i];
if (doors[door]) write(door, " ");
}
} |
http://rosettacode.org/wiki/Arrays | Arrays | This task is about arrays.
For hashes or associative arrays, please see Creating an Associative Array.
For a definition and in-depth discussion of what an array is, see Array.
Task
Show basic array syntax in your language.
Basically, create an array, assign a value to it, and retrieve an element (if available, show both fixed-length arrays and
dynamic arrays, pushing a value into it).
Please discuss at Village Pump: Arrays.
Please merge code in from these obsolete tasks:
Creating an Array
Assigning Values to an Array
Retrieving an Element of an Array
Related tasks
Collections
Creating an Associative Array
Two-dimensional array (runtime)
| #Halon | Halon | $array = [];
$array[] = 1;
$array["key"] = 3;
$array[0] = 2;
echo $array[0];
echo $array["key"]; |
http://rosettacode.org/wiki/Arithmetic/Complex | Arithmetic/Complex | A complex number is a number which can be written as:
a
+
b
×
i
{\displaystyle a+b\times i}
(sometimes shown as:
b
+
a
×
i
{\displaystyle b+a\times i}
where
a
{\displaystyle a}
and
b
{\displaystyle b}
are real numbers, and
i
{\displaystyle i}
is √ -1
Typically, complex numbers are represented as a pair of real numbers called the "imaginary part" and "real part", where the imaginary part is the number to be multiplied by
i
{\displaystyle i}
.
Task
Show addition, multiplication, negation, and inversion of complex numbers in separate functions. (Subtraction and division operations can be made with pairs of these operations.)
Print the results for each operation tested.
Optional: Show complex conjugation.
By definition, the complex conjugate of
a
+
b
i
{\displaystyle a+bi}
is
a
−
b
i
{\displaystyle a-bi}
Some languages have complex number libraries available. If your language does, show the operations. If your language does not, also show the definition of this type.
| #Vlang | Vlang | import math.complex
fn main() {
a := complex.complex(1, 1)
b := complex.complex(3.14159, 1.25)
println("a: $a")
println("b: $b")
println("a + b: ${a+b}")
println("a * b: ${a*b}")
println("-a: ${a.addinv()}")
println("1 / a: ${complex.complex(1,0)/a}")
println("a̅: ${a.conjugate()}")
} |
http://rosettacode.org/wiki/Arithmetic/Complex | Arithmetic/Complex | A complex number is a number which can be written as:
a
+
b
×
i
{\displaystyle a+b\times i}
(sometimes shown as:
b
+
a
×
i
{\displaystyle b+a\times i}
where
a
{\displaystyle a}
and
b
{\displaystyle b}
are real numbers, and
i
{\displaystyle i}
is √ -1
Typically, complex numbers are represented as a pair of real numbers called the "imaginary part" and "real part", where the imaginary part is the number to be multiplied by
i
{\displaystyle i}
.
Task
Show addition, multiplication, negation, and inversion of complex numbers in separate functions. (Subtraction and division operations can be made with pairs of these operations.)
Print the results for each operation tested.
Optional: Show complex conjugation.
By definition, the complex conjugate of
a
+
b
i
{\displaystyle a+bi}
is
a
−
b
i
{\displaystyle a-bi}
Some languages have complex number libraries available. If your language does, show the operations. If your language does not, also show the definition of this type.
| #Wortel | Wortel | @class Complex {
&[r i] @: {
^r || r 0
^i || i 0
^m +@sq^r @sq^i
}
add &o @new Complex[+ ^r o.r + ^i o.i]
mul &o @new Complex[-* ^r o.r * ^i o.i +* ^r o.i * ^i o.r]
neg &^ @new Complex[@-^r @-^i]
inv &^ @new Complex[/ ^r ^m / @-^i ^m]
toString &^?{
=^i 0 "{^r}"
=^r 0 "{^i}i"
>^i 0 "{^r} + {^i}i"
"{^r} - {@-^i}i"
}
}
@vars {
a @new Complex[5 3]
b @new Complex[4 3N]
}
@each &x !console.log x [
"({a}) + ({b}) = {!a.add b}"
"({a}) * ({b}) = {!a.mul b}"
"-1 * ({b}) = {b.neg.}"
"({a}) - ({b}) = {!a.add b.neg.}"
"1 / ({b}) = {b.inv.}"
"({!a.mul b}) / ({b}) = {`!.mul b.inv. !a.mul b}"
] |
http://rosettacode.org/wiki/Zero_to_the_zero_power | Zero to the zero power | Some computer programming languages are not exactly consistent (with other computer programming languages)
when raising zero to the zeroth power: 00
Task
Show the results of raising zero to the zeroth power.
If your computer language objects to 0**0 or 0^0 at compile time, you may also try something like:
x = 0
y = 0
z = x**y
say 'z=' z
Show the result here.
And of course use any symbols or notation that is supported in your computer programming language for exponentiation.
See also
The Wiki entry: Zero to the power of zero.
The Wiki entry: History of differing points of view.
The MathWorld™ entry: exponent laws.
Also, in the above MathWorld™ entry, see formula (9):
x
0
=
1
{\displaystyle x^{0}=1}
.
The OEIS entry: The special case of zero to the zeroth power
| #Quackery | Quackery | /O> 0 0 **
...
Stack: 1
|
http://rosettacode.org/wiki/Zero_to_the_zero_power | Zero to the zero power | Some computer programming languages are not exactly consistent (with other computer programming languages)
when raising zero to the zeroth power: 00
Task
Show the results of raising zero to the zeroth power.
If your computer language objects to 0**0 or 0^0 at compile time, you may also try something like:
x = 0
y = 0
z = x**y
say 'z=' z
Show the result here.
And of course use any symbols or notation that is supported in your computer programming language for exponentiation.
See also
The Wiki entry: Zero to the power of zero.
The Wiki entry: History of differing points of view.
The MathWorld™ entry: exponent laws.
Also, in the above MathWorld™ entry, see formula (9):
x
0
=
1
{\displaystyle x^{0}=1}
.
The OEIS entry: The special case of zero to the zeroth power
| #R | R | print(0^0) |
http://rosettacode.org/wiki/Zero_to_the_zero_power | Zero to the zero power | Some computer programming languages are not exactly consistent (with other computer programming languages)
when raising zero to the zeroth power: 00
Task
Show the results of raising zero to the zeroth power.
If your computer language objects to 0**0 or 0^0 at compile time, you may also try something like:
x = 0
y = 0
z = x**y
say 'z=' z
Show the result here.
And of course use any symbols or notation that is supported in your computer programming language for exponentiation.
See also
The Wiki entry: Zero to the power of zero.
The Wiki entry: History of differing points of view.
The MathWorld™ entry: exponent laws.
Also, in the above MathWorld™ entry, see formula (9):
x
0
=
1
{\displaystyle x^{0}=1}
.
The OEIS entry: The special case of zero to the zeroth power
| #Racket | Racket | #lang racket
;; as many zeros as I can think of...
(define zeros (list
0 ; unspecified number type
0. ; hinted as float
#e0 ; explicitly exact
#i0 ; explicitly inexact
0+0i ; exact complex
0.+0.i ; float inexact
))
(for*((z zeros) (p zeros))
(printf "(~a)^(~a) = ~s~%" z p
(with-handlers [(exn:fail:contract:divide-by-zero? exn-message)]
(expt z p)))) |
http://rosettacode.org/wiki/Zebra_puzzle | Zebra puzzle | Zebra puzzle
You are encouraged to solve this task according to the task description, using any language you may know.
The Zebra puzzle, a.k.a. Einstein's Riddle,
is a logic puzzle which is to be solved programmatically.
It has several variants, one of them this:
There are five houses.
The English man lives in the red house.
The Swede has a dog.
The Dane drinks tea.
The green house is immediately to the left of the white house.
They drink coffee in the green house.
The man who smokes Pall Mall has birds.
In the yellow house they smoke Dunhill.
In the middle house they drink milk.
The Norwegian lives in the first house.
The man who smokes Blend lives in the house next to the house with cats.
In a house next to the house where they have a horse, they smoke Dunhill.
The man who smokes Blue Master drinks beer.
The German smokes Prince.
The Norwegian lives next to the blue house.
They drink water in a house next to the house where they smoke Blend.
The question is, who owns the zebra?
Additionally, list the solution for all the houses.
Optionally, show the solution is unique.
Related tasks
Dinesman's multiple-dwelling problem
Twelve statements
| #FormulaOne | FormulaOne |
// First, let's give some type-variables some values:
Nationality = Englishman | Swede | Dane | Norwegian | German
Colour = Red | Green | Yellow | Blue | White
Cigarette = PallMall | Dunhill | BlueMaster | Blend | Prince
Domestic = Dog | Bird | Cat | Zebra | Horse
Beverage = Tea | Coffee | Milk | Beer | Water
HouseRow = First | Second | Third | Fourth | Fifth
{
We use injections to make the array-elements unique.
Example: 'Pet' is an array of unique elements of type 'Domestic', indexed by 'Nationality'.
In the predicate 'Zebra', we use this injection 'Pet' to define the array-variable 'pet'
as a parameter of the 'Zebra'-predicate.
The symbol used is the '->>'. 'Nationality->>Domestic' can be read as 'Domestic(Nationality)'
in "plain array-speak";
the difference being that the elements are by definition unique (cf. 'injective function').
So, in FormulaOne we use a formula like: 'pet(Swede) = Dog', which simply means that the 'Swede'
(type 'Nationality') has a 'pet' (type 'Pet', of type 'Domestic', indexed by 'Nationality'),
which appears to be a 'Dog' (type 'Domestic').
Or, one could say that the 'Swede' has been mapped to the 'Dog' (Oh, well...).
}
Pet = Nationality->>Domestic
Drink = Nationality->>Beverage
HouseColour = Nationality->>Colour
Smoke = Nationality->>Cigarette
HouseOrder = HouseRow->>Nationality
pred Zebra(house_olour::HouseColour, pet::Pet, smoke::Smoke, drink::Drink, house_order::HouseOrder) iff
// For convenience sake, some temporary place_holder variables are used.
// An underscore distinguishes them:
house_colour(green_house) = Green &
house_colour(white_house) = White &
house_colour(yellow_house) = Yellow &
smoke(pallmall_smoker) = PallMall &
smoke(blend_smoker) = Blend &
smoke(dunhill_smoker) = Dunhill &
smoke(bluemaster_smoker) = BlueMaster &
pet(cat_keeper) = Cat &
pet(neighbour_dunhill_smoker) = Horse &
{ 2. The English man lives in the red house: }
house_colour(Englishman) = Red &
{ 3. The Swede has a dog: }
pet(Swede) = Dog &
{ 4. The Dane drinks tea: }
drink(Dane) = Tea &
{ 'smoke' and 'drink' are both nouns, like the other variables.
One could read the formulas like: 'the colour of the Englishman's house is Red' ->
'the Swede's pet is a dog' -> 'the Dane's drink is tea'.
}
{ 5. The green house is immediately to the left of the white house.
The local predicate 'LeftOf' (see below) determines the house order: }
LeftOf(green_house, white_house, house_order) &
{ 6. They drink coffee in the green house: }
drink(green_house) = Coffee &
{ 7. The man who smokes Pall Mall has birds: }
pet(pallmall_smoker) = Bird &
{ 8. In the yellow house they smoke Dunhill: }
smoke(yellow_house) = Dunhill &
{ 9. In the middle house (third in the row) they drink milk: }
drink(house_order(Third)) = Milk &
{10. The Norwegian lives in the first house: }
house_order(First) = Norwegian &
{11. The man who smokes Blend lives in the house next to the house with cats.
Another local predicate 'Neighbour' makes them neighbours: }
Neighbour(blend_smoker, cat_keeper, house_order) &
{12. In a house next to the house where they have a horse, they smoke Dunhill: }
Neighbour(dunhill_smoker, neighbour_dunhill_smoker, house_order) &
{13. The man who smokes Blue Master drinks beer: }
drink(bluemaster_smoker) = Beer &
{14. The German smokes Prince: }
smoke(German) = Prince &
{15. The Norwegian lives next to the blue house
Cf. 10. "The Norwegian lives in the first house", so the blue house is the second house: }
house_colour(house_order(Second)) = Blue &
{16. They drink water in a house next to the house where they smoke Blend: }
drink(neighbour_blend_smoker) = Water &
Neighbour(blend_smoker, neighbour_blend_smoker, house_order)
{ A simplified solution would number the houses 1, 2, 3, 4, 5
which makes it easier to order the houses.
'right in the center' would become 3; 'in the first house', 1
But we stick to the original puzzle and use some local predicates.
}
local pred Neighbour(neighbour1::Nationality, neighbour2::Nationality, house_order::HouseOrder)iff
neighbour1 <> neighbour2 &
house_order(house1) = neighbour1 &
house_order(house2) = neighbour2 &
( house1 = house2 + 1 |
house1 = house2 - 1 )
local pred LeftOf(neighbour1::Nationality, neighbour2::Nationality, house_order::HouseOrder) iff
neighbour1 <> neighbour2 &
house_order(house1) = neighbour1 &
house_order(house2) = neighbour2 &
house1 = house2 - 1
{
The 'all'-query in FormulaOne:
all Zebra(house_colour, pet, smokes, drinks, house_order)
gives, of course, only one solution, so it can be replaced by:
one Zebra(house_colour, pet, smokes, drinks, house_order)
}
// The compacted version:
Nationality = Englishman | Swede | Dane | Norwegian | German
Colour = Red | Green | Yellow | Blue | White
Cigarette = PallMall | Dunhill | BlueMaster | Blend | Prince
Domestic = Dog | Bird | Cat | Zebra | Horse
Beverage = Tea | Coffee | Milk | Beer | Water
HouseRow = First | Second | Third | Fourth | Fifth
Pet = Nationality->>Domestic
Drink = Nationality->>Beverage
HouseColour = Nationality->>Colour
Smoke = Nationality->>Cigarette
HouseOrder = HouseRow->>Nationality
pred Zebra(house_colour::HouseColour, pet::Pet, smoke::Smoke, drink::Drink, house_order::HouseOrder) iff
house-colour(green_house) = Green &
house-colour(white_house) = White &
house-colour(yellow_house) = Yellow &
smoke(pallmall_smoker) = PallMall &
smoke(blend_smoker) = Blend &
smoke(dunhill_smoker) = Dunhill &
smoke(bluemaster_smoker) = BlueMaster &
pet(cat_keeper) = Cat &
pet(neighbour_dunhill_smoker) = Horse &
house_colour(Englishman) = Red &
pet(Swede) = Dog &
drink(Dane) = Tea &
LeftOf(green_house, white_house, house_order) &
drink(green_house) = Coffee &
pet(pallmall_smoker) = Bird &
smoke(yellow_house) = Dunhill &
drink(house_order(Third)) = Milk &
house_order(First) = Norwegian &
Neighbour(blend_smoker, cat_keeper, house_order) &
Neighbour(dunhill_smoker, neighbour_dunhill_smoker, house_order) &
drink(bluemaster_smoker) = Beer &
smoke(German) = Prince &
house_colour(house_order(Second)) = Blue &
drink(neighbour_blend_smoker) = Water &
Neighbour(blend_smoker, neighbour_blend_smoker, house_order)
local pred Neighbour(neighbour1::Nationality, neighbour2::Nationality, house_order::HouseOrder)iff
neighbour1 <> neighbour2 &
house_order(house1) = neighbour1 & house_order(house2) = neighbour2 &
( house1 = house2 + 1 | house1 = house2 - 1 )
local pred LeftOf(neighbour1::Nationality, neighbour2::Nationality, house_::HouseOrder) iff
neighbour1 <> neighbour2 &
house_order(house1) = neighbour1 & house_order(house2) = neighbour2 &
house1 = house2 - 1
|
http://rosettacode.org/wiki/XML/XPath | XML/XPath | Perform the following three XPath queries on the XML Document below:
//item[1]: Retrieve the first "item" element
//price/text(): Perform an action on each "price" element (print it out)
//name: Get an array of all the "name" elements
XML Document:
<inventory title="OmniCorp Store #45x10^3">
<section name="health">
<item upc="123456789" stock="12">
<name>Invisibility Cream</name>
<price>14.50</price>
<description>Makes you invisible</description>
</item>
<item upc="445322344" stock="18">
<name>Levitation Salve</name>
<price>23.99</price>
<description>Levitate yourself for up to 3 hours per application</description>
</item>
</section>
<section name="food">
<item upc="485672034" stock="653">
<name>Blork and Freen Instameal</name>
<price>4.95</price>
<description>A tasty meal in a tablet; just add water</description>
</item>
<item upc="132957764" stock="44">
<name>Grob winglets</name>
<price>3.56</price>
<description>Tender winglets of Grob. Just add water</description>
</item>
</section>
</inventory>
| #HicEst | HicEst | CHARACTER xml*1000, output*1000
READ(ClipBoard) xml
EDIT(Text=xml, Right='<item', Right=5, GetPosition=a, Right='</item>', Left, GetPosition=z)
WRITE(Text=output) xml( a : z), $CRLF
i = 1
1 EDIT(Text=xml, SetPosition=i, SePaRators='<>', Right='<price>', Word=1, Parse=price, GetPosition=i, ERror=99)
IF(i > 0) THEN
WRITE(Text=output, APPend) 'Price element = ', price, $CRLF
GOTO 1 ! HicEst does not have a "WHILE"
ENDIF
EDIT(Text=xml, SPR='<>', R='<name>', W=1, WordEnd=$CR, APpendTo=output, DO=999)
WRITE(ClipBoard) TRIM(output) |
http://rosettacode.org/wiki/XML/XPath | XML/XPath | Perform the following three XPath queries on the XML Document below:
//item[1]: Retrieve the first "item" element
//price/text(): Perform an action on each "price" element (print it out)
//name: Get an array of all the "name" elements
XML Document:
<inventory title="OmniCorp Store #45x10^3">
<section name="health">
<item upc="123456789" stock="12">
<name>Invisibility Cream</name>
<price>14.50</price>
<description>Makes you invisible</description>
</item>
<item upc="445322344" stock="18">
<name>Levitation Salve</name>
<price>23.99</price>
<description>Levitate yourself for up to 3 hours per application</description>
</item>
</section>
<section name="food">
<item upc="485672034" stock="653">
<name>Blork and Freen Instameal</name>
<price>4.95</price>
<description>A tasty meal in a tablet; just add water</description>
</item>
<item upc="132957764" stock="44">
<name>Grob winglets</name>
<price>3.56</price>
<description>Tender winglets of Grob. Just add water</description>
</item>
</section>
</inventory>
| #Java | Java | import java.io.StringReader;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.xpath.XPath;
import javax.xml.xpath.XPathConstants;
import javax.xml.xpath.XPathFactory;
import org.w3c.dom.Document;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;
import org.xml.sax.InputSource;
public class XMLParser {
final static String xmlStr =
"<inventory title=\"OmniCorp Store #45x10^3\">"
+ " <section name=\"health\">"
+ " <item upc=\"123456789\" stock=\"12\">"
+ " <name>Invisibility Cream</name>"
+ " <price>14.50</price>"
+ " <description>Makes you invisible</description>"
+ " </item>"
+ " <item upc=\"445322344\" stock=\"18\">"
+ " <name>Levitation Salve</name>"
+ " <price>23.99</price>"
+ " <description>Levitate yourself for up to 3 hours per application</description>"
+ " </item>"
+ " </section>"
+ " <section name=\"food\">"
+ " <item upc=\"485672034\" stock=\"653\">"
+ " <name>Blork and Freen Instameal</name>"
+ " <price>4.95</price>"
+ " <description>A tasty meal in a tablet; just add water</description>"
+ " </item>"
+ " <item upc=\"132957764\" stock=\"44\">"
+ " <name>Grob winglets</name>"
+ " <price>3.56</price>"
+ " <description>Tender winglets of Grob. Just add priwater</description>"
+ " </item>"
+ " </section>"
+ "</inventory>";
public static void main(String[] args) {
try {
Document doc = DocumentBuilderFactory.newInstance()
.newDocumentBuilder()
.parse(new InputSource(new StringReader(xmlStr)));
XPath xpath = XPathFactory.newInstance().newXPath();
// 1
System.out.println(((Node) xpath.evaluate(
"/inventory/section/item[1]", doc, XPathConstants.NODE))
.getAttributes().getNamedItem("upc"));
// 2, 3
NodeList nodes = (NodeList) xpath.evaluate(
"/inventory/section/item/price", doc,
XPathConstants.NODESET);
for (int i = 0; i < nodes.getLength(); i++)
System.out.println(nodes.item(i).getTextContent());
} catch (Exception e) {
System.out.println("Error ocurred while parsing XML.");
}
}
} |
http://rosettacode.org/wiki/Yin_and_yang | Yin and yang | One well-known symbol of the philosophy of duality known as yin and yang is the taijitu.
Task
Create a function that, given a parameter representing size, generates such a symbol scaled to the requested size.
Generate and display the symbol for two different (small) sizes.
| #J | J | yinyang=:3 :0
radii=. y*1 3 6
ranges=. i:each radii
squares=. ,"0/~each ranges
circles=. radii ([ >: +/"1&.:*:@])each squares
cInds=. ({:radii) +each circles #&(,/)each squares
M=. ' *.' {~ circles (* 1 + 0 >: {:"1)&(_1&{::) squares
offset=. 3*y,0
M=. '*' ((_2 {:: cInds) <@:+"1 offset)} M
M=. '.' ((_2 {:: cInds) <@:-"1 offset)} M
M=. '.' ((_3 {:: cInds) <@:+"1 offset)} M
M=. '*' ((_3 {:: cInds) <@:-"1 offset)} M
) |
http://rosettacode.org/wiki/Y_combinator | Y combinator | In strict functional programming and the lambda calculus, functions (lambda expressions) don't have state and are only allowed to refer to arguments of enclosing functions.
This rules out the usual definition of a recursive function wherein a function is associated with the state of a variable and this variable's state is used in the body of the function.
The Y combinator is itself a stateless function that, when applied to another stateless function, returns a recursive version of the function.
The Y combinator is the simplest of the class of such functions, called fixed-point combinators.
Task
Define the stateless Y combinator and use it to compute factorials and Fibonacci numbers from other stateless functions or lambda expressions.
Cf
Jim Weirich: Adventures in Functional Programming
| #D.C3.A9j.C3.A0_Vu | Déjà Vu | Y f:
labda y:
labda:
call y @y
f
labda x:
x @x
call
labda f:
labda n:
if < 1 n:
* n f -- n
else:
1
set :fac Y
labda f:
labda n:
if < 1 n:
+ f - n 2 f -- n
else:
1
set :fib Y
!. fac 6
!. fib 6 |
http://rosettacode.org/wiki/Zig-zag_matrix | Zig-zag matrix | Task
Produce a zig-zag array.
A zig-zag array is a square arrangement of the first N2 natural numbers, where the
numbers increase sequentially as you zig-zag along the array's anti-diagonals.
For a graphical representation, see JPG zigzag (JPG uses such arrays to encode images).
For example, given 5, produce this array:
0 1 5 6 14
2 4 7 13 15
3 8 12 16 21
9 11 17 20 22
10 18 19 23 24
Related tasks
Spiral matrix
Identity matrix
Ulam spiral (for primes)
See also
Wiktionary entry: anti-diagonals
| #Erlang | Erlang |
-module( zigzag ).
-export( [matrix/1, task/0] ).
matrix( N ) ->
{{_X_Y, N}, Proplist} = lists:foldl( fun matrix_as_proplist/2, {{{0, 0}, N}, []}, lists:seq(0, (N * N) - 1) ),
[columns( X, Proplist ) || X <- lists:seq(0, N - 1)].
task() -> matrix( 5 ).
columns( Column, Proplist ) -> lists:sort( [Value || {{_X, Y}, Value} <- Proplist, Y =:= Column] ).
matrix_as_proplist( N, {{X_Y, Max}, Acc} ) ->
Next = next_indexes( X_Y, Max ),
{{Next, Max}, [{X_Y, N} | Acc]}.
next_indexes( {X, Y}, Max ) when Y + 1 =:= Max, (X + Y) rem 2 =:= 0 -> {X + 1, Y - 1};
next_indexes( {X, Y}, Max ) when Y + 1 =:= Max, (X + Y) rem 2 =:= 1 -> {X + 1, Y};
next_indexes( {X, Y}, Max ) when X + 1 =:= Max, (X + Y) rem 2 =:= 0 -> {X, Y + 1};
next_indexes( {X, Y}, Max ) when X + 1 =:= Max, (X + Y) rem 2 =:= 1 -> {X - 1, Y + 1};
next_indexes( {X, 0}, _Max ) when X rem 2 =:= 0 -> {X + 1, 0};
next_indexes( {X, 0}, _Max ) when X rem 2 =:= 1 -> {X - 1, 1};
next_indexes( {0, Y}, _Max ) when Y rem 2 =:= 0 -> {1, Y - 1};
next_indexes( {0, Y}, _Max ) when Y rem 2 =:= 1 -> {0, Y + 1};
next_indexes( {X, Y}, _Max ) when (X + Y) rem 2 =:= 0 -> {X + 1, Y - 1};
next_indexes( {X, Y}, _Max ) when (X + Y) rem 2 =:= 1 -> {X - 1, Y + 1}.
|
http://rosettacode.org/wiki/Arbitrary-precision_integers_(included) | Arbitrary-precision integers (included) | Using the in-built capabilities of your language, calculate the integer value of:
5
4
3
2
{\displaystyle 5^{4^{3^{2}}}}
Confirm that the first and last twenty digits of the answer are:
62060698786608744707...92256259918212890625
Find and show the number of decimal digits in the answer.
Note: Do not submit an implementation of arbitrary precision arithmetic. The intention is to show the capabilities of the language as supplied. If a language has a single, overwhelming, library of varied modules that is endorsed by its home site – such as CPAN for Perl or Boost for C++ – then that may be used instead.
Strictly speaking, this should not be solved by fixed-precision numeric libraries where the precision has to be manually set to a large value; although if this is the only recourse then it may be used with a note explaining that the precision must be set manually to a large enough value.
Related tasks
Long multiplication
Exponentiation order
exponentiation operator
Exponentiation with infix operators in (or operating on) the base
| #Oz | Oz | declare
Pow5432 = {Pow 5 {Pow 4 {Pow 3 2}}}
S = {Int.toString Pow5432}
Len = {Length S}
in
{System.showInfo
{List.take S 20}#"..."#
{List.drop S Len-20}#" ("#Len#" Digits)"} |
http://rosettacode.org/wiki/Arbitrary-precision_integers_(included) | Arbitrary-precision integers (included) | Using the in-built capabilities of your language, calculate the integer value of:
5
4
3
2
{\displaystyle 5^{4^{3^{2}}}}
Confirm that the first and last twenty digits of the answer are:
62060698786608744707...92256259918212890625
Find and show the number of decimal digits in the answer.
Note: Do not submit an implementation of arbitrary precision arithmetic. The intention is to show the capabilities of the language as supplied. If a language has a single, overwhelming, library of varied modules that is endorsed by its home site – such as CPAN for Perl or Boost for C++ – then that may be used instead.
Strictly speaking, this should not be solved by fixed-precision numeric libraries where the precision has to be manually set to a large value; although if this is the only recourse then it may be used with a note explaining that the precision must be set manually to a large enough value.
Related tasks
Long multiplication
Exponentiation order
exponentiation operator
Exponentiation with infix operators in (or operating on) the base
| #PARI.2FGP | PARI/GP | num_first_last_digits(a=5,b=4^3^2,n=20)={ my(L = b*log(a)/log(10), m=Mod(a,10^n)^b);
[L\1+1, 10^frac(L)\10^(1-n), lift(m)] \\ where x\y = floor(x/y) but more efficient
}
print("Length, first and last 20 digits of 5^4^3^2: ", num_first_last_digits()) \\ uses default values a=5, b=4^3^2, n=20 |
http://rosettacode.org/wiki/Zeckendorf_number_representation | Zeckendorf number representation | Just as numbers can be represented in a positional notation as sums of multiples of the powers of ten (decimal) or two (binary); all the positive integers can be represented as the sum of one or zero times the distinct members of the Fibonacci series.
Recall that the first six distinct Fibonacci numbers are: 1, 2, 3, 5, 8, 13.
The decimal number eleven can be written as 0*13 + 1*8 + 0*5 + 1*3 + 0*2 + 0*1 or 010100 in positional notation where the columns represent multiplication by a particular member of the sequence. Leading zeroes are dropped so that 11 decimal becomes 10100.
10100 is not the only way to make 11 from the Fibonacci numbers however; 0*13 + 1*8 + 0*5 + 0*3 + 1*2 + 1*1 or 010011 would also represent decimal 11. For a true Zeckendorf number there is the added restriction that no two consecutive Fibonacci numbers can be used which leads to the former unique solution.
Task
Generate and show here a table of the Zeckendorf number representations of the decimal numbers zero to twenty, in order.
The intention in this task to find the Zeckendorf form of an arbitrary integer. The Zeckendorf form can be iterated by some bit twiddling rather than calculating each value separately but leave that to another separate task.
Also see
OEIS A014417 for the the sequence of required results.
Brown's Criterion - Numberphile
Related task
Fibonacci sequence
| #PureBasic | PureBasic | Procedure.s zeck(n.i)
Dim f.i(1) : Define i.i=1, o$
f(0)=1 : f(1)=1
While f(i)<n
i+1 : ReDim f(ArraySize(f())+1) : f(i)=f(i-1)+f(i-2)
Wend
For i=i To 1 Step -1
If n>=f(i) : o$+"1" : n-f(i) : Else : o$+"0" : EndIf
Next
If Len(o$)>1 : o$=LTrim(o$,"0") : EndIf
ProcedureReturn o$
EndProcedure
Define n.i, t$
OpenConsole("Zeckendorf number representation")
PrintN(~"\tNr.\tZeckendorf")
For n=0 To 20
t$=zeck(n)
If FindString(t$,"11")
PrintN("Error: n= "+Str(n)+~"\tZeckendorf= "+t$)
Break
Else
PrintN(~"\t"+RSet(Str(n),3," ")+~"\t"+RSet(t$,7," "))
EndIf
Next
Input() |
http://rosettacode.org/wiki/100_doors | 100 doors | There are 100 doors in a row that are all initially closed.
You make 100 passes by the doors.
The first time through, visit every door and toggle the door (if the door is closed, open it; if it is open, close it).
The second time, only visit every 2nd door (door #2, #4, #6, ...), and toggle it.
The third time, visit every 3rd door (door #3, #6, #9, ...), etc, until you only visit the 100th door.
Task
Answer the question: what state are the doors in after the last pass? Which are open, which are closed?
Alternate:
As noted in this page's discussion page, the only doors that remain open are those whose numbers are perfect squares.
Opening only those doors is an optimization that may also be expressed;
however, as should be obvious, this defeats the intent of comparing implementations across programming languages.
| #Dafny | Dafny |
datatype Door = Closed | Open
method InitializeDoors(n:int) returns (doors:array<Door>)
// Precondition: n must be a valid array size.
requires n >= 0
// Postcondition: doors is an array, which is not an alias for any other
// object, with a length of n, all of whose elements are Closed. The "fresh"
// (non-alias) condition is needed to allow doors to be modified by the
// remaining code.
ensures doors != null && fresh(doors) && doors.Length == n
ensures forall j :: 0 <= j < doors.Length ==> doors[j] == Closed;
{
doors := new Door[n];
var i := 0;
// Invariant: i is always a valid index inside the loop, and all doors less
// than i are Closed. These invariants are needed to ensure the second
// postcondition.
while i < doors.Length
invariant i <= doors.Length
invariant forall j :: 0 <= j < i ==> doors[j] == Closed;
{
doors[i] := Closed;
i := i + 1;
}
}
method Main ()
{
var doors := InitializeDoors(100);
var pass := 1;
while pass <= doors.Length
{
var door := pass;
while door < doors.Length
{
doors[door] := if doors[door] == Closed then Open else Closed;
door := door + pass;
}
pass := pass + 1;
}
var i := 0;
while i < doors.Length
{
print i, " is ", if doors[i] == Closed then "closed\n" else "open\n";
i := i + 1;
}
}
|
http://rosettacode.org/wiki/Arrays | Arrays | This task is about arrays.
For hashes or associative arrays, please see Creating an Associative Array.
For a definition and in-depth discussion of what an array is, see Array.
Task
Show basic array syntax in your language.
Basically, create an array, assign a value to it, and retrieve an element (if available, show both fixed-length arrays and
dynamic arrays, pushing a value into it).
Please discuss at Village Pump: Arrays.
Please merge code in from these obsolete tasks:
Creating an Array
Assigning Values to an Array
Retrieving an Element of an Array
Related tasks
Collections
Creating an Associative Array
Two-dimensional array (runtime)
| #Harbour | Harbour | // Declare and initialize two-dimensional array
local arr1 := { { "NITEM", "N", 10, 0 }, { "CONTENT", "C", 60, 0 } }
// Create an empty array
local arr2 := {}
// Declare three-dimensional array
local arr3[ 2, 100, 3 ]
// Create an array
local arr4 := Array( 50 )
// Array can be dynamically resized:
arr4 := ASize( arr4, 80 ) |
http://rosettacode.org/wiki/Arithmetic/Complex | Arithmetic/Complex | A complex number is a number which can be written as:
a
+
b
×
i
{\displaystyle a+b\times i}
(sometimes shown as:
b
+
a
×
i
{\displaystyle b+a\times i}
where
a
{\displaystyle a}
and
b
{\displaystyle b}
are real numbers, and
i
{\displaystyle i}
is √ -1
Typically, complex numbers are represented as a pair of real numbers called the "imaginary part" and "real part", where the imaginary part is the number to be multiplied by
i
{\displaystyle i}
.
Task
Show addition, multiplication, negation, and inversion of complex numbers in separate functions. (Subtraction and division operations can be made with pairs of these operations.)
Print the results for each operation tested.
Optional: Show complex conjugation.
By definition, the complex conjugate of
a
+
b
i
{\displaystyle a+bi}
is
a
−
b
i
{\displaystyle a-bi}
Some languages have complex number libraries available. If your language does, show the operations. If your language does not, also show the definition of this type.
| #Wren | Wren | import "/complex" for Complex
var x = Complex.new(1, 3)
var y = Complex.new(5, 2)
System.print("x = %(x)")
System.print("y = %(y)")
System.print("x + y = %(x + y)")
System.print("x - y = %(x - y)")
System.print("x * y = %(x * y)")
System.print("x / y = %(x / y)")
System.print("-x = %(-x)")
System.print("1 / x = %(x.inverse)")
System.print("x* = %(x.conj)") |
http://rosettacode.org/wiki/Zero_to_the_zero_power | Zero to the zero power | Some computer programming languages are not exactly consistent (with other computer programming languages)
when raising zero to the zeroth power: 00
Task
Show the results of raising zero to the zeroth power.
If your computer language objects to 0**0 or 0^0 at compile time, you may also try something like:
x = 0
y = 0
z = x**y
say 'z=' z
Show the result here.
And of course use any symbols or notation that is supported in your computer programming language for exponentiation.
See also
The Wiki entry: Zero to the power of zero.
The Wiki entry: History of differing points of view.
The MathWorld™ entry: exponent laws.
Also, in the above MathWorld™ entry, see formula (9):
x
0
=
1
{\displaystyle x^{0}=1}
.
The OEIS entry: The special case of zero to the zeroth power
| #Raku | Raku | say ' type n n**n exp(n,n)';
say '-------- -------- -------- --------';
for 0, 0.0, FatRat.new(0), 0e0, 0+0i {
printf "%8s %8s %8s %8s\n", .^name, $_, $_**$_, exp($_,$_);
} |
http://rosettacode.org/wiki/Zero_to_the_zero_power | Zero to the zero power | Some computer programming languages are not exactly consistent (with other computer programming languages)
when raising zero to the zeroth power: 00
Task
Show the results of raising zero to the zeroth power.
If your computer language objects to 0**0 or 0^0 at compile time, you may also try something like:
x = 0
y = 0
z = x**y
say 'z=' z
Show the result here.
And of course use any symbols or notation that is supported in your computer programming language for exponentiation.
See also
The Wiki entry: Zero to the power of zero.
The Wiki entry: History of differing points of view.
The MathWorld™ entry: exponent laws.
Also, in the above MathWorld™ entry, see formula (9):
x
0
=
1
{\displaystyle x^{0}=1}
.
The OEIS entry: The special case of zero to the zeroth power
| #Red | Red | Red[]
print 0 ** 0
print power 0 0
print math [0 ** 0] |
http://rosettacode.org/wiki/Zebra_puzzle | Zebra puzzle | Zebra puzzle
You are encouraged to solve this task according to the task description, using any language you may know.
The Zebra puzzle, a.k.a. Einstein's Riddle,
is a logic puzzle which is to be solved programmatically.
It has several variants, one of them this:
There are five houses.
The English man lives in the red house.
The Swede has a dog.
The Dane drinks tea.
The green house is immediately to the left of the white house.
They drink coffee in the green house.
The man who smokes Pall Mall has birds.
In the yellow house they smoke Dunhill.
In the middle house they drink milk.
The Norwegian lives in the first house.
The man who smokes Blend lives in the house next to the house with cats.
In a house next to the house where they have a horse, they smoke Dunhill.
The man who smokes Blue Master drinks beer.
The German smokes Prince.
The Norwegian lives next to the blue house.
They drink water in a house next to the house where they smoke Blend.
The question is, who owns the zebra?
Additionally, list the solution for all the houses.
Optionally, show the solution is unique.
Related tasks
Dinesman's multiple-dwelling problem
Twelve statements
| #GAP | GAP | leftOf :=function(setA, vA, setB, vB)
local i;
for i in [1..4] do
if ( setA[i] = vA) and (setB[i+1] = vB) then return true ;fi;
od;
return false;
end;
nextTo :=function(setA, vA, setB, vB)
local i;
for i in [1..4] do
if ( setA[i] = vA) and (setB[i+1] = vB) then return true ;fi;
if ( setB[i] = vB) and (setA[i+1] = vA) then return true ;fi;
od;
return false;
end;
requires := function(setA, vA, setB, vB)
local i;
for i in [1..5] do
if ( setA[i] = vA) and (setB[i] = vB) then return true ;fi;
od;
return false;
end;
pcolors :=PermutationsList(["white" ,"yellow" ,"blue" ,"red" ,"green"]);
pcigars :=PermutationsList(["blends", "pall_mall", "prince", "bluemasters", "dunhill"]);
pnats:=PermutationsList(["german", "swedish", "british", "norwegian", "danish"]);
pdrinks :=PermutationsList(["beer", "water", "tea", "milk", "coffee"]);
ppets :=PermutationsList(["birds", "cats", "horses", "fish", "dogs"]);
for colors in pcolors do
if not (leftOf(colors,"green",colors,"white")) then continue ;fi;
for nats in pnats do
if not (requires(nats,"british",colors,"red")) then continue ;fi;
if not (nats[1]="norwegian") then continue ;fi;
if not (nextTo(nats,"norwegian",colors,"blue")) then continue ;fi;
for pets in ppets do
if not (requires(nats,"swedish",pets,"dogs")) then continue ;fi;
for drinks in pdrinks do
if not (drinks[3]="milk") then continue ;fi;
if not (requires(colors,"green",drinks,"coffee")) then continue ;fi;
if not (requires(nats,"danish",drinks,"tea")) then continue ;fi;
for cigars in pcigars do
if not (nextTo(pets,"horses",cigars,"dunhill")) then continue ;fi;
if not (requires(cigars,"pall_mall",pets,"birds")) then continue ;fi;
if not (nextTo(cigars,"blends",drinks,"water")) then continue ;fi;
if not (nextTo(cigars,"blends",pets,"cats")) then continue ;fi;
if not (requires(nats,"german",cigars,"prince")) then continue ;fi;
if not (requires(colors,"yellow",cigars,"dunhill")) then continue ;fi;
if not (requires(cigars,"bluemasters",drinks,"beer")) then continue ;fi;
Print(colors,"\n");
Print(nats,"\n");
Print(drinks,"\n");
Print(pets,"\n");
Print(cigars,"\n");
od;od;od;od;od;
|
http://rosettacode.org/wiki/XML/XPath | XML/XPath | Perform the following three XPath queries on the XML Document below:
//item[1]: Retrieve the first "item" element
//price/text(): Perform an action on each "price" element (print it out)
//name: Get an array of all the "name" elements
XML Document:
<inventory title="OmniCorp Store #45x10^3">
<section name="health">
<item upc="123456789" stock="12">
<name>Invisibility Cream</name>
<price>14.50</price>
<description>Makes you invisible</description>
</item>
<item upc="445322344" stock="18">
<name>Levitation Salve</name>
<price>23.99</price>
<description>Levitate yourself for up to 3 hours per application</description>
</item>
</section>
<section name="food">
<item upc="485672034" stock="653">
<name>Blork and Freen Instameal</name>
<price>4.95</price>
<description>A tasty meal in a tablet; just add water</description>
</item>
<item upc="132957764" stock="44">
<name>Grob winglets</name>
<price>3.56</price>
<description>Tender winglets of Grob. Just add water</description>
</item>
</section>
</inventory>
| #JavaScript | JavaScript | //create XMLDocument object from file
var xhr = new XMLHttpRequest();
xhr.open('GET', 'file.xml', false);
xhr.send(null);
var doc = xhr.responseXML;
//get first <item> element
var firstItem = doc.evaluate( '//item[1]', doc, null, XPathResult.FIRST_ORDERED_NODE_TYPE, null ).singleNodeValue;
alert( firstItem.textContent );
//output contents of <price> elements
var prices = doc.evaluate( '//price', doc, null, XPathResult.ANY_TYPE, null );
for( var price = prices.iterateNext(); price != null; price = prices.iterateNext() ) {
alert( price.textContent );
}
//add <name> elements to array
var names = doc.evaluate( '//name', doc, null, XPathResult.ANY_TYPE, null);
var namesArray = [];
for( var name = names.iterateNext(); name != null; name = names.iterateNext() ) {
namesArray.push( name );
}
alert( namesArray ); |
http://rosettacode.org/wiki/Yin_and_yang | Yin and yang | One well-known symbol of the philosophy of duality known as yin and yang is the taijitu.
Task
Create a function that, given a parameter representing size, generates such a symbol scaled to the requested size.
Generate and display the symbol for two different (small) sizes.
| #Java | Java | package org.rosettacode.yinandyang;
import java.awt.Color;
import java.awt.Graphics;
import java.awt.Image;
import java.awt.image.BufferedImage;
import javax.swing.ImageIcon;
import javax.swing.JFrame;
import javax.swing.JLabel;
public class YinYangGenerator
{
private final int size;
public YinYangGenerator(final int size)
{
this.size = size;
}
/**
* Draw a yin yang symbol on the given graphics context.
*/
public void drawYinYang(final Graphics graphics)
{
// Preserve the color for the caller
final Color colorSave = graphics.getColor();
graphics.setColor(Color.WHITE);
// Use fillOval to draw a filled in circle
graphics.fillOval(0, 0, size-1, size-1);
graphics.setColor(Color.BLACK);
// Use fillArc to draw part of a filled in circle
graphics.fillArc(0, 0, size-1, size-1, 270, 180);
graphics.fillOval(size/4, size/2, size/2, size/2);
graphics.setColor(Color.WHITE);
graphics.fillOval(size/4, 0, size/2, size/2);
graphics.fillOval(7*size/16, 11*size/16, size/8, size/8);
graphics.setColor(Color.BLACK);
graphics.fillOval(7*size/16, 3*size/16, size/8, size/8);
// Use drawOval to draw an empty circle for the outside border
graphics.drawOval(0, 0, size-1, size-1);
// Restore the color for the caller
graphics.setColor(colorSave);
}
/**
* Create an image containing a yin yang symbol.
*/
public Image createImage(final Color bg)
{
// A BufferedImage creates the image in memory
final BufferedImage image = new BufferedImage(size, size, BufferedImage.TYPE_INT_RGB);
// Get the graphics object for the image; note in many
// applications you actually use Graphics2D for the
// additional API calls
final Graphics graphics = image.getGraphics();
// Color in the background of the image
graphics.setColor(bg);
graphics.fillRect(0,0,size,size);
drawYinYang(graphics);
return image;
}
public static void main(final String args[])
{
final int size = Integer.parseInt(args[0]);
final YinYangGenerator generator = new YinYangGenerator(size);
final JFrame frame = new JFrame("Yin Yang Generator");
frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
final Image yinYang = generator.createImage(frame.getBackground());
// Use JLabel to display an image
frame.add(new JLabel(new ImageIcon(yinYang)));
frame.pack();
frame.setVisible(true);
}
} |
http://rosettacode.org/wiki/Y_combinator | Y combinator | In strict functional programming and the lambda calculus, functions (lambda expressions) don't have state and are only allowed to refer to arguments of enclosing functions.
This rules out the usual definition of a recursive function wherein a function is associated with the state of a variable and this variable's state is used in the body of the function.
The Y combinator is itself a stateless function that, when applied to another stateless function, returns a recursive version of the function.
The Y combinator is the simplest of the class of such functions, called fixed-point combinators.
Task
Define the stateless Y combinator and use it to compute factorials and Fibonacci numbers from other stateless functions or lambda expressions.
Cf
Jim Weirich: Adventures in Functional Programming
| #E | E | def y := fn f { fn x { x(x) }(fn y { f(fn a { y(y)(a) }) }) }
def fac := fn f { fn n { if (n<2) {1} else { n*f(n-1) } }}
def fib := fn f { fn n { if (n == 0) {0} else if (n == 1) {1} else { f(n-1) + f(n-2) } }} |
http://rosettacode.org/wiki/Zig-zag_matrix | Zig-zag matrix | Task
Produce a zig-zag array.
A zig-zag array is a square arrangement of the first N2 natural numbers, where the
numbers increase sequentially as you zig-zag along the array's anti-diagonals.
For a graphical representation, see JPG zigzag (JPG uses such arrays to encode images).
For example, given 5, produce this array:
0 1 5 6 14
2 4 7 13 15
3 8 12 16 21
9 11 17 20 22
10 18 19 23 24
Related tasks
Spiral matrix
Identity matrix
Ulam spiral (for primes)
See also
Wiktionary entry: anti-diagonals
| #ERRE | ERRE | PROGRAM ZIG_ZAG
!$DYNAMIC
DIM ARRAY%[0,0]
BEGIN
SIZE%=5
!$DIM ARRAY%[SIZE%-1,SIZE%-1]
I%=1
J%=1
FOR E%=0 TO SIZE%^2-1 DO
ARRAY%[I%-1,J%-1]=E%
IF ((I%+J%) AND 1)=0 THEN
IF J%<SIZE% THEN J%+=1 ELSE I%+=2 END IF
IF I%>1 THEN I%-=1 END IF
ELSE
IF I%<SIZE% THEN I%+=1 ELSE J%+=2 END IF
IF J%>1 THEN J%-=1 END IF
END IF
END FOR
FOR ROW%=0 TO SIZE%-1 DO
FOR COL%=0 TO SIZE%-1 DO
WRITE("###";ARRAY%[ROW%,COL%];)
END FOR
PRINT
END FOR
END PROGRAM |
http://rosettacode.org/wiki/Arbitrary-precision_integers_(included) | Arbitrary-precision integers (included) | Using the in-built capabilities of your language, calculate the integer value of:
5
4
3
2
{\displaystyle 5^{4^{3^{2}}}}
Confirm that the first and last twenty digits of the answer are:
62060698786608744707...92256259918212890625
Find and show the number of decimal digits in the answer.
Note: Do not submit an implementation of arbitrary precision arithmetic. The intention is to show the capabilities of the language as supplied. If a language has a single, overwhelming, library of varied modules that is endorsed by its home site – such as CPAN for Perl or Boost for C++ – then that may be used instead.
Strictly speaking, this should not be solved by fixed-precision numeric libraries where the precision has to be manually set to a large value; although if this is the only recourse then it may be used with a note explaining that the precision must be set manually to a large enough value.
Related tasks
Long multiplication
Exponentiation order
exponentiation operator
Exponentiation with infix operators in (or operating on) the base
| #Pascal | Pascal | program GMP_Demo;
uses
math, gmp;
var
a: mpz_t;
out: pchar;
len: longint;
i: longint;
begin
mpz_init_set_ui(a, 5);
mpz_pow_ui(a, a, 4 ** (3 ** 2));
len := mpz_sizeinbase(a, 10);
writeln('GMP says size is: ', len);
out := mpz_get_str(NIL, 10, a);
writeln('Actual size is: ', length(out));
write('Digits: ');
for i := 0 to 19 do
write(out[i]);
write ('...');
for i := len - 20 to len do
write(out[i]);
writeln;
end. |
http://rosettacode.org/wiki/Zeckendorf_number_representation | Zeckendorf number representation | Just as numbers can be represented in a positional notation as sums of multiples of the powers of ten (decimal) or two (binary); all the positive integers can be represented as the sum of one or zero times the distinct members of the Fibonacci series.
Recall that the first six distinct Fibonacci numbers are: 1, 2, 3, 5, 8, 13.
The decimal number eleven can be written as 0*13 + 1*8 + 0*5 + 1*3 + 0*2 + 0*1 or 010100 in positional notation where the columns represent multiplication by a particular member of the sequence. Leading zeroes are dropped so that 11 decimal becomes 10100.
10100 is not the only way to make 11 from the Fibonacci numbers however; 0*13 + 1*8 + 0*5 + 0*3 + 1*2 + 1*1 or 010011 would also represent decimal 11. For a true Zeckendorf number there is the added restriction that no two consecutive Fibonacci numbers can be used which leads to the former unique solution.
Task
Generate and show here a table of the Zeckendorf number representations of the decimal numbers zero to twenty, in order.
The intention in this task to find the Zeckendorf form of an arbitrary integer. The Zeckendorf form can be iterated by some bit twiddling rather than calculating each value separately but leave that to another separate task.
Also see
OEIS A014417 for the the sequence of required results.
Brown's Criterion - Numberphile
Related task
Fibonacci sequence
| #Python | Python | def fib():
memo = [1, 2]
while True:
memo.append(sum(memo))
yield memo.pop(0)
def sequence_down_from_n(n, seq_generator):
seq = []
for s in seq_generator():
seq.append(s)
if s >= n: break
return seq[::-1]
def zeckendorf(n):
if n == 0: return [0]
seq = sequence_down_from_n(n, fib)
digits, nleft = [], n
for s in seq:
if s <= nleft:
digits.append(1)
nleft -= s
else:
digits.append(0)
assert nleft == 0, 'Check all of n is accounted for'
assert sum(x*y for x,y in zip(digits, seq)) == n, 'Assert digits are correct'
while digits[0] == 0:
# Remove any zeroes padding L.H.S.
digits.pop(0)
return digits
n = 20
print('Fibonacci digit multipliers: %r' % sequence_down_from_n(n, fib))
for i in range(n + 1):
print('%3i: %8s' % (i, ''.join(str(d) for d in zeckendorf(i)))) |
http://rosettacode.org/wiki/100_doors | 100 doors | There are 100 doors in a row that are all initially closed.
You make 100 passes by the doors.
The first time through, visit every door and toggle the door (if the door is closed, open it; if it is open, close it).
The second time, only visit every 2nd door (door #2, #4, #6, ...), and toggle it.
The third time, visit every 3rd door (door #3, #6, #9, ...), etc, until you only visit the 100th door.
Task
Answer the question: what state are the doors in after the last pass? Which are open, which are closed?
Alternate:
As noted in this page's discussion page, the only doors that remain open are those whose numbers are perfect squares.
Opening only those doors is an optimization that may also be expressed;
however, as should be obvious, this defeats the intent of comparing implementations across programming languages.
| #Dart | Dart | main() {
for (var k = 1, x = new List(101); k <= 100; k++) {
for (int i = k; i <= 100; i += k)
x[i] = !x[i];
if (x[k]) print("$k open");
}
} |
http://rosettacode.org/wiki/Arrays | Arrays | This task is about arrays.
For hashes or associative arrays, please see Creating an Associative Array.
For a definition and in-depth discussion of what an array is, see Array.
Task
Show basic array syntax in your language.
Basically, create an array, assign a value to it, and retrieve an element (if available, show both fixed-length arrays and
dynamic arrays, pushing a value into it).
Please discuss at Village Pump: Arrays.
Please merge code in from these obsolete tasks:
Creating an Array
Assigning Values to an Array
Retrieving an Element of an Array
Related tasks
Collections
Creating an Associative Array
Two-dimensional array (runtime)
| #Haskell | Haskell | import Data.Array.IO
main = do arr <- newArray (1,10) 37 :: IO (IOArray Int Int)
a <- readArray arr 1
writeArray arr 1 64
b <- readArray arr 1
print (a,b) |
http://rosettacode.org/wiki/Arithmetic/Complex | Arithmetic/Complex | A complex number is a number which can be written as:
a
+
b
×
i
{\displaystyle a+b\times i}
(sometimes shown as:
b
+
a
×
i
{\displaystyle b+a\times i}
where
a
{\displaystyle a}
and
b
{\displaystyle b}
are real numbers, and
i
{\displaystyle i}
is √ -1
Typically, complex numbers are represented as a pair of real numbers called the "imaginary part" and "real part", where the imaginary part is the number to be multiplied by
i
{\displaystyle i}
.
Task
Show addition, multiplication, negation, and inversion of complex numbers in separate functions. (Subtraction and division operations can be made with pairs of these operations.)
Print the results for each operation tested.
Optional: Show complex conjugation.
By definition, the complex conjugate of
a
+
b
i
{\displaystyle a+bi}
is
a
−
b
i
{\displaystyle a-bi}
Some languages have complex number libraries available. If your language does, show the operations. If your language does not, also show the definition of this type.
| #XPL0 | XPL0 | include c:\cxpl\codes;
func real CAdd(A, B, C); \Return complex sum of two complex numbers
real A, B, C;
[C(0):= A(0) + B(0);
C(1):= A(1) + B(1);
return C;
];
func real CMul(A, B, C); \Return complex product of two complex numbers
real A, B, C;
[C(0):= A(0)*B(0) - A(1)*B(1);
C(1):= A(1)*B(0) + A(0)*B(1);
return C;
];
func real CNeg(A, C); \Return negative of a complex number
real A, C;
[C(0):= -A(0);
C(1):= -A(1);
return C;
];
func real CInv(A, C); \Return inversion (reciprical) of complex number
real A, C;
real D;
[D:= sq(A(0)) + sq(A(1));
C(0):= A(0)/D;
C(1):=-A(1)/D;
return C;
];
func real Conj(A, C); \Return conjugate of a complex number
real A, C;
[C(0):= A(0);
C(1):=-A(1);
return C;
];
proc COut(D, A); \Output a complex number to specified device
int D; real A;
[RlOut(D, A(0));
Text(D, if A(1)>=0.0 then " +" else " -");
RlOut(D, abs(A(1)));
ChOut(D, ^i);
];
real U, V, W(2);
[Format(2,2);
U:= [1.0, 1.0];
V:= [3.14, 1.2];
COut(0, CAdd(U,V,W)); CrLf(0);
COut(0, CMul(U,V,W)); CrLf(0);
COut(0, CNeg(U,W)); CrLf(0);
COut(0, CInv(U,W)); CrLf(0);
COut(0, Conj(U,W)); CrLf(0);
] |
http://rosettacode.org/wiki/Arithmetic/Complex | Arithmetic/Complex | A complex number is a number which can be written as:
a
+
b
×
i
{\displaystyle a+b\times i}
(sometimes shown as:
b
+
a
×
i
{\displaystyle b+a\times i}
where
a
{\displaystyle a}
and
b
{\displaystyle b}
are real numbers, and
i
{\displaystyle i}
is √ -1
Typically, complex numbers are represented as a pair of real numbers called the "imaginary part" and "real part", where the imaginary part is the number to be multiplied by
i
{\displaystyle i}
.
Task
Show addition, multiplication, negation, and inversion of complex numbers in separate functions. (Subtraction and division operations can be made with pairs of these operations.)
Print the results for each operation tested.
Optional: Show complex conjugation.
By definition, the complex conjugate of
a
+
b
i
{\displaystyle a+bi}
is
a
−
b
i
{\displaystyle a-bi}
Some languages have complex number libraries available. If your language does, show the operations. If your language does not, also show the definition of this type.
| #Yabasic | Yabasic | rem ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
rem CADDI/CADDR addition of complex numbers Z1 + Z2 with Z1 = a1 + b1 *i Z2 = a2 + b2*i
rem CADDI returns imaginary part and CADDR the real part
rem ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
export sub caddi( a1 , b1 , a2 , b2)
return (b1 + b2)
end sub
export sub caddr( a1 , b1 , a2 , b2)
return (a1 + a2)
end sub
rem ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
rem CDIVI/CDIVR division of complex numbers Z1 / Z2 with Z1 = r + s *i Z2 = t + u*i
rem CDIVI returns imaginary part and CDIVR the real part
rem ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
export sub cdivi(r,s,t,u)
return ((s*t- u*r) / (t^2 + u^2))
end sub
export sub cdivr( r , s , t , u)
return ((r*t- s*u) / (t^2 + u^2))
end sub
rem ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
rem CMULI/CMULR multiplication of complex numbers Z1 * Z2, with Z1 = r + s *i Z2 = t + u*i
rem CMULI returns imaginary part and CMULR the real part
rem ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
export sub cmuli( r , s , t , u)
return (r * u + s * t)
end sub
export sub cmulr( r , s , t , u)
return (r * t - s * u)
end sub
rem ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
rem CSUBI/CSUBR subtraction of complex numbers Z1 - Z2 with Z1 = a1 + b1 *i Z2 = a2 + b2*i
rem CSUBI returns imaginary part and CSUBR the real part
rem ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
export sub csubi( a1 , b1 , a2 , b2)
return (b1 - b2)
end sub
export sub csubr( a1 , b1 , a2 , b2)
return (a1 - a2)
end sub
if (peek$("library") = "main") then
print "Example: Z1 + Z2 with Z1 = 3 +2i , Z2 = 1-3i: Z1 + Z2 = 4 -1i"
print caddr(3,2,1,-2), "/", caddi(3,2,1,-3) // 4/-1
end if |
http://rosettacode.org/wiki/Zero_to_the_zero_power | Zero to the zero power | Some computer programming languages are not exactly consistent (with other computer programming languages)
when raising zero to the zeroth power: 00
Task
Show the results of raising zero to the zeroth power.
If your computer language objects to 0**0 or 0^0 at compile time, you may also try something like:
x = 0
y = 0
z = x**y
say 'z=' z
Show the result here.
And of course use any symbols or notation that is supported in your computer programming language for exponentiation.
See also
The Wiki entry: Zero to the power of zero.
The Wiki entry: History of differing points of view.
The MathWorld™ entry: exponent laws.
Also, in the above MathWorld™ entry, see formula (9):
x
0
=
1
{\displaystyle x^{0}=1}
.
The OEIS entry: The special case of zero to the zeroth power
| #Relation | Relation |
echo pow(0,0)
// 1
|
http://rosettacode.org/wiki/Zero_to_the_zero_power | Zero to the zero power | Some computer programming languages are not exactly consistent (with other computer programming languages)
when raising zero to the zeroth power: 00
Task
Show the results of raising zero to the zeroth power.
If your computer language objects to 0**0 or 0^0 at compile time, you may also try something like:
x = 0
y = 0
z = x**y
say 'z=' z
Show the result here.
And of course use any symbols or notation that is supported in your computer programming language for exponentiation.
See also
The Wiki entry: Zero to the power of zero.
The Wiki entry: History of differing points of view.
The MathWorld™ entry: exponent laws.
Also, in the above MathWorld™ entry, see formula (9):
x
0
=
1
{\displaystyle x^{0}=1}
.
The OEIS entry: The special case of zero to the zeroth power
| #REXX | REXX | /*REXX program shows the results of raising zero to the zeroth power.*/
say '0 ** 0 (zero to the zeroth power) ───► ' 0**0 |
http://rosettacode.org/wiki/Zebra_puzzle | Zebra puzzle | Zebra puzzle
You are encouraged to solve this task according to the task description, using any language you may know.
The Zebra puzzle, a.k.a. Einstein's Riddle,
is a logic puzzle which is to be solved programmatically.
It has several variants, one of them this:
There are five houses.
The English man lives in the red house.
The Swede has a dog.
The Dane drinks tea.
The green house is immediately to the left of the white house.
They drink coffee in the green house.
The man who smokes Pall Mall has birds.
In the yellow house they smoke Dunhill.
In the middle house they drink milk.
The Norwegian lives in the first house.
The man who smokes Blend lives in the house next to the house with cats.
In a house next to the house where they have a horse, they smoke Dunhill.
The man who smokes Blue Master drinks beer.
The German smokes Prince.
The Norwegian lives next to the blue house.
They drink water in a house next to the house where they smoke Blend.
The question is, who owns the zebra?
Additionally, list the solution for all the houses.
Optionally, show the solution is unique.
Related tasks
Dinesman's multiple-dwelling problem
Twelve statements
| #Go | Go | package main
import (
"fmt"
"log"
"strings"
)
// Define some types
type HouseSet [5]*House
type House struct {
n Nationality
c Colour
a Animal
d Drink
s Smoke
}
type Nationality int8
type Colour int8
type Animal int8
type Drink int8
type Smoke int8
// Define the possible values
const (
English Nationality = iota
Swede
Dane
Norwegian
German
)
const (
Red Colour = iota
Green
White
Yellow
Blue
)
const (
Dog Animal = iota
Birds
Cats
Horse
Zebra
)
const (
Tea Drink = iota
Coffee
Milk
Beer
Water
)
const (
PallMall Smoke = iota
Dunhill
Blend
BlueMaster
Prince
)
// And how to print them
var nationalities = [...]string{"English", "Swede", "Dane", "Norwegian", "German"}
var colours = [...]string{"red", "green", "white", "yellow", "blue"}
var animals = [...]string{"dog", "birds", "cats", "horse", "zebra"}
var drinks = [...]string{"tea", "coffee", "milk", "beer", "water"}
var smokes = [...]string{"Pall Mall", "Dunhill", "Blend", "Blue Master", "Prince"}
func (n Nationality) String() string { return nationalities[n] }
func (c Colour) String() string { return colours[c] }
func (a Animal) String() string { return animals[a] }
func (d Drink) String() string { return drinks[d] }
func (s Smoke) String() string { return smokes[s] }
func (h House) String() string {
return fmt.Sprintf("%-9s %-6s %-5s %-6s %s", h.n, h.c, h.a, h.d, h.s)
}
func (hs HouseSet) String() string {
lines := make([]string, 0, len(hs))
for i, h := range hs {
s := fmt.Sprintf("%d %s", i, h)
lines = append(lines, s)
}
return strings.Join(lines, "\n")
}
// Simple brute force solution
func simpleBruteForce() (int, HouseSet) {
var v []House
for n := range nationalities {
for c := range colours {
for a := range animals {
for d := range drinks {
for s := range smokes {
h := House{
n: Nationality(n),
c: Colour(c),
a: Animal(a),
d: Drink(d),
s: Smoke(s),
}
if !h.Valid() {
continue
}
v = append(v, h)
}
}
}
}
}
n := len(v)
log.Println("Generated", n, "valid houses")
combos := 0
first := 0
valid := 0
var validSet HouseSet
for a := 0; a < n; a++ {
if v[a].n != Norwegian { // Condition 10:
continue
}
for b := 0; b < n; b++ {
if b == a {
continue
}
if v[b].anyDups(&v[a]) {
continue
}
for c := 0; c < n; c++ {
if c == b || c == a {
continue
}
if v[c].d != Milk { // Condition 9:
continue
}
if v[c].anyDups(&v[b], &v[a]) {
continue
}
for d := 0; d < n; d++ {
if d == c || d == b || d == a {
continue
}
if v[d].anyDups(&v[c], &v[b], &v[a]) {
continue
}
for e := 0; e < n; e++ {
if e == d || e == c || e == b || e == a {
continue
}
if v[e].anyDups(&v[d], &v[c], &v[b], &v[a]) {
continue
}
combos++
set := HouseSet{&v[a], &v[b], &v[c], &v[d], &v[e]}
if set.Valid() {
valid++
if valid == 1 {
first = combos
}
validSet = set
//return set
}
}
}
}
}
}
log.Println("Tested", first, "different combinations of valid houses before finding solution")
log.Println("Tested", combos, "different combinations of valid houses in total")
return valid, validSet
}
// anyDups returns true if h as any duplicate attributes with any of the specified houses
func (h *House) anyDups(list ...*House) bool {
for _, b := range list {
if h.n == b.n || h.c == b.c || h.a == b.a || h.d == b.d || h.s == b.s {
return true
}
}
return false
}
func (h *House) Valid() bool {
// Condition 2:
if h.n == English && h.c != Red || h.n != English && h.c == Red {
return false
}
// Condition 3:
if h.n == Swede && h.a != Dog || h.n != Swede && h.a == Dog {
return false
}
// Condition 4:
if h.n == Dane && h.d != Tea || h.n != Dane && h.d == Tea {
return false
}
// Condition 6:
if h.c == Green && h.d != Coffee || h.c != Green && h.d == Coffee {
return false
}
// Condition 7:
if h.a == Birds && h.s != PallMall || h.a != Birds && h.s == PallMall {
return false
}
// Condition 8:
if h.c == Yellow && h.s != Dunhill || h.c != Yellow && h.s == Dunhill {
return false
}
// Condition 11:
if h.a == Cats && h.s == Blend {
return false
}
// Condition 12:
if h.a == Horse && h.s == Dunhill {
return false
}
// Condition 13:
if h.d == Beer && h.s != BlueMaster || h.d != Beer && h.s == BlueMaster {
return false
}
// Condition 14:
if h.n == German && h.s != Prince || h.n != German && h.s == Prince {
return false
}
// Condition 15:
if h.n == Norwegian && h.c == Blue {
return false
}
// Condition 16:
if h.d == Water && h.s == Blend {
return false
}
return true
}
func (hs *HouseSet) Valid() bool {
ni := make(map[Nationality]int, 5)
ci := make(map[Colour]int, 5)
ai := make(map[Animal]int, 5)
di := make(map[Drink]int, 5)
si := make(map[Smoke]int, 5)
for i, h := range hs {
ni[h.n] = i
ci[h.c] = i
ai[h.a] = i
di[h.d] = i
si[h.s] = i
}
// Condition 5:
if ci[Green]+1 != ci[White] {
return false
}
// Condition 11:
if dist(ai[Cats], si[Blend]) != 1 {
return false
}
// Condition 12:
if dist(ai[Horse], si[Dunhill]) != 1 {
return false
}
// Condition 15:
if dist(ni[Norwegian], ci[Blue]) != 1 {
return false
}
// Condition 16:
if dist(di[Water], si[Blend]) != 1 {
return false
}
// Condition 9: (already tested elsewhere)
if hs[2].d != Milk {
return false
}
// Condition 10: (already tested elsewhere)
if hs[0].n != Norwegian {
return false
}
return true
}
func dist(a, b int) int {
if a > b {
return a - b
}
return b - a
}
func main() {
log.SetFlags(0)
n, sol := simpleBruteForce()
fmt.Println(n, "solution found")
fmt.Println(sol)
} |
http://rosettacode.org/wiki/XML/XPath | XML/XPath | Perform the following three XPath queries on the XML Document below:
//item[1]: Retrieve the first "item" element
//price/text(): Perform an action on each "price" element (print it out)
//name: Get an array of all the "name" elements
XML Document:
<inventory title="OmniCorp Store #45x10^3">
<section name="health">
<item upc="123456789" stock="12">
<name>Invisibility Cream</name>
<price>14.50</price>
<description>Makes you invisible</description>
</item>
<item upc="445322344" stock="18">
<name>Levitation Salve</name>
<price>23.99</price>
<description>Levitate yourself for up to 3 hours per application</description>
</item>
</section>
<section name="food">
<item upc="485672034" stock="653">
<name>Blork and Freen Instameal</name>
<price>4.95</price>
<description>A tasty meal in a tablet; just add water</description>
</item>
<item upc="132957764" stock="44">
<name>Grob winglets</name>
<price>3.56</price>
<description>Tender winglets of Grob. Just add water</description>
</item>
</section>
</inventory>
| #Julia | Julia | using LibExpat
xdoc = raw"""<inventory title="OmniCorp Store #45x10^3">
<section name="health">
<item upc="123456789" stock="12">
<name>Invisibility Cream</name>
<price>14.50</price>
<description>Makes you invisible</description>
</item>
<item upc="445322344" stock="18">
<name>Levitation Salve</name>
<price>23.99</price>
<description>Levitate yourself for up to 3 hours per application</description>
</item>
</section>
<section name="food">
<item upc="485672034" stock="653">
<name>Blork and Freen Instameal</name>
<price>4.95</price>
<description>A tasty meal in a tablet; just add water</description>
</item>
<item upc="132957764" stock="44">
<name>Grob winglets</name>
<price>3.56</price>
<description>Tender winglets of Grob. Just add water</description>
</item>
</section>
</inventory>
"""
debracket(s) = replace(s, r".+\>(.+)\<.+" => s"\1")
etree = xp_parse(xdoc)
firstshow = LibExpat.find(etree, "//item")[1]
println("The first item's node XML entry is:\n", firstshow, "\n\n")
prices = LibExpat.find(etree, "//price")
println("Prices:")
for p in prices
println("\t", debracket(string(p)))
end
println("\n")
namearray = LibExpat.find(etree, "//name")
println("Array of names of items:\n\t", map(s -> debracket(string(s)), namearray))
|
http://rosettacode.org/wiki/XML/XPath | XML/XPath | Perform the following three XPath queries on the XML Document below:
//item[1]: Retrieve the first "item" element
//price/text(): Perform an action on each "price" element (print it out)
//name: Get an array of all the "name" elements
XML Document:
<inventory title="OmniCorp Store #45x10^3">
<section name="health">
<item upc="123456789" stock="12">
<name>Invisibility Cream</name>
<price>14.50</price>
<description>Makes you invisible</description>
</item>
<item upc="445322344" stock="18">
<name>Levitation Salve</name>
<price>23.99</price>
<description>Levitate yourself for up to 3 hours per application</description>
</item>
</section>
<section name="food">
<item upc="485672034" stock="653">
<name>Blork and Freen Instameal</name>
<price>4.95</price>
<description>A tasty meal in a tablet; just add water</description>
</item>
<item upc="132957764" stock="44">
<name>Grob winglets</name>
<price>3.56</price>
<description>Tender winglets of Grob. Just add water</description>
</item>
</section>
</inventory>
| #Kotlin | Kotlin | // version 1.1.3
import javax.xml.parsers.DocumentBuilderFactory
import org.xml.sax.InputSource
import java.io.StringReader
import javax.xml.xpath.XPathFactory
import javax.xml.xpath.XPathConstants
import org.w3c.dom.Node
import org.w3c.dom.NodeList
val xml =
"""
<inventory title="OmniCorp Store #45x10^3">
<section name="health">
<item upc="123456789" stock="12">
<name>Invisibility Cream</name>
<price>14.50</price>
<description>Makes you invisible</description>
</item>
<item upc="445322344" stock="18">
<name>Levitation Salve</name>
<price>23.99</price>
<description>Levitate yourself for up to 3 hours per application</description>
</item>
</section>
<section name="food">
<item upc="485672034" stock="653">
<name>Blork and Freen Instameal</name>
<price>4.95</price>
<description>A tasty meal in a tablet; just add water</description>
</item>
<item upc="132957764" stock="44">
<name>Grob winglets</name>
<price>3.56</price>
<description>Tender winglets of Grob. Just add water</description>
</item>
</section>
</inventory>
"""
fun main(args: Array<String>) {
val dbFactory = DocumentBuilderFactory.newInstance()
val dBuilder = dbFactory.newDocumentBuilder()
val xmlInput = InputSource(StringReader(xml))
val doc = dBuilder.parse(xmlInput)
val xpFactory = XPathFactory.newInstance()
val xPath = xpFactory.newXPath()
val qNode = xPath.evaluate("/inventory/section/item[1]", doc, XPathConstants.NODE) as Node
val upc = qNode.attributes.getNamedItem("upc")
val stock = qNode.attributes.getNamedItem("stock")
println("For the first item : upc = ${upc.textContent} and stock = ${stock.textContent}")
val qNodes = xPath.evaluate("/inventory/section/item/price", doc, XPathConstants.NODESET) as NodeList
print("\nThe prices of each item are : ")
for (i in 0 until qNodes.length) print("${qNodes.item(i).textContent} ")
println()
val qNodes2 = xPath.evaluate("/inventory/section/item/name", doc, XPathConstants.NODESET) as NodeList
val names = Array<String>(qNodes2.length) { qNodes2.item(it).textContent }
println("\nThe names of each item are as follows :")
println(" ${names.joinToString("\n ")}")
} |
http://rosettacode.org/wiki/Yin_and_yang | Yin and yang | One well-known symbol of the philosophy of duality known as yin and yang is the taijitu.
Task
Create a function that, given a parameter representing size, generates such a symbol scaled to the requested size.
Generate and display the symbol for two different (small) sizes.
| #JavaScript | JavaScript |
function Arc(posX,posY,radius,startAngle,endAngle,color){//Angle in radians.
this.posX=posX;
this.posY=posY;
this.radius=radius;
this.startAngle=startAngle;
this.endAngle=endAngle;
this.color=color;
}
//0,0 is the top left of the screen
var YingYang=[
new Arc(0.5,0.5,1,0.5*Math.PI,1.5*Math.PI,"white"),//Half white semi-circle
new Arc(0.5,0.5,1,1.5*Math.PI,0.5*Math.PI,"black"),//Half black semi-circle
new Arc(0.5,0.25,.5,0,2*Math.PI,"black"),//black circle
new Arc(0.5,0.75,.5,0,2*Math.PI,"white"),//white circle
new Arc(0.5,0.25,1/6,0,2*Math.PI,"white"),//small white circle
new Arc(0.5,0.75,1/6,0,2*Math.PI,"black")//small black circle
]
//Ying Yang is DONE!
//Now we'll have to draw it.
//We'll draw it in a matrix that way we can get results graphically or by text!
function Array2D(width,height){
this.height=height;
this.width=width;
this.array2d=[];
for(var i=0;i<this.height;i++){
this.array2d.push(new Array(this.width));
}
}
Array2D.prototype.resize=function(width,height){//This is expensive
//nheight and nwidth is the difference of the new and old height
var nheight=height-this.height,nwidth=width-this.width;
if(nwidth>0){
for(var i=0;i<this.height;i++){
if(i<height)
Array.prototype.push.apply(this.array2d[i],new Array(nwidth));
}
}
else if(nwidth<0){
for(var i=0;i<this.height;i++){
if(i<height)
this.array2d[i].splice(width,nwidth);
}
}
if(nheight>0){
Array.prototype.push.apply(this.array2d,new Array(width));
}
else if(nheight<0){
this.array2d.splice(height,nheight)
}
}
Array2D.prototype.loop=function(callback){
for(var i=0;i<this.height;i++)
for(var i2=0;i2<this.width;i++)
callback.call(this,this.array2d[i][i2],i,i2);
}
var mat=new Array2D(100,100);//this sounds fine;
YingYang[0];
//In construction.
|
http://rosettacode.org/wiki/Y_combinator | Y combinator | In strict functional programming and the lambda calculus, functions (lambda expressions) don't have state and are only allowed to refer to arguments of enclosing functions.
This rules out the usual definition of a recursive function wherein a function is associated with the state of a variable and this variable's state is used in the body of the function.
The Y combinator is itself a stateless function that, when applied to another stateless function, returns a recursive version of the function.
The Y combinator is the simplest of the class of such functions, called fixed-point combinators.
Task
Define the stateless Y combinator and use it to compute factorials and Fibonacci numbers from other stateless functions or lambda expressions.
Cf
Jim Weirich: Adventures in Functional Programming
| #EchoLisp | EchoLisp |
;; Ref : http://www.ece.uc.edu/~franco/C511/html/Scheme/ycomb.html
(define Y
(lambda (X)
((lambda (procedure)
(X (lambda (arg) ((procedure procedure) arg))))
(lambda (procedure)
(X (lambda (arg) ((procedure procedure) arg)))))))
; Fib
(define Fib* (lambda (func-arg)
(lambda (n) (if (< n 2) n (+ (func-arg (- n 1)) (func-arg (- n 2)))))))
(define fib (Y Fib*))
(fib 6)
→ 8
; Fact
(define F*
(lambda (func-arg) (lambda (n) (if (zero? n) 1 (* n (func-arg (- n 1)))))))
(define fact (Y F*))
(fact 10)
→ 3628800
|
http://rosettacode.org/wiki/Zig-zag_matrix | Zig-zag matrix | Task
Produce a zig-zag array.
A zig-zag array is a square arrangement of the first N2 natural numbers, where the
numbers increase sequentially as you zig-zag along the array's anti-diagonals.
For a graphical representation, see JPG zigzag (JPG uses such arrays to encode images).
For example, given 5, produce this array:
0 1 5 6 14
2 4 7 13 15
3 8 12 16 21
9 11 17 20 22
10 18 19 23 24
Related tasks
Spiral matrix
Identity matrix
Ulam spiral (for primes)
See also
Wiktionary entry: anti-diagonals
| #Euphoria | Euphoria | function zigzag(integer size)
sequence s
integer i, j, d, max
s = repeat(repeat(0,size),size)
i = 1 j = 1 d = -1
max = size*size
for n = 1 to floor(max/2)+1 do
s[i][j] = n
s[size-i+1][size-j+1] = max-n+1
i += d j-= d
if i < 1 then
i += 1 d = -d
elsif j < 1 then
j += 1 d = -d
end if
end for
return s
end function
? zigzag(5) |
http://rosettacode.org/wiki/Arbitrary-precision_integers_(included) | Arbitrary-precision integers (included) | Using the in-built capabilities of your language, calculate the integer value of:
5
4
3
2
{\displaystyle 5^{4^{3^{2}}}}
Confirm that the first and last twenty digits of the answer are:
62060698786608744707...92256259918212890625
Find and show the number of decimal digits in the answer.
Note: Do not submit an implementation of arbitrary precision arithmetic. The intention is to show the capabilities of the language as supplied. If a language has a single, overwhelming, library of varied modules that is endorsed by its home site – such as CPAN for Perl or Boost for C++ – then that may be used instead.
Strictly speaking, this should not be solved by fixed-precision numeric libraries where the precision has to be manually set to a large value; although if this is the only recourse then it may be used with a note explaining that the precision must be set manually to a large enough value.
Related tasks
Long multiplication
Exponentiation order
exponentiation operator
Exponentiation with infix operators in (or operating on) the base
| #Perl | Perl | use Math::BigInt;
my $x = Math::BigInt->new('5') ** Math::BigInt->new('4') ** Math::BigInt->new('3') ** Math::BigInt->new('2');
my $y = "$x";
printf("5**4**3**2 = %s...%s and has %i digits\n", substr($y,0,20), substr($y,-20), length($y)); |
http://rosettacode.org/wiki/Arbitrary-precision_integers_(included) | Arbitrary-precision integers (included) | Using the in-built capabilities of your language, calculate the integer value of:
5
4
3
2
{\displaystyle 5^{4^{3^{2}}}}
Confirm that the first and last twenty digits of the answer are:
62060698786608744707...92256259918212890625
Find and show the number of decimal digits in the answer.
Note: Do not submit an implementation of arbitrary precision arithmetic. The intention is to show the capabilities of the language as supplied. If a language has a single, overwhelming, library of varied modules that is endorsed by its home site – such as CPAN for Perl or Boost for C++ – then that may be used instead.
Strictly speaking, this should not be solved by fixed-precision numeric libraries where the precision has to be manually set to a large value; although if this is the only recourse then it may be used with a note explaining that the precision must be set manually to a large enough value.
Related tasks
Long multiplication
Exponentiation order
exponentiation operator
Exponentiation with infix operators in (or operating on) the base
| #Phix | Phix | with javascript_semantics
include mpfr.e
atom t0 = time()
mpz res = mpz_init()
mpz_ui_pow_ui(res,5,power(4,power(3,2)))
string s = mpz_get_short_str(res),
e = elapsed(time()-t0)
printf(1,"5^4^3^2 = %s (%s)\n", {s,e})
|
http://rosettacode.org/wiki/Zeckendorf_number_representation | Zeckendorf number representation | Just as numbers can be represented in a positional notation as sums of multiples of the powers of ten (decimal) or two (binary); all the positive integers can be represented as the sum of one or zero times the distinct members of the Fibonacci series.
Recall that the first six distinct Fibonacci numbers are: 1, 2, 3, 5, 8, 13.
The decimal number eleven can be written as 0*13 + 1*8 + 0*5 + 1*3 + 0*2 + 0*1 or 010100 in positional notation where the columns represent multiplication by a particular member of the sequence. Leading zeroes are dropped so that 11 decimal becomes 10100.
10100 is not the only way to make 11 from the Fibonacci numbers however; 0*13 + 1*8 + 0*5 + 0*3 + 1*2 + 1*1 or 010011 would also represent decimal 11. For a true Zeckendorf number there is the added restriction that no two consecutive Fibonacci numbers can be used which leads to the former unique solution.
Task
Generate and show here a table of the Zeckendorf number representations of the decimal numbers zero to twenty, in order.
The intention in this task to find the Zeckendorf form of an arbitrary integer. The Zeckendorf form can be iterated by some bit twiddling rather than calculating each value separately but leave that to another separate task.
Also see
OEIS A014417 for the the sequence of required results.
Brown's Criterion - Numberphile
Related task
Fibonacci sequence
| #Quackery | Quackery | [ 2 base put
echo
base release ] is binecho ( n --> )
[ 0 swap ' [ 2 1 ]
[ 2dup 0 peek < iff
[ behead drop ]
done
dup 0 peek
over 1 peek
+ swap join again ]
witheach
[ rot 1 << unrot
2dup < iff drop
else
[ -
dip
[ 1 | ] ] ]
drop ] is n->z ( n --> z )
[ 0 temp put
1 1 rot
[ dup while
dup 1 & if
[ over
temp tally ]
1 >>
dip [ tuck + ]
again ]
2drop drop
temp take ] is z->n ( z --> n )
21 times
[ i^ dup echo
say " -> "
n->z dup binecho
say " -> "
z->n echo cr ] |
http://rosettacode.org/wiki/100_doors | 100 doors | There are 100 doors in a row that are all initially closed.
You make 100 passes by the doors.
The first time through, visit every door and toggle the door (if the door is closed, open it; if it is open, close it).
The second time, only visit every 2nd door (door #2, #4, #6, ...), and toggle it.
The third time, visit every 3rd door (door #3, #6, #9, ...), etc, until you only visit the 100th door.
Task
Answer the question: what state are the doors in after the last pass? Which are open, which are closed?
Alternate:
As noted in this page's discussion page, the only doors that remain open are those whose numbers are perfect squares.
Opening only those doors is an optimization that may also be expressed;
however, as should be obvious, this defeats the intent of comparing implementations across programming languages.
| #Dc | Dc |
## NB: This code uses the dc command "r" via register "r".
## You may comment out the unwanted version.
[SxSyLxLy]sr # this should work with every "dc"
[r]sr # GNU dc can exchange top 2 stack values by "r"
## Now use "lrx" instead of "r" ...
0k # we work without decimal places
[q]sq # useful e.g. as loop termination
## (x)(y)>R == if (y)>(x) eval R
## isle x y --> (x <= y)
[
[1q]S. [ !<. 0 ]x s.L.
]sl
## l: isle
[
100 llx
]sL
## L: isle100
## for initcode condcode incrcode body
## [1] [2] [3] [4]
[
[q]S. 4:. 3:. 2:. 1:. 1;.x [2;.x 0=. 4;.x 3;.x 0;.x]d0:.x Os.L.o
]sf
## f: for
##----------------------------------------------------------------------------
## for( i=1 ; i<=100 ; ++i ) {
## door[i] = 0;
## }
#[init ...]P []ps-
[1si] [li lLx] [li1+si] [
li 0:d
]lfx
## for( s=1 ; s<=100 ; ++s ) {
## for( i=s ; i<=100 ; i+=s ) {
## door[i] = 1 - door[i]
## }
## }
[1ss] [ls lLx] [ls1+ss] [
#[step ]P lsn [ ...]ps-
[lssi] [li lLx] [lils+si] [
1 li;d - li:d
]lfx
]lfx
## long output:
## for( i=1 ; i<=100 ; ++i ) {
## print "door #", i, " is ", (door[i] ? "open" : "closed")), NL
## }
[
[1si] [li lLx] [li1+si] [
[door #]P
li n
[ is ]P
[closed]
[open]
li;d 0=r lrx s- n
[.]ps-
]lfx
]
## terse output:
## for( i=1 ; i<=100 ; ++i ) {
## if( door[i] ) {
## print i
## }
## print NL
## }
[
[1si] [li lLx] [li1+si] [
[] [ [ ]n lin ]
li;d 0=r lrx s- x
]lfx
[]ps-
]
lrx # comment out for the long output version
s- x
#[stack rest...]P []ps- f
|
http://rosettacode.org/wiki/Arrays | Arrays | This task is about arrays.
For hashes or associative arrays, please see Creating an Associative Array.
For a definition and in-depth discussion of what an array is, see Array.
Task
Show basic array syntax in your language.
Basically, create an array, assign a value to it, and retrieve an element (if available, show both fixed-length arrays and
dynamic arrays, pushing a value into it).
Please discuss at Village Pump: Arrays.
Please merge code in from these obsolete tasks:
Creating an Array
Assigning Values to an Array
Retrieving an Element of an Array
Related tasks
Collections
Creating an Associative Array
Two-dimensional array (runtime)
| #hexiscript | hexiscript | let a arr 2 # fixed size
let a[0] 123 # index starting at 0
let a[1] "test" # can hold different types
println a[1] |
http://rosettacode.org/wiki/Arithmetic/Complex | Arithmetic/Complex | A complex number is a number which can be written as:
a
+
b
×
i
{\displaystyle a+b\times i}
(sometimes shown as:
b
+
a
×
i
{\displaystyle b+a\times i}
where
a
{\displaystyle a}
and
b
{\displaystyle b}
are real numbers, and
i
{\displaystyle i}
is √ -1
Typically, complex numbers are represented as a pair of real numbers called the "imaginary part" and "real part", where the imaginary part is the number to be multiplied by
i
{\displaystyle i}
.
Task
Show addition, multiplication, negation, and inversion of complex numbers in separate functions. (Subtraction and division operations can be made with pairs of these operations.)
Print the results for each operation tested.
Optional: Show complex conjugation.
By definition, the complex conjugate of
a
+
b
i
{\displaystyle a+bi}
is
a
−
b
i
{\displaystyle a-bi}
Some languages have complex number libraries available. If your language does, show the operations. If your language does not, also show the definition of this type.
| #zkl | zkl | var [const] GSL=Import("zklGSL"); // libGSL (GNU Scientific Library)
(GSL.Z(3,4) + GSL.Z(1,2)).println(); // (4.00+6.00i)
(GSL.Z(3,4) - GSL.Z(1,2)).println(); // (2.00+2.00i)
(GSL.Z(3,4) * GSL.Z(1,2)).println(); // (-5.00+10.00i)
(GSL.Z(3,4) / GSL.Z(1,2)).println(); // (2.20-0.40i)
(GSL.Z(1,0) / GSL.Z(1,1)).println(); // (0.50-0.50i) // inversion
(-GSL.Z(3,4)).println(); // (-3.00-4.00i)
GSL.Z(3,4).conjugate().println(); // (3.00-4.00i) |
http://rosettacode.org/wiki/Arithmetic/Complex | Arithmetic/Complex | A complex number is a number which can be written as:
a
+
b
×
i
{\displaystyle a+b\times i}
(sometimes shown as:
b
+
a
×
i
{\displaystyle b+a\times i}
where
a
{\displaystyle a}
and
b
{\displaystyle b}
are real numbers, and
i
{\displaystyle i}
is √ -1
Typically, complex numbers are represented as a pair of real numbers called the "imaginary part" and "real part", where the imaginary part is the number to be multiplied by
i
{\displaystyle i}
.
Task
Show addition, multiplication, negation, and inversion of complex numbers in separate functions. (Subtraction and division operations can be made with pairs of these operations.)
Print the results for each operation tested.
Optional: Show complex conjugation.
By definition, the complex conjugate of
a
+
b
i
{\displaystyle a+bi}
is
a
−
b
i
{\displaystyle a-bi}
Some languages have complex number libraries available. If your language does, show the operations. If your language does not, also show the definition of this type.
| #zonnon | zonnon |
module Numbers;
type
{public,immutable}
Complex = record
re,im: real;
end Complex;
operator {public} "+" (a,b: Complex): Complex;
var
r: Complex;
begin
r.re := a.re + b.re;
r.im := a.im + b.im;
return r
end "+";
operator {public} "-" (a,b: Complex): Complex;
var
r: Complex;
begin
r.re := a.re - b.re;
r.im := a.im - b.im;
return r
end "-";
operator {public} "*" (a,b: Complex): Complex;
var
r: Complex;
begin
r.re := a.re*b.re - a.im*b.im;
r.im := a.re*b.im + a.im*b.re;
return r
end "*";
operator {public} "/" (a,b: Complex): Complex;
var
r: Complex;
d: real;
begin
d := b.re * b.re + b.im * b.im;
r.re := (a.re * b.re + a.im * b.im)/d;
r.im := (a.im * b.re - a.re * b.im)/d;
return r
end "/";
operator {public} "-" (a: Complex): Complex;
begin
a.im := -1 * a.im;
return a
end "-";
operator {public} "~" (a: Complex): Complex;
var
d: real;
c: Complex;
begin
d := a.re * a.re + a.im * a.im;
c.re := a.re/d;
c.im := (-1.0 * a.im)/d;
return c
end "~";
end Numbers.
module Main;
import Numbers;
var
a,b,c: Numbers.Complex;
procedure Writeln(c: Numbers.Complex);
begin
writeln("(",c.re:4:2,";",c.im:4:2,"i)");
end Writeln;
procedure NewComplex(x,y: real): Numbers.Complex;
var
r: Numbers.Complex;
begin
r.re := x;r.im := y;
return r
end NewComplex;
begin
a := NewComplex(1.5,3.0);
b := NewComplex(1.0,1.0);
Writeln(a + b);
Writeln(a - b);
Writeln(a * b);
Writeln(a / b);
Writeln(-a);
Writeln(~b);
end Main.
|
http://rosettacode.org/wiki/Zero_to_the_zero_power | Zero to the zero power | Some computer programming languages are not exactly consistent (with other computer programming languages)
when raising zero to the zeroth power: 00
Task
Show the results of raising zero to the zeroth power.
If your computer language objects to 0**0 or 0^0 at compile time, you may also try something like:
x = 0
y = 0
z = x**y
say 'z=' z
Show the result here.
And of course use any symbols or notation that is supported in your computer programming language for exponentiation.
See also
The Wiki entry: Zero to the power of zero.
The Wiki entry: History of differing points of view.
The MathWorld™ entry: exponent laws.
Also, in the above MathWorld™ entry, see formula (9):
x
0
=
1
{\displaystyle x^{0}=1}
.
The OEIS entry: The special case of zero to the zeroth power
| #Ring | Ring |
x = 0
y = 0
z = pow(x,y)
see "z=" + z + nl # z=1
|
http://rosettacode.org/wiki/Zero_to_the_zero_power | Zero to the zero power | Some computer programming languages are not exactly consistent (with other computer programming languages)
when raising zero to the zeroth power: 00
Task
Show the results of raising zero to the zeroth power.
If your computer language objects to 0**0 or 0^0 at compile time, you may also try something like:
x = 0
y = 0
z = x**y
say 'z=' z
Show the result here.
And of course use any symbols or notation that is supported in your computer programming language for exponentiation.
See also
The Wiki entry: Zero to the power of zero.
The Wiki entry: History of differing points of view.
The MathWorld™ entry: exponent laws.
Also, in the above MathWorld™ entry, see formula (9):
x
0
=
1
{\displaystyle x^{0}=1}
.
The OEIS entry: The special case of zero to the zeroth power
| #Ruby | Ruby | require 'bigdecimal'
[0, 0.0, Complex(0), Rational(0), BigDecimal("0")].each do |n|
printf "%10s: ** -> %s\n" % [n.class, n**n]
end |
http://rosettacode.org/wiki/Zero_to_the_zero_power | Zero to the zero power | Some computer programming languages are not exactly consistent (with other computer programming languages)
when raising zero to the zeroth power: 00
Task
Show the results of raising zero to the zeroth power.
If your computer language objects to 0**0 or 0^0 at compile time, you may also try something like:
x = 0
y = 0
z = x**y
say 'z=' z
Show the result here.
And of course use any symbols or notation that is supported in your computer programming language for exponentiation.
See also
The Wiki entry: Zero to the power of zero.
The Wiki entry: History of differing points of view.
The MathWorld™ entry: exponent laws.
Also, in the above MathWorld™ entry, see formula (9):
x
0
=
1
{\displaystyle x^{0}=1}
.
The OEIS entry: The special case of zero to the zeroth power
| #Rust | Rust | fn main() {
println!("{}",0u32.pow(0));
} |
http://rosettacode.org/wiki/Zebra_puzzle | Zebra puzzle | Zebra puzzle
You are encouraged to solve this task according to the task description, using any language you may know.
The Zebra puzzle, a.k.a. Einstein's Riddle,
is a logic puzzle which is to be solved programmatically.
It has several variants, one of them this:
There are five houses.
The English man lives in the red house.
The Swede has a dog.
The Dane drinks tea.
The green house is immediately to the left of the white house.
They drink coffee in the green house.
The man who smokes Pall Mall has birds.
In the yellow house they smoke Dunhill.
In the middle house they drink milk.
The Norwegian lives in the first house.
The man who smokes Blend lives in the house next to the house with cats.
In a house next to the house where they have a horse, they smoke Dunhill.
The man who smokes Blue Master drinks beer.
The German smokes Prince.
The Norwegian lives next to the blue house.
They drink water in a house next to the house where they smoke Blend.
The question is, who owns the zebra?
Additionally, list the solution for all the houses.
Optionally, show the solution is unique.
Related tasks
Dinesman's multiple-dwelling problem
Twelve statements
| #Haskell | Haskell | module Main where
import Control.Applicative ((<$>), (<*>))
import Control.Monad (foldM, forM_)
import Data.List ((\\))
-- types
data House = House
{ color :: Color -- <trait> :: House -> <Trait>
, man :: Man
, pet :: Pet
, drink :: Drink
, smoke :: Smoke
}
deriving (Eq, Show)
data Color = Red | Green | Blue | Yellow | White
deriving (Eq, Show, Enum, Bounded)
data Man = Eng | Swe | Dan | Nor | Ger
deriving (Eq, Show, Enum, Bounded)
data Pet = Dog | Birds | Cats | Horse | Zebra
deriving (Eq, Show, Enum, Bounded)
data Drink = Coffee | Tea | Milk | Beer | Water
deriving (Eq, Show, Enum, Bounded)
data Smoke = PallMall | Dunhill | Blend | BlueMaster | Prince
deriving (Eq, Show, Enum, Bounded)
type Solution = [House]
main :: IO ()
main = do
forM_ solutions $ \sol -> mapM_ print sol
>> putStrLn "----"
putStrLn "No More Solutions"
solutions :: [Solution]
solutions = filter finalCheck . map reverse $ foldM next [] [1..5]
where
-- NOTE: list of houses is generated in reversed order
next :: Solution -> Int -> [Solution]
next sol pos = [h:sol | h <- newHouses sol, consistent h pos]
newHouses :: Solution -> Solution
newHouses sol = -- all combinations of traits not yet used
House <$> new color <*> new man <*> new pet <*> new drink <*> new smoke
where
new trait = [minBound ..] \\ map trait sol -- :: [<Trait>]
consistent :: House -> Int -> Bool
consistent house pos = and -- consistent with the rules:
[ man `is` Eng <=> color `is` Red -- 2
, man `is` Swe <=> pet `is` Dog -- 3
, man `is` Dan <=> drink `is` Tea -- 4
, color `is` Green <=> drink `is` Coffee -- 6
, pet `is` Birds <=> smoke `is` PallMall -- 7
, color `is` Yellow <=> smoke `is` Dunhill -- 8
, const (pos == 3) <=> drink `is` Milk -- 9
, const (pos == 1) <=> man `is` Nor -- 10
, drink `is` Beer <=> smoke `is` BlueMaster -- 13
, man `is` Ger <=> smoke `is` Prince -- 14
]
where
infix 4 <=>
p <=> q = p house == q house -- both True or both False
is :: Eq a => (House -> a) -> a -> House -> Bool
(trait `is` value) house = trait house == value
finalCheck :: [House] -> Bool
finalCheck solution = and -- fulfills the rules:
[ (color `is` Green) `leftOf` (color `is` White) -- 5
, (smoke `is` Blend ) `nextTo` (pet `is` Cats ) -- 11
, (smoke `is` Dunhill) `nextTo` (pet `is` Horse) -- 12
, (color `is` Blue ) `nextTo` (man `is` Nor ) -- 15
, (smoke `is` Blend ) `nextTo` (drink `is` Water) -- 16
]
where
nextTo :: (House -> Bool) -> (House -> Bool) -> Bool
nextTo p q = leftOf p q || leftOf q p
leftOf p q
| (_:h:_) <- dropWhile (not . p) solution = q h
| otherwise = False |
http://rosettacode.org/wiki/XML/XPath | XML/XPath | Perform the following three XPath queries on the XML Document below:
//item[1]: Retrieve the first "item" element
//price/text(): Perform an action on each "price" element (print it out)
//name: Get an array of all the "name" elements
XML Document:
<inventory title="OmniCorp Store #45x10^3">
<section name="health">
<item upc="123456789" stock="12">
<name>Invisibility Cream</name>
<price>14.50</price>
<description>Makes you invisible</description>
</item>
<item upc="445322344" stock="18">
<name>Levitation Salve</name>
<price>23.99</price>
<description>Levitate yourself for up to 3 hours per application</description>
</item>
</section>
<section name="food">
<item upc="485672034" stock="653">
<name>Blork and Freen Instameal</name>
<price>4.95</price>
<description>A tasty meal in a tablet; just add water</description>
</item>
<item upc="132957764" stock="44">
<name>Grob winglets</name>
<price>3.56</price>
<description>Tender winglets of Grob. Just add water</description>
</item>
</section>
</inventory>
| #Ksh | Ksh |
#!/bin/ksh
# Perform XPath queries on a XML Document
# # Variables:
#
typeset -T Xml_t=(
typeset -h 'UPC' upc
typeset -i -h 'num in stock' stock=0
typeset -h 'name' name
typeset -F2 -h 'price' price
typeset -h 'description' description
function init_item {
typeset key ; key="$1"
typeset val ; val="${2%\<\/${key}*}"
case ${key} in
upc) _.upc="${val//@(\D)/}"
;;
stock) _.stock="${val//@(\D)/}"
;;
name) _.name="${val%\<\/${key}*}"
;;
price) _.price="${val}"
;;
description) _.description=$(echo ${val})
;;
esac
}
function prt_item {
print "upc= ${_.upc}"
print "stock= ${_.stock}"
print "name= ${_.name}"
print "price= ${_.price}"
print "description= ${_.description}"
}
)
# # Functions:
#
######
# main #
######
integer i=0
typeset -a Item_t
buff=$(< xmldoc) # read xmldoc
item=${buff%%'</item>'*} ; buff=${.sh.match}
while [[ -n ${item} ]]; do
Xml_t Item_t[i]
item=${item#*'<item'} ; item=$(echo ${item})
for word in ${item}; do
if [[ ${word} == *=* ]]; then
Item_t[i].init_item ${word%\=*} ${word#*\=}
else
if [[ ${word} == \<* ]]; then # Beginning
key=${word%%\>*} ; key=${key#*\<}
val=${word#*\>}
fi
[[ ${word} != \<* && ${word} != *\> ]] && val+=" ${word} "
if [[ ${word} == *\> ]]; then # End
val+=" ${word%\<${key}\>*}"
Item_t[i].init_item "${key}" "${val}"
fi
fi
done
(( i++ ))
item=${buff#*'</item>'} ; item=${item%%'</item>'*} ; buff=${.sh.match}
done
print "First Item element:"
Item_t[0].prt_item
typeset -a names
printf "\nList of prices:\n"
for ((i=0; i<${#Item_t[*]}-1; i++)); do
print ${Item_t[i].price}
names[i]=${Item_t[i].name}
done
printf "\nArray of names:\n"
for (( i=0; i<${#names[*]}; i++)); do
print "names[$i] = ${names[i]}"
done |
http://rosettacode.org/wiki/XML/XPath | XML/XPath | Perform the following three XPath queries on the XML Document below:
//item[1]: Retrieve the first "item" element
//price/text(): Perform an action on each "price" element (print it out)
//name: Get an array of all the "name" elements
XML Document:
<inventory title="OmniCorp Store #45x10^3">
<section name="health">
<item upc="123456789" stock="12">
<name>Invisibility Cream</name>
<price>14.50</price>
<description>Makes you invisible</description>
</item>
<item upc="445322344" stock="18">
<name>Levitation Salve</name>
<price>23.99</price>
<description>Levitate yourself for up to 3 hours per application</description>
</item>
</section>
<section name="food">
<item upc="485672034" stock="653">
<name>Blork and Freen Instameal</name>
<price>4.95</price>
<description>A tasty meal in a tablet; just add water</description>
</item>
<item upc="132957764" stock="44">
<name>Grob winglets</name>
<price>3.56</price>
<description>Tender winglets of Grob. Just add water</description>
</item>
</section>
</inventory>
| #Lasso | Lasso | // makes extracting attribute values easier
define xml_attrmap(in::xml_namedNodeMap_attr) => {
local(out = map)
with attr in #in
do #out->insert(#attr->name = #attr->value)
return #out
}
local(
text = '<inventory title="OmniCorp Store #45x10^3">
<section name="health">
<item upc="123456789" stock="12">
<name>Invisibility Cream</name>
<price>14.50</price>
<description>Makes you invisible</description>
</item>
<item upc="445322344" stock="18">
<name>Levitation Salve</name>
<price>23.99</price>
<description>Levitate yourself for up to 3 hours per application</description>
</item>
</section>
<section name="food">
<item upc="485672034" stock="653">
<name>Blork and Freen Instameal</name>
<price>4.95</price>
<description>A tasty meal in a tablet; just add water</description>
</item>
<item upc="132957764" stock="44">
<name>Grob winglets</name>
<price>3.56</price>
<description>Tender winglets of Grob. Just add water</description>
</item>
</section>
</inventory>
',
xml = xml(#text)
)
local(
items = #xml -> extract('//item'),
firstitem = #items -> first,
itemattr = xml_attrmap(#firstitem -> attributes),
newprices = array
)
'<strong>First item:</strong><br />
UPC: '
#itemattr -> find('upc')
' (stock: '
#itemattr -> find('stock')
')<br />'
#firstitem -> extractone('name') -> nodevalue
' ['
#firstitem -> extractone('price') -> nodevalue
'] ('
#firstitem -> extractone('description') -> nodevalue
')<br /><br />'
with item in #items
let name = #item -> extractone('name') -> nodevalue
let price = #item -> extractone('price') -> nodevalue
do {
#newprices -> insert(#name + ': ' + (decimal(#price) * 1.10) -> asstring(-precision = 2) + ' (' + #price + ')')
}
'<strong>Adjusted prices:</strong><br />'
#newprices -> join('<br />')
'<br /><br />'
'<strong>Array with all names:</strong><br />'
#xml -> extract('//name') -> asstaticarray |
http://rosettacode.org/wiki/Yin_and_yang | Yin and yang | One well-known symbol of the philosophy of duality known as yin and yang is the taijitu.
Task
Create a function that, given a parameter representing size, generates such a symbol scaled to the requested size.
Generate and display the symbol for two different (small) sizes.
| #jq | jq |
def svg:
"<svg width='100%' height='100%' version='1.1'
xmlns='http://www.w3.org/2000/svg'
xmlns:xlink='http://www.w3.org/1999/xlink'>" ;
def draw_yinyang(x; scale):
"<use xlink:href='#y' transform='translate(\(x),\(x)) scale(\(scale))'/>";
def define_yinyang:
"<defs>
<g id='y'>
<circle cx='0' cy='0' r='200' stroke='black'
fill='white' stroke-width='1'/>
<path d='M0 -200 A 200 200 0 0 0 0 200
100 100 0 0 0 0 0 100 100 0 0 1 0 -200
z' fill='black'/>
<circle cx='0' cy='100' r='33' fill='white'/>
<circle cx='0' cy='-100' r='33' fill='black'/>
</g>
</defs>" ;
def draw:
svg,
define_yinyang,
draw_yinyang(20; .05),
draw_yinyang(8 ; .02),
"</svg>" ;
draw |
http://rosettacode.org/wiki/Yin_and_yang | Yin and yang | One well-known symbol of the philosophy of duality known as yin and yang is the taijitu.
Task
Create a function that, given a parameter representing size, generates such a symbol scaled to the requested size.
Generate and display the symbol for two different (small) sizes.
| #Julia | Julia | function yinyang(n::Int=3)
radii = (i * n for i in (1, 3, 6))
ranges = collect(collect(-r:r) for r in radii)
squares = collect(collect((x, y) for x in rnge, y in rnge) for rnge in ranges)
circles = collect(collect((x, y) for (x,y) in sqrpoints if hypot(x, y) ≤ radius)
for (sqrpoints, radius) in zip(squares, radii))
m = Dict((x, y) => ' ' for (x, y) in squares[end])
for (x, y) in circles[end] m[(x, y)] = x > 0 ? '·' : '*' end
for (x, y) in circles[end-1]
m[(x, y + 3n)] = '*'
m[(x, y - 3n)] = '·'
end
for (x, y) in circles[end-2]
m[(x, y + 3n)] = '·'
m[(x, y - 3n)] = '*'
end
return join((join(m[(x, y)] for x in reverse(ranges[end])) for y in ranges[end]), '\n')
end
println(yinyang(4))
|
http://rosettacode.org/wiki/Y_combinator | Y combinator | In strict functional programming and the lambda calculus, functions (lambda expressions) don't have state and are only allowed to refer to arguments of enclosing functions.
This rules out the usual definition of a recursive function wherein a function is associated with the state of a variable and this variable's state is used in the body of the function.
The Y combinator is itself a stateless function that, when applied to another stateless function, returns a recursive version of the function.
The Y combinator is the simplest of the class of such functions, called fixed-point combinators.
Task
Define the stateless Y combinator and use it to compute factorials and Fibonacci numbers from other stateless functions or lambda expressions.
Cf
Jim Weirich: Adventures in Functional Programming
| #Eero | Eero | #import <Foundation/Foundation.h>
typedef int (^Func)(int)
typedef Func (^FuncFunc)(Func)
typedef Func (^RecursiveFunc)(id) // hide recursive typing behind dynamic typing
Func fix(FuncFunc f)
Func r(RecursiveFunc g)
int s(int x)
return g(g)(x)
return f(s)
return r(r)
int main(int argc, const char *argv[])
autoreleasepool
Func almost_fac(Func f)
return (int n | return n <= 1 ? 1 : n * f(n - 1))
Func almost_fib(Func f)
return (int n | return n <= 2 ? 1 : f(n - 1) + f(n - 2))
fib := fix(almost_fib)
fac := fix(almost_fac)
Log('fib(10) = %d', fib(10))
Log('fac(10) = %d', fac(10))
return 0 |
http://rosettacode.org/wiki/Zig-zag_matrix | Zig-zag matrix | Task
Produce a zig-zag array.
A zig-zag array is a square arrangement of the first N2 natural numbers, where the
numbers increase sequentially as you zig-zag along the array's anti-diagonals.
For a graphical representation, see JPG zigzag (JPG uses such arrays to encode images).
For example, given 5, produce this array:
0 1 5 6 14
2 4 7 13 15
3 8 12 16 21
9 11 17 20 22
10 18 19 23 24
Related tasks
Spiral matrix
Identity matrix
Ulam spiral (for primes)
See also
Wiktionary entry: anti-diagonals
| #F.23 | F# |
//Produce a zig zag matrix - Nigel Galloway: April 7th., 2015
let zz l a =
let N = Array2D.create l a 0
let rec gng (n, i, g, e) =
N.[n,i] <- g
match e with
| _ when i=a-1 && n=l-1 -> N
| 1 when n = l-1 -> gng (n, i+1, g+1, 2)
| 2 when i = a-1 -> gng (n+1, i, g+1, 1)
| 1 when i = 0 -> gng (n+1, 0, g+1, 2)
| 2 when n = 0 -> gng (0, i+1, g+1, 1)
| 1 -> gng (n+1, i-1, g+1, 1)
| _ -> gng (n-1, i+1, g+1, 2)
gng (0, 0, 0, 2)
|
http://rosettacode.org/wiki/Arbitrary-precision_integers_(included) | Arbitrary-precision integers (included) | Using the in-built capabilities of your language, calculate the integer value of:
5
4
3
2
{\displaystyle 5^{4^{3^{2}}}}
Confirm that the first and last twenty digits of the answer are:
62060698786608744707...92256259918212890625
Find and show the number of decimal digits in the answer.
Note: Do not submit an implementation of arbitrary precision arithmetic. The intention is to show the capabilities of the language as supplied. If a language has a single, overwhelming, library of varied modules that is endorsed by its home site – such as CPAN for Perl or Boost for C++ – then that may be used instead.
Strictly speaking, this should not be solved by fixed-precision numeric libraries where the precision has to be manually set to a large value; although if this is the only recourse then it may be used with a note explaining that the precision must be set manually to a large enough value.
Related tasks
Long multiplication
Exponentiation order
exponentiation operator
Exponentiation with infix operators in (or operating on) the base
| #PHP | PHP | <?php
$y = bcpow('5', bcpow('4', bcpow('3', '2')));
printf("5**4**3**2 = %s...%s and has %d digits\n", substr($y,0,20), substr($y,-20), strlen($y));
?> |
http://rosettacode.org/wiki/Arbitrary-precision_integers_(included) | Arbitrary-precision integers (included) | Using the in-built capabilities of your language, calculate the integer value of:
5
4
3
2
{\displaystyle 5^{4^{3^{2}}}}
Confirm that the first and last twenty digits of the answer are:
62060698786608744707...92256259918212890625
Find and show the number of decimal digits in the answer.
Note: Do not submit an implementation of arbitrary precision arithmetic. The intention is to show the capabilities of the language as supplied. If a language has a single, overwhelming, library of varied modules that is endorsed by its home site – such as CPAN for Perl or Boost for C++ – then that may be used instead.
Strictly speaking, this should not be solved by fixed-precision numeric libraries where the precision has to be manually set to a large value; although if this is the only recourse then it may be used with a note explaining that the precision must be set manually to a large enough value.
Related tasks
Long multiplication
Exponentiation order
exponentiation operator
Exponentiation with infix operators in (or operating on) the base
| #PicoLisp | PicoLisp | (let L (chop (** 5 (** 4 (** 3 2))))
(prinl (head 20 L) "..." (tail 20 L))
(length L) ) |
http://rosettacode.org/wiki/Zeckendorf_number_representation | Zeckendorf number representation | Just as numbers can be represented in a positional notation as sums of multiples of the powers of ten (decimal) or two (binary); all the positive integers can be represented as the sum of one or zero times the distinct members of the Fibonacci series.
Recall that the first six distinct Fibonacci numbers are: 1, 2, 3, 5, 8, 13.
The decimal number eleven can be written as 0*13 + 1*8 + 0*5 + 1*3 + 0*2 + 0*1 or 010100 in positional notation where the columns represent multiplication by a particular member of the sequence. Leading zeroes are dropped so that 11 decimal becomes 10100.
10100 is not the only way to make 11 from the Fibonacci numbers however; 0*13 + 1*8 + 0*5 + 0*3 + 1*2 + 1*1 or 010011 would also represent decimal 11. For a true Zeckendorf number there is the added restriction that no two consecutive Fibonacci numbers can be used which leads to the former unique solution.
Task
Generate and show here a table of the Zeckendorf number representations of the decimal numbers zero to twenty, in order.
The intention in this task to find the Zeckendorf form of an arbitrary integer. The Zeckendorf form can be iterated by some bit twiddling rather than calculating each value separately but leave that to another separate task.
Also see
OEIS A014417 for the the sequence of required results.
Brown's Criterion - Numberphile
Related task
Fibonacci sequence
| #R | R | zeckendorf <- function(number) {
# Get an upper limit on Fibonacci numbers needed to cover number
indexOfFibonacciNumber <- function(n) {
if (n < 1) {
2
} else {
Phi <- (1 + sqrt(5)) / 2
invertClosedFormula <- log(n * sqrt(5)) / log(Phi)
ceiling(invertClosedFormula)
}
}
upperLimit <- indexOfFibonacciNumber(number)
# Return the sequence as digits, sorted descending
fibonacciSequenceDigits <- function(n) {
fibGenerator <- function(f, ...) { c(f[2], sum(f)) }
fibSeq <- Reduce(fibGenerator, 1:n, c(0,1), accumulate=TRUE)
fibNums <- unlist(lapply(fibSeq, head, n=1))
# drop last F0 and F1 and reverse sequence
rev(fibNums[-2:-1])
}
digits <- fibonacciSequenceDigits(upperLimit)
isInNumber <- function(digit) {
if (number >= digit) {
number <<- number - digit
1
} else {
0
}
}
zeckSeq <- Map(isInNumber, digits)
# drop leading 0 and convert to String
gsub("^0+1", "1", paste(zeckSeq, collapse=""))
}
print(unlist(lapply(0:20, zeckendorf))) |
http://rosettacode.org/wiki/100_doors | 100 doors | There are 100 doors in a row that are all initially closed.
You make 100 passes by the doors.
The first time through, visit every door and toggle the door (if the door is closed, open it; if it is open, close it).
The second time, only visit every 2nd door (door #2, #4, #6, ...), and toggle it.
The third time, visit every 3rd door (door #3, #6, #9, ...), etc, until you only visit the 100th door.
Task
Answer the question: what state are the doors in after the last pass? Which are open, which are closed?
Alternate:
As noted in this page's discussion page, the only doors that remain open are those whose numbers are perfect squares.
Opening only those doors is an optimization that may also be expressed;
however, as should be obvious, this defeats the intent of comparing implementations across programming languages.
| #DCL | DCL |
$! doors.com
$! Excecute by running @doors at prompt.
$ square = 1
$ incr = 3
$ count2 = 0
$ d = 1
$ LOOP2:
$ count2 = count2 + 1
$ IF (d .NE. square)
$ THEN WRITE SYS$OUTPUT "door ''d' is closed"
$ ELSE WRITE SYS$OUTPUT "door ''d' is open"
$ square = incr + square
$ incr = incr + 2
$ ENDIF
$ d = d + 1
$ IF (count2 .LT. 100) THEN GOTO LOOP2
|
http://rosettacode.org/wiki/Arrays | Arrays | This task is about arrays.
For hashes or associative arrays, please see Creating an Associative Array.
For a definition and in-depth discussion of what an array is, see Array.
Task
Show basic array syntax in your language.
Basically, create an array, assign a value to it, and retrieve an element (if available, show both fixed-length arrays and
dynamic arrays, pushing a value into it).
Please discuss at Village Pump: Arrays.
Please merge code in from these obsolete tasks:
Creating an Array
Assigning Values to an Array
Retrieving an Element of an Array
Related tasks
Collections
Creating an Associative Array
Two-dimensional array (runtime)
| #HicEst | HicEst | REAL :: n = 3, Astat(n), Bdyn(1, 1)
Astat(2) = 2.22222222
WRITE(Messagebox, Name) Astat(2)
ALLOCATE(Bdyn, 2*n, 3*n)
Bdyn(n-1, n) = -123
WRITE(Row=27) Bdyn(n-1, n)
ALIAS(Astat, n-1, last2ofAstat, 2)
WRITE(ClipBoard) last2ofAstat ! 2.22222222 0 |
http://rosettacode.org/wiki/Arithmetic/Complex | Arithmetic/Complex | A complex number is a number which can be written as:
a
+
b
×
i
{\displaystyle a+b\times i}
(sometimes shown as:
b
+
a
×
i
{\displaystyle b+a\times i}
where
a
{\displaystyle a}
and
b
{\displaystyle b}
are real numbers, and
i
{\displaystyle i}
is √ -1
Typically, complex numbers are represented as a pair of real numbers called the "imaginary part" and "real part", where the imaginary part is the number to be multiplied by
i
{\displaystyle i}
.
Task
Show addition, multiplication, negation, and inversion of complex numbers in separate functions. (Subtraction and division operations can be made with pairs of these operations.)
Print the results for each operation tested.
Optional: Show complex conjugation.
By definition, the complex conjugate of
a
+
b
i
{\displaystyle a+bi}
is
a
−
b
i
{\displaystyle a-bi}
Some languages have complex number libraries available. If your language does, show the operations. If your language does not, also show the definition of this type.
| #ZX_Spectrum_Basic | ZX Spectrum Basic | 5 LET complex=2: LET r=1: LET i=2
10 DIM a(complex): LET a(r)=1.0: LET a(i)=1.0
20 DIM b(complex): LET b(r)=PI: LET b(i)=1.2
30 DIM o(complex)
40 REM add
50 LET o(r)=a(r)+b(r)
60 LET o(i)=a(i)+b(i)
70 PRINT "Result of addition is:": GO SUB 1000
80 REM mult
90 LET o(r)=a(r)*b(r)-a(i)*b(i)
100 LET o(i)=a(i)*b(r)+a(r)*b(i)
110 PRINT "Result of multiplication is:": GO SUB 1000
120 REM neg
130 LET o(r)=-a(r)
140 LET o(i)=-a(i)
150 PRINT "Result of negation is:": GO SUB 1000
160 LET denom=a(r)^2+a(i)^2
170 LET o(r)=a(r)/denom
180 LET o(i)=-a(i)/denom
190 PRINT "Result of inversion is:": GO SUB 1000
200 STOP
1000 IF o(i)>=0 THEN PRINT o(r);" + ";o(i);"i": RETURN
1010 PRINT o(r);" - ";-o(i);"i": RETURN
|
http://rosettacode.org/wiki/Zero_to_the_zero_power | Zero to the zero power | Some computer programming languages are not exactly consistent (with other computer programming languages)
when raising zero to the zeroth power: 00
Task
Show the results of raising zero to the zeroth power.
If your computer language objects to 0**0 or 0^0 at compile time, you may also try something like:
x = 0
y = 0
z = x**y
say 'z=' z
Show the result here.
And of course use any symbols or notation that is supported in your computer programming language for exponentiation.
See also
The Wiki entry: Zero to the power of zero.
The Wiki entry: History of differing points of view.
The MathWorld™ entry: exponent laws.
Also, in the above MathWorld™ entry, see formula (9):
x
0
=
1
{\displaystyle x^{0}=1}
.
The OEIS entry: The special case of zero to the zeroth power
| #S-lang | S-lang | print(0^0); |
http://rosettacode.org/wiki/Zero_to_the_zero_power | Zero to the zero power | Some computer programming languages are not exactly consistent (with other computer programming languages)
when raising zero to the zeroth power: 00
Task
Show the results of raising zero to the zeroth power.
If your computer language objects to 0**0 or 0^0 at compile time, you may also try something like:
x = 0
y = 0
z = x**y
say 'z=' z
Show the result here.
And of course use any symbols or notation that is supported in your computer programming language for exponentiation.
See also
The Wiki entry: Zero to the power of zero.
The Wiki entry: History of differing points of view.
The MathWorld™ entry: exponent laws.
Also, in the above MathWorld™ entry, see formula (9):
x
0
=
1
{\displaystyle x^{0}=1}
.
The OEIS entry: The special case of zero to the zeroth power
| #Scala | Scala | assert(math.pow(0, 0) == 1, "Scala blunder, should go back to school !") |
http://rosettacode.org/wiki/Zero_to_the_zero_power | Zero to the zero power | Some computer programming languages are not exactly consistent (with other computer programming languages)
when raising zero to the zeroth power: 00
Task
Show the results of raising zero to the zeroth power.
If your computer language objects to 0**0 or 0^0 at compile time, you may also try something like:
x = 0
y = 0
z = x**y
say 'z=' z
Show the result here.
And of course use any symbols or notation that is supported in your computer programming language for exponentiation.
See also
The Wiki entry: Zero to the power of zero.
The Wiki entry: History of differing points of view.
The MathWorld™ entry: exponent laws.
Also, in the above MathWorld™ entry, see formula (9):
x
0
=
1
{\displaystyle x^{0}=1}
.
The OEIS entry: The special case of zero to the zeroth power
| #Scheme | Scheme | (display (expt 0 0)) (newline)
(display (expt 0.0 0.0)) (newline)
(display (expt 0+0i 0+0i)) (newline) |
http://rosettacode.org/wiki/Zebra_puzzle | Zebra puzzle | Zebra puzzle
You are encouraged to solve this task according to the task description, using any language you may know.
The Zebra puzzle, a.k.a. Einstein's Riddle,
is a logic puzzle which is to be solved programmatically.
It has several variants, one of them this:
There are five houses.
The English man lives in the red house.
The Swede has a dog.
The Dane drinks tea.
The green house is immediately to the left of the white house.
They drink coffee in the green house.
The man who smokes Pall Mall has birds.
In the yellow house they smoke Dunhill.
In the middle house they drink milk.
The Norwegian lives in the first house.
The man who smokes Blend lives in the house next to the house with cats.
In a house next to the house where they have a horse, they smoke Dunhill.
The man who smokes Blue Master drinks beer.
The German smokes Prince.
The Norwegian lives next to the blue house.
They drink water in a house next to the house where they smoke Blend.
The question is, who owns the zebra?
Additionally, list the solution for all the houses.
Optionally, show the solution is unique.
Related tasks
Dinesman's multiple-dwelling problem
Twelve statements
| #J | J | ehs=: 5$a:
cr=: (('English';'red') 0 3} ehs);<('Dane';'tea') 0 2}ehs
cr=: cr, (('German';'Prince') 0 4}ehs);<('Swede';'dog') 0 1 }ehs
cs=: <('PallMall';'birds') 4 1}ehs
cs=: cs, (('yellow';'Dunhill') 3 4}ehs);<('BlueMaster';'beer') 4 2}ehs
lof=: (('coffee';'green')2 3}ehs);<(<'white')3}ehs
next=: <((<'Blend') 4 }ehs);<(<'water')2}ehs
next=: next,<((<'Blend') 4 }ehs);<(<'cats')1}ehs
next=: next,<((<'Dunhill') 4}ehs);<(<'horse')1}ehs |
http://rosettacode.org/wiki/XML/XPath | XML/XPath | Perform the following three XPath queries on the XML Document below:
//item[1]: Retrieve the first "item" element
//price/text(): Perform an action on each "price" element (print it out)
//name: Get an array of all the "name" elements
XML Document:
<inventory title="OmniCorp Store #45x10^3">
<section name="health">
<item upc="123456789" stock="12">
<name>Invisibility Cream</name>
<price>14.50</price>
<description>Makes you invisible</description>
</item>
<item upc="445322344" stock="18">
<name>Levitation Salve</name>
<price>23.99</price>
<description>Levitate yourself for up to 3 hours per application</description>
</item>
</section>
<section name="food">
<item upc="485672034" stock="653">
<name>Blork and Freen Instameal</name>
<price>4.95</price>
<description>A tasty meal in a tablet; just add water</description>
</item>
<item upc="132957764" stock="44">
<name>Grob winglets</name>
<price>3.56</price>
<description>Tender winglets of Grob. Just add water</description>
</item>
</section>
</inventory>
| #LiveCode | LiveCode | put revXMLCreateTree(fld "FieldXML",true,true,false) into xmltree
// task 1
put revXMLEvaluateXPath(xmltree,"//item[1]") into nodepath
put revXMLText(xmltree,nodepath,true)
// task 2
put revXMLDataFromXPathQuery(xmltree,"//item/price",,comma)
// task 3
put revXMLDataFromXPathQuery(xmltree,"//name") into namenodes
filter namenodes without empty
split namenodes using cr
put namenodes is an array |
http://rosettacode.org/wiki/XML/XPath | XML/XPath | Perform the following three XPath queries on the XML Document below:
//item[1]: Retrieve the first "item" element
//price/text(): Perform an action on each "price" element (print it out)
//name: Get an array of all the "name" elements
XML Document:
<inventory title="OmniCorp Store #45x10^3">
<section name="health">
<item upc="123456789" stock="12">
<name>Invisibility Cream</name>
<price>14.50</price>
<description>Makes you invisible</description>
</item>
<item upc="445322344" stock="18">
<name>Levitation Salve</name>
<price>23.99</price>
<description>Levitate yourself for up to 3 hours per application</description>
</item>
</section>
<section name="food">
<item upc="485672034" stock="653">
<name>Blork and Freen Instameal</name>
<price>4.95</price>
<description>A tasty meal in a tablet; just add water</description>
</item>
<item upc="132957764" stock="44">
<name>Grob winglets</name>
<price>3.56</price>
<description>Tender winglets of Grob. Just add water</description>
</item>
</section>
</inventory>
| #Lua | Lua | require 'lxp'
data = [[<inventory title="OmniCorp Store #45x10^3">
<section name="health">
<item upc="123456789" stock="12">
<name>Invisibility Cream</name>
<price>14.50</price>
<description>Makes you invisible</description>
</item>
<item upc="445322344" stock="18">
<name>Levitation Salve</name>
<price>23.99</price>
<description>Levitate yourself for up to 3 hours per application</description>
</item>
</section>
<section name="food">
<item upc="485672034" stock="653">
<name>Blork and Freen Instameal</name>
<price>4.95</price>
<description>A tasty meal in a tablet; just add water</description>
</item>
<item upc="132957764" stock="44">
<name>Grob winglets</name>
<price>3.56</price>
<description>Tender winglets of Grob. Just add water</description>
</item>
</section>
</inventory>]]
local first = true
local names, prices = {}, {}
p = lxp.new({StartElement = function (parser, name)
local a, b, c = parser:pos() --line, offset, pos
if name == 'item' and first then
print(data:match('.-</item>', c - b + 1))
first = false
end
if name == 'name' then names[#names+1] = data:match('>(.-)<', c) end
if name == 'price' then prices[#prices+1] = data:match('>(.-)<', c) end
end})
p:parse(data)
p:close()
print('Name: ', table.concat(names, ', '))
print('Price: ', table.concat(prices, ', ')) |
http://rosettacode.org/wiki/Yin_and_yang | Yin and yang | One well-known symbol of the philosophy of duality known as yin and yang is the taijitu.
Task
Create a function that, given a parameter representing size, generates such a symbol scaled to the requested size.
Generate and display the symbol for two different (small) sizes.
| #Kotlin | Kotlin | // version 1.1.2
import java.awt.Color
import java.awt.Graphics
import java.awt.Image
import java.awt.image.BufferedImage
import javax.swing.ImageIcon
import javax.swing.JFrame
import javax.swing.JPanel
import javax.swing.JLabel
class YinYangGenerator {
private fun drawYinYang(size: Int, g: Graphics) {
with(g) {
// Preserve the color for the caller
val colorSave = color
color = Color.WHITE
// Use fillOval to draw a filled in circle
fillOval(0, 0, size - 1, size - 1)
color = Color.BLACK
// Use fillArc to draw part of a filled in circle
fillArc(0, 0, size - 1, size - 1, 270, 180)
fillOval(size / 4, size / 2, size / 2, size / 2)
color = Color.WHITE
fillOval(size / 4, 0, size / 2, size / 2)
fillOval(7 * size / 16, 11 * size / 16, size /8, size / 8)
color = Color.BLACK
fillOval(7 * size / 16, 3 * size / 16, size / 8, size / 8)
// Use drawOval to draw an empty circle for the outside border
drawOval(0, 0, size - 1, size - 1)
// Restore the color for the caller
color = colorSave
}
}
fun createImage(size: Int, bg: Color): Image {
// A BufferedImage creates the image in memory
val image = BufferedImage(size, size, BufferedImage.TYPE_INT_RGB)
// Get the graphics object for the image
val g = image.graphics
// Color in the background of the image
g.color = bg
g.fillRect(0, 0, size, size)
drawYinYang(size, g)
return image
}
}
fun main(args: Array<String>) {
val gen = YinYangGenerator()
val size = 400 // say
val p = JPanel()
val yinYang = gen.createImage(size, p.background)
p.add(JLabel(ImageIcon(yinYang)))
val size2 = size / 2 // say
val yinYang2 = gen.createImage(size2, p.background)
p.add(JLabel(ImageIcon(yinYang2)))
val f = JFrame("Big and Small Yin Yang")
with (f) {
defaultCloseOperation = JFrame.EXIT_ON_CLOSE
add(p)
pack()
isVisible = true
}
} |
http://rosettacode.org/wiki/Y_combinator | Y combinator | In strict functional programming and the lambda calculus, functions (lambda expressions) don't have state and are only allowed to refer to arguments of enclosing functions.
This rules out the usual definition of a recursive function wherein a function is associated with the state of a variable and this variable's state is used in the body of the function.
The Y combinator is itself a stateless function that, when applied to another stateless function, returns a recursive version of the function.
The Y combinator is the simplest of the class of such functions, called fixed-point combinators.
Task
Define the stateless Y combinator and use it to compute factorials and Fibonacci numbers from other stateless functions or lambda expressions.
Cf
Jim Weirich: Adventures in Functional Programming
| #Ela | Ela | fix = \f -> (\x -> & f (x x)) (\x -> & f (x x))
fac _ 0 = 1
fac f n = n * f (n - 1)
fib _ 0 = 0
fib _ 1 = 1
fib f n = f (n - 1) + f (n - 2)
(fix fac 12, fix fib 12) |
http://rosettacode.org/wiki/Zig-zag_matrix | Zig-zag matrix | Task
Produce a zig-zag array.
A zig-zag array is a square arrangement of the first N2 natural numbers, where the
numbers increase sequentially as you zig-zag along the array's anti-diagonals.
For a graphical representation, see JPG zigzag (JPG uses such arrays to encode images).
For example, given 5, produce this array:
0 1 5 6 14
2 4 7 13 15
3 8 12 16 21
9 11 17 20 22
10 18 19 23 24
Related tasks
Spiral matrix
Identity matrix
Ulam spiral (for primes)
See also
Wiktionary entry: anti-diagonals
| #Factor | Factor | USING: columns fry kernel make math math.ranges prettyprint
sequences sequences.cords sequences.extras ;
IN: rosetta-code.zig-zag-matrix
: [1,b,1] ( n -- seq )
[1,b] dup but-last-slice <reversed> cord-append ;
: <reversed-evens> ( seq -- seq' )
[ even? [ <reversed> ] when ] map-index ;
: diagonals ( n -- seq )
[ sq <iota> ] [ [1,b,1] ] bi
[ [ cut [ , ] dip ] each ] { } make nip <reversed-evens> ;
: zig-zag-matrix ( n -- seq )
[ diagonals ] [ dup ] bi '[
[
dup 0 <column> _ head ,
[ _ < [ rest-slice ] when ] map-index harvest
] until-empty
] { } make ;
: zig-zag-demo ( -- ) 5 zig-zag-matrix simple-table. ;
MAIN: zig-zag-demo |
http://rosettacode.org/wiki/Arbitrary-precision_integers_(included) | Arbitrary-precision integers (included) | Using the in-built capabilities of your language, calculate the integer value of:
5
4
3
2
{\displaystyle 5^{4^{3^{2}}}}
Confirm that the first and last twenty digits of the answer are:
62060698786608744707...92256259918212890625
Find and show the number of decimal digits in the answer.
Note: Do not submit an implementation of arbitrary precision arithmetic. The intention is to show the capabilities of the language as supplied. If a language has a single, overwhelming, library of varied modules that is endorsed by its home site – such as CPAN for Perl or Boost for C++ – then that may be used instead.
Strictly speaking, this should not be solved by fixed-precision numeric libraries where the precision has to be manually set to a large value; although if this is the only recourse then it may be used with a note explaining that the precision must be set manually to a large enough value.
Related tasks
Long multiplication
Exponentiation order
exponentiation operator
Exponentiation with infix operators in (or operating on) the base
| #Pike | Pike | > string res = (string)pow(5,pow(4,pow(3,2)));
> res[..19] == "62060698786608744707";
Result: 1
> res[<19..] == "92256259918212890625";
Result: 1
> sizeof(result);
Result: 183231 |
http://rosettacode.org/wiki/Arbitrary-precision_integers_(included) | Arbitrary-precision integers (included) | Using the in-built capabilities of your language, calculate the integer value of:
5
4
3
2
{\displaystyle 5^{4^{3^{2}}}}
Confirm that the first and last twenty digits of the answer are:
62060698786608744707...92256259918212890625
Find and show the number of decimal digits in the answer.
Note: Do not submit an implementation of arbitrary precision arithmetic. The intention is to show the capabilities of the language as supplied. If a language has a single, overwhelming, library of varied modules that is endorsed by its home site – such as CPAN for Perl or Boost for C++ – then that may be used instead.
Strictly speaking, this should not be solved by fixed-precision numeric libraries where the precision has to be manually set to a large value; although if this is the only recourse then it may be used with a note explaining that the precision must be set manually to a large enough value.
Related tasks
Long multiplication
Exponentiation order
exponentiation operator
Exponentiation with infix operators in (or operating on) the base
| #PowerShell | PowerShell | # Perform calculation
$BigNumber = [BigInt]::Pow( 5, [BigInt]::Pow( 4, [BigInt]::Pow( 3, 2 ) ) )
# Display first and last 20 digits
$BigNumberString = [string]$BigNumber
$BigNumberString.Substring( 0, 20 ) + "..." + $BigNumberString.Substring( $BigNumberString.Length - 20, 20 )
# Display number of digits
$BigNumberString.Length |
http://rosettacode.org/wiki/Zeckendorf_number_representation | Zeckendorf number representation | Just as numbers can be represented in a positional notation as sums of multiples of the powers of ten (decimal) or two (binary); all the positive integers can be represented as the sum of one or zero times the distinct members of the Fibonacci series.
Recall that the first six distinct Fibonacci numbers are: 1, 2, 3, 5, 8, 13.
The decimal number eleven can be written as 0*13 + 1*8 + 0*5 + 1*3 + 0*2 + 0*1 or 010100 in positional notation where the columns represent multiplication by a particular member of the sequence. Leading zeroes are dropped so that 11 decimal becomes 10100.
10100 is not the only way to make 11 from the Fibonacci numbers however; 0*13 + 1*8 + 0*5 + 0*3 + 1*2 + 1*1 or 010011 would also represent decimal 11. For a true Zeckendorf number there is the added restriction that no two consecutive Fibonacci numbers can be used which leads to the former unique solution.
Task
Generate and show here a table of the Zeckendorf number representations of the decimal numbers zero to twenty, in order.
The intention in this task to find the Zeckendorf form of an arbitrary integer. The Zeckendorf form can be iterated by some bit twiddling rather than calculating each value separately but leave that to another separate task.
Also see
OEIS A014417 for the the sequence of required results.
Brown's Criterion - Numberphile
Related task
Fibonacci sequence
| #Racket | Racket |
#lang racket (require math)
(define (fibs n)
(reverse
(for/list ([i (in-naturals 2)] #:break (> (fibonacci i) n))
(fibonacci i))))
(define (zechendorf n)
(match/values
(for/fold ([n n] [xs '()]) ([f (fibs n)])
(if (> f n)
(values n (cons 0 xs))
(values (- n f) (cons 1 xs))))
[(_ xs) (reverse xs)]))
(for/list ([n 21])
(list n (zechendorf n)))
|
http://rosettacode.org/wiki/100_doors | 100 doors | There are 100 doors in a row that are all initially closed.
You make 100 passes by the doors.
The first time through, visit every door and toggle the door (if the door is closed, open it; if it is open, close it).
The second time, only visit every 2nd door (door #2, #4, #6, ...), and toggle it.
The third time, visit every 3rd door (door #3, #6, #9, ...), etc, until you only visit the 100th door.
Task
Answer the question: what state are the doors in after the last pass? Which are open, which are closed?
Alternate:
As noted in this page's discussion page, the only doors that remain open are those whose numbers are perfect squares.
Opening only those doors is an optimization that may also be expressed;
however, as should be obvious, this defeats the intent of comparing implementations across programming languages.
| #Delphi | Delphi | proc nonrec main() void:
byte DOORS = 100;
[DOORS+1] bool door_open;
unsigned DOORS i, j;
/* make sure all doors are closed */
for i from 1 upto DOORS do door_open[i] := false od;
/* pass through the doors */
for i from 1 upto DOORS do
for j from i by i upto DOORS do
door_open[j] := not door_open[j]
od
od;
/* show the open doors */
for i from 1 upto DOORS do
if door_open[i] then
writeln("Door ", i, " is open.")
fi
od
corp |
http://rosettacode.org/wiki/Arrays | Arrays | This task is about arrays.
For hashes or associative arrays, please see Creating an Associative Array.
For a definition and in-depth discussion of what an array is, see Array.
Task
Show basic array syntax in your language.
Basically, create an array, assign a value to it, and retrieve an element (if available, show both fixed-length arrays and
dynamic arrays, pushing a value into it).
Please discuss at Village Pump: Arrays.
Please merge code in from these obsolete tasks:
Creating an Array
Assigning Values to an Array
Retrieving an Element of an Array
Related tasks
Collections
Creating an Associative Array
Two-dimensional array (runtime)
| #HolyC | HolyC | // Create an array of fixed size
U8 array[10] = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10;
// The first element of a HolyC array is indexed at 0. To set a value:
array[0] = 123;
// Access an element
Print("%d\n", array[0]); |
http://rosettacode.org/wiki/Zero_to_the_zero_power | Zero to the zero power | Some computer programming languages are not exactly consistent (with other computer programming languages)
when raising zero to the zeroth power: 00
Task
Show the results of raising zero to the zeroth power.
If your computer language objects to 0**0 or 0^0 at compile time, you may also try something like:
x = 0
y = 0
z = x**y
say 'z=' z
Show the result here.
And of course use any symbols or notation that is supported in your computer programming language for exponentiation.
See also
The Wiki entry: Zero to the power of zero.
The Wiki entry: History of differing points of view.
The MathWorld™ entry: exponent laws.
Also, in the above MathWorld™ entry, see formula (9):
x
0
=
1
{\displaystyle x^{0}=1}
.
The OEIS entry: The special case of zero to the zeroth power
| #Seed7 | Seed7 | $ include "seed7_05.s7i";
include "float.s7i";
include "complex.s7i";
const proc: main is func
begin
writeln("0 ** 0 = " <& 0 ** 0);
writeln("0.0 ** 0 = " <& 0.0 ** 0);
writeln("0.0 ** 0.0 = " <& 0.0 ** 0.0);
writeln("0.0+0i ** 0 = " <& complex(0.0) ** 0);
end func;
|
http://rosettacode.org/wiki/Zero_to_the_zero_power | Zero to the zero power | Some computer programming languages are not exactly consistent (with other computer programming languages)
when raising zero to the zeroth power: 00
Task
Show the results of raising zero to the zeroth power.
If your computer language objects to 0**0 or 0^0 at compile time, you may also try something like:
x = 0
y = 0
z = x**y
say 'z=' z
Show the result here.
And of course use any symbols or notation that is supported in your computer programming language for exponentiation.
See also
The Wiki entry: Zero to the power of zero.
The Wiki entry: History of differing points of view.
The MathWorld™ entry: exponent laws.
Also, in the above MathWorld™ entry, see formula (9):
x
0
=
1
{\displaystyle x^{0}=1}
.
The OEIS entry: The special case of zero to the zeroth power
| #SenseTalk | SenseTalk | set a to 0
set b to 0
put a to the power of b
// Prints: 1 |
http://rosettacode.org/wiki/Zebra_puzzle | Zebra puzzle | Zebra puzzle
You are encouraged to solve this task according to the task description, using any language you may know.
The Zebra puzzle, a.k.a. Einstein's Riddle,
is a logic puzzle which is to be solved programmatically.
It has several variants, one of them this:
There are five houses.
The English man lives in the red house.
The Swede has a dog.
The Dane drinks tea.
The green house is immediately to the left of the white house.
They drink coffee in the green house.
The man who smokes Pall Mall has birds.
In the yellow house they smoke Dunhill.
In the middle house they drink milk.
The Norwegian lives in the first house.
The man who smokes Blend lives in the house next to the house with cats.
In a house next to the house where they have a horse, they smoke Dunhill.
The man who smokes Blue Master drinks beer.
The German smokes Prince.
The Norwegian lives next to the blue house.
They drink water in a house next to the house where they smoke Blend.
The question is, who owns the zebra?
Additionally, list the solution for all the houses.
Optionally, show the solution is unique.
Related tasks
Dinesman's multiple-dwelling problem
Twelve statements
| #Java | Java | package org.rosettacode.zebra;
import java.util.Arrays;
import java.util.Iterator;
import java.util.LinkedHashSet;
import java.util.Objects;
import java.util.Set;
public class Zebra {
private static final int[] orders = {1, 2, 3, 4, 5};
private static final String[] nations = {"English", "Danish", "German", "Swedish", "Norwegian"};
private static final String[] animals = {"Zebra", "Horse", "Birds", "Dog", "Cats"};
private static final String[] drinks = {"Coffee", "Tea", "Beer", "Water", "Milk"};
private static final String[] cigarettes = {"Pall Mall", "Blend", "Blue Master", "Prince", "Dunhill"};
private static final String[] colors = {"Red", "Green", "White", "Blue", "Yellow"};
static class Solver {
private final PossibleLines puzzleTable = new PossibleLines();
void solve() {
PossibleLines constraints = new PossibleLines();
constraints.add(new PossibleLine(null, "English", "Red", null, null, null));
constraints.add(new PossibleLine(null, "Swedish", null, "Dog", null, null));
constraints.add(new PossibleLine(null, "Danish", null, null, "Tea", null));
constraints.add(new PossibleLine(null, null, "Green", null, "Coffee", null));
constraints.add(new PossibleLine(null, null, null, "Birds", null, "Pall Mall"));
constraints.add(new PossibleLine(null, null, "Yellow", null, null, "Dunhill"));
constraints.add(new PossibleLine(3, null, null, null, "Milk", null));
constraints.add(new PossibleLine(1, "Norwegian", null, null, null, null));
constraints.add(new PossibleLine(null, null, null, null, "Beer", "Blue Master"));
constraints.add(new PossibleLine(null, "German", null, null, null, "Prince"));
constraints.add(new PossibleLine(2, null, "Blue", null, null, null));
//Creating all possible combination of a puzzle line.
//The maximum number of lines is 5^^6 (15625).
//Each combination line is checked against a set of knowing facts, thus
//only a small number of line result at the end.
for (Integer orderId : Zebra.orders) {
for (String nation : Zebra.nations) {
for (String color : Zebra.colors) {
for (String animal : Zebra.animals) {
for (String drink : Zebra.drinks) {
for (String cigarette : Zebra.cigarettes) {
addPossibleNeighbors(constraints, orderId, nation, color, animal, drink, cigarette);
}
}
}
}
}
}
System.out.println("After general rule set validation, remains " +
puzzleTable.size() + " lines.");
for (Iterator<PossibleLine> it = puzzleTable.iterator(); it.hasNext(); ) {
boolean validLine = true;
PossibleLine possibleLine = it.next();
if (possibleLine.leftNeighbor != null) {
PossibleLine neighbor = possibleLine.leftNeighbor;
if (neighbor.order < 1 || neighbor.order > 5) {
validLine = false;
it.remove();
}
}
if (validLine && possibleLine.rightNeighbor != null) {
PossibleLine neighbor = possibleLine.rightNeighbor;
if (neighbor.order < 1 || neighbor.order > 5) {
it.remove();
}
}
}
System.out.println("After removing out of bound neighbors, remains " +
puzzleTable.size() + " lines.");
//Setting left and right neighbors
for (PossibleLine puzzleLine : puzzleTable) {
for (PossibleLine leftNeighbor : puzzleLine.neighbors) {
PossibleLine rightNeighbor = leftNeighbor.copy();
//make it left neighbor
leftNeighbor.order = puzzleLine.order - 1;
if (puzzleTable.contains(leftNeighbor)) {
if (puzzleLine.leftNeighbor != null)
puzzleLine.leftNeighbor.merge(leftNeighbor);
else
puzzleLine.setLeftNeighbor(leftNeighbor);
}
rightNeighbor.order = puzzleLine.order + 1;
if (puzzleTable.contains(rightNeighbor)) {
if (puzzleLine.rightNeighbor != null)
puzzleLine.rightNeighbor.merge(rightNeighbor);
else
puzzleLine.setRightNeighbor(rightNeighbor);
}
}
}
int iteration = 1;
int lastSize = 0;
//Recursively validate against neighbor rules
while (puzzleTable.size() > 5 && lastSize != puzzleTable.size()) {
lastSize = puzzleTable.size();
puzzleTable.clearLineCountFlags();
recursiveSearch(null, puzzleTable, -1);
constraints.clear();
// Assuming we'll get at leas one valid line each iteration, we create
// a set of new rules with lines which have no more then one instance of same OrderId.
for (int i = 1; i < 6; i++) {
if (puzzleTable.getLineCountByOrderId(i) == 1)
constraints.addAll(puzzleTable.getSimilarLines(new PossibleLine(i, null, null, null, null,
null)));
}
puzzleTable.removeIf(puzzleLine -> !constraints.accepts(puzzleLine));
System.out.println("After " + iteration + " recursive iteration, remains "
+ puzzleTable.size() + " lines");
iteration++;
}
// Print the results
System.out.println("-------------------------------------------");
if (puzzleTable.size() == 5) {
for (PossibleLine puzzleLine : puzzleTable) {
System.out.println(puzzleLine.getWholeLine());
}
} else
System.out.println("Sorry, solution not found!");
}
private void addPossibleNeighbors(
PossibleLines constraints, Integer orderId, String nation,
String color, String animal, String drink, String cigarette) {
boolean validLine = true;
PossibleLine pzlLine = new PossibleLine(orderId,
nation,
color,
animal,
drink,
cigarette);
// Checking against a set of knowing facts
if (constraints.accepts(pzlLine)) {
// Adding rules of neighbors
if (cigarette.equals("Blend")
&& (animal.equals("Cats") || drink.equals("Water")))
validLine = false;
if (cigarette.equals("Dunhill")
&& animal.equals("Horse"))
validLine = false;
if (validLine) {
puzzleTable.add(pzlLine);
//set neighbors constraints
if (color.equals("Green")) {
pzlLine.setRightNeighbor(
new PossibleLine(null, null, "White", null, null, null));
}
if (color.equals("White")) {
pzlLine.setLeftNeighbor(
new PossibleLine(null, null, "Green", null, null, null));
}
//
if (animal.equals("Cats") && !cigarette.equals("Blend")) {
pzlLine.neighbors.add(new PossibleLine(null, null, null, null, null,
"Blend"));
}
if (cigarette.equals("Blend") && !animal.equals("Cats")) {
pzlLine.neighbors.add(new PossibleLine(null, null, null, "Cats", null
, null));
}
//
if (drink.equals("Water")
&& !animal.equals("Cats")
&& !cigarette.equals("Blend")) {
pzlLine.neighbors.add(new PossibleLine(null, null, null, null, null,
"Blend"));
}
if (cigarette.equals("Blend") && !drink.equals("Water")) {
pzlLine.neighbors.add(new PossibleLine(null, null, null, null, "Water"
, null));
}
//
if (animal.equals("Horse") && !cigarette.equals("Dunhill")) {
pzlLine.neighbors.add(new PossibleLine(null, null, null, null, null,
"Dunhill"));
}
if (cigarette.equals("Dunhill") && !animal.equals("Horse")) {
pzlLine.neighbors.add(new PossibleLine(null, null, null, "Horse",
null, null));
}
}
}
}
// Recursively checks the input set to ensure each line has right neighbor.
// Neighbors can be of three type, left, right or undefined.
// Direction: -1 left, 0 undefined, 1 right
private boolean recursiveSearch(PossibleLine pzzlNodeLine,
PossibleLines possibleLines, int direction) {
boolean validLeaf = false;
boolean hasNeighbor;
PossibleLines puzzleSubSet;
for (Iterator<PossibleLine> it = possibleLines.iterator(); it.hasNext(); ) {
PossibleLine pzzlLeafLine = it.next();
validLeaf = false;
hasNeighbor = pzzlLeafLine.hasNeighbor(direction);
if (hasNeighbor) {
puzzleSubSet = puzzleTable.getSimilarLines(pzzlLeafLine.getNeighbor(direction));
if (puzzleSubSet != null) {
if (pzzlNodeLine != null)
validLeaf = puzzleSubSet.contains(pzzlNodeLine);
else
validLeaf = recursiveSearch(pzzlLeafLine, puzzleSubSet, -1 * direction);
}
}
if (!validLeaf && pzzlLeafLine.hasNeighbor(-1 * direction)) {
hasNeighbor = true;
puzzleSubSet = puzzleTable.getSimilarLines(pzzlLeafLine.getNeighbor(-1 * direction));
if (puzzleSubSet != null) {
if (pzzlNodeLine != null)
validLeaf = puzzleSubSet.contains(pzzlNodeLine);
else
validLeaf = recursiveSearch(pzzlLeafLine, puzzleSubSet, direction);
}
}
if (pzzlNodeLine != null && validLeaf)
return true;
if (pzzlNodeLine == null && hasNeighbor && !validLeaf) {
it.remove();
}
if (pzzlNodeLine == null) {
if (hasNeighbor && validLeaf) {
possibleLines.riseLineCountFlags(pzzlLeafLine.order);
}
if (!hasNeighbor) {
possibleLines.riseLineCountFlags(pzzlLeafLine.order);
}
}
}
return validLeaf;
}
}
public static void main(String[] args) {
Solver solver = new Solver();
solver.solve();
}
static class PossibleLines extends LinkedHashSet<PossibleLine> {
private final int[] count = new int[5];
public PossibleLine get(int index) {
return ((PossibleLine) toArray()[index]);
}
public PossibleLines getSimilarLines(PossibleLine searchLine) {
PossibleLines puzzleSubSet = new PossibleLines();
for (PossibleLine possibleLine : this) {
if (possibleLine.getCommonFactsCount(searchLine) == searchLine.getFactsCount())
puzzleSubSet.add(possibleLine);
}
if (puzzleSubSet.isEmpty())
return null;
return puzzleSubSet;
}
public boolean contains(PossibleLine searchLine) {
for (PossibleLine puzzleLine : this) {
if (puzzleLine.getCommonFactsCount(searchLine) == searchLine.getFactsCount())
return true;
}
return false;
}
public boolean accepts(PossibleLine searchLine) {
int passed = 0;
int notpassed = 0;
for (PossibleLine puzzleSetLine : this) {
int lineFactsCnt = puzzleSetLine.getFactsCount();
int comnFactsCnt = puzzleSetLine.getCommonFactsCount(searchLine);
if (lineFactsCnt != comnFactsCnt && lineFactsCnt != 0 && comnFactsCnt != 0) {
notpassed++;
}
if (lineFactsCnt == comnFactsCnt)
passed++;
}
return passed >= 0 && notpassed == 0;
}
public void riseLineCountFlags(int lineOrderId) {
count[lineOrderId - 1]++;
}
public void clearLineCountFlags() {
Arrays.fill(count, 0);
}
public int getLineCountByOrderId(int lineOrderId) {
return count[lineOrderId - 1];
}
}
static class PossibleLine {
Integer order;
String nation;
String color;
String animal;
String drink;
String cigarette;
PossibleLine rightNeighbor;
PossibleLine leftNeighbor;
Set<PossibleLine> neighbors = new LinkedHashSet<>();
public PossibleLine(Integer order, String nation, String color,
String animal, String drink, String cigarette) {
this.animal = animal;
this.cigarette = cigarette;
this.color = color;
this.drink = drink;
this.nation = nation;
this.order = order;
}
@Override
public boolean equals(Object obj) {
return obj instanceof PossibleLine
&& getWholeLine().equals(((PossibleLine) obj).getWholeLine());
}
public int getFactsCount() {
int facts = 0;
facts += order != null ? 1 : 0;
facts += nation != null ? 1 : 0;
facts += color != null ? 1 : 0;
facts += animal != null ? 1 : 0;
facts += cigarette != null ? 1 : 0;
facts += drink != null ? 1 : 0;
return facts;
}
private static int common(Object a, Object b) {
return a != null && Objects.equals(a, b) ? 1 : 0;
}
public int getCommonFactsCount(PossibleLine facts) {
return common(order, facts.order)
+ common(nation, facts.nation)
+ common(color, facts.color)
+ common(animal, facts.animal)
+ common(cigarette, facts.cigarette)
+ common(drink, facts.drink);
}
public void setLeftNeighbor(PossibleLine leftNeighbor) {
this.leftNeighbor = leftNeighbor;
this.leftNeighbor.order = order - 1;
}
public void setRightNeighbor(PossibleLine rightNeighbor) {
this.rightNeighbor = rightNeighbor;
this.rightNeighbor.order = order + 1;
}
public boolean hasNeighbor(int direction) {
return getNeighbor(direction) != null;
}
public PossibleLine getNeighbor(int direction) {
if (direction < 0)
return leftNeighbor;
else
return rightNeighbor;
}
public String getWholeLine() {
return order + " - " +
nation + " - " +
color + " - " +
animal + " - " +
drink + " - " +
cigarette;
}
@Override
public int hashCode() {
return Objects.hash(order, nation, color, animal, drink, cigarette);
}
public void merge(PossibleLine mergedLine) {
if (order == null) order = mergedLine.order;
if (nation == null) nation = mergedLine.nation;
if (color == null) color = mergedLine.color;
if (animal == null) animal = mergedLine.animal;
if (drink == null) drink = mergedLine.drink;
if (cigarette == null) cigarette = mergedLine.cigarette;
}
public PossibleLine copy() {
PossibleLine clone = new PossibleLine(order, nation, color, animal, drink, cigarette);
clone.leftNeighbor = leftNeighbor;
clone.rightNeighbor = rightNeighbor;
clone.neighbors = neighbors; // shallow copy
return clone;
}
}
} |
http://rosettacode.org/wiki/XML/XPath | XML/XPath | Perform the following three XPath queries on the XML Document below:
//item[1]: Retrieve the first "item" element
//price/text(): Perform an action on each "price" element (print it out)
//name: Get an array of all the "name" elements
XML Document:
<inventory title="OmniCorp Store #45x10^3">
<section name="health">
<item upc="123456789" stock="12">
<name>Invisibility Cream</name>
<price>14.50</price>
<description>Makes you invisible</description>
</item>
<item upc="445322344" stock="18">
<name>Levitation Salve</name>
<price>23.99</price>
<description>Levitate yourself for up to 3 hours per application</description>
</item>
</section>
<section name="food">
<item upc="485672034" stock="653">
<name>Blork and Freen Instameal</name>
<price>4.95</price>
<description>A tasty meal in a tablet; just add water</description>
</item>
<item upc="132957764" stock="44">
<name>Grob winglets</name>
<price>3.56</price>
<description>Tender winglets of Grob. Just add water</description>
</item>
</section>
</inventory>
| #Mathematica.2FWolfram_Language | Mathematica/Wolfram Language | example = Import["test.txt", "XML"];
Cases[example, XMLElement["item", _ , _] , Infinity] // First
Cases[example, XMLElement["price", _, List[n_]] -> n, Infinity] // Column
Cases[example, XMLElement["name", _, List[n_]] -> n, Infinity] // Column |
http://rosettacode.org/wiki/XML/XPath | XML/XPath | Perform the following three XPath queries on the XML Document below:
//item[1]: Retrieve the first "item" element
//price/text(): Perform an action on each "price" element (print it out)
//name: Get an array of all the "name" elements
XML Document:
<inventory title="OmniCorp Store #45x10^3">
<section name="health">
<item upc="123456789" stock="12">
<name>Invisibility Cream</name>
<price>14.50</price>
<description>Makes you invisible</description>
</item>
<item upc="445322344" stock="18">
<name>Levitation Salve</name>
<price>23.99</price>
<description>Levitate yourself for up to 3 hours per application</description>
</item>
</section>
<section name="food">
<item upc="485672034" stock="653">
<name>Blork and Freen Instameal</name>
<price>4.95</price>
<description>A tasty meal in a tablet; just add water</description>
</item>
<item upc="132957764" stock="44">
<name>Grob winglets</name>
<price>3.56</price>
<description>Tender winglets of Grob. Just add water</description>
</item>
</section>
</inventory>
| #NetRexx | NetRexx | /* NetRexx */
options replace format comments java symbols binary
import javax.xml.parsers.
import javax.xml.xpath.
import org.w3c.dom.
import org.w3c.dom.Node
import org.xml.sax.
xmlStr = '' -
|| '<inventory title="OmniCorp Store #45x10^3">' -
|| ' <section name="health">' -
|| ' <item upc="123456789" stock="12">' -
|| ' <name>Invisibility Cream</name>' -
|| ' <price>14.50</price>' -
|| ' <description>Makes you invisible</description>' -
|| ' </item>' -
|| ' <item upc="445322344" stock="18">' -
|| ' <name>Levitation Salve</name>' -
|| ' <price>23.99</price>' -
|| ' <description>Levitate yourself for up to 3 hours per application</description>' -
|| ' </item>' -
|| ' </section>' -
|| ' <section name="food">' -
|| ' <item upc="485672034" stock="653">' -
|| ' <name>Blork and Freen Instameal</name>' -
|| ' <price>4.95</price>' -
|| ' <description>A tasty meal in a tablet; just add water</description>' -
|| ' </item>' -
|| ' <item upc="132957764" stock="44">' -
|| ' <name>Grob winglets</name>' -
|| ' <price>3.56</price>' -
|| ' <description>Tender winglets of Grob. Just add priwater</description>' -
|| ' </item>' -
|| ' </section>' -
|| '</inventory>'
expr1 = '/inventory/section/item[1]'
expr2 = '/inventory/section/item/price'
expr3 = '/inventory/section/item/name'
attr1 = 'upc'
do
doc = DocumentBuilderFactory.newInstance().newDocumentBuilder().parse(InputSource(StringReader(xmlStr)))
xpath = XPathFactory.newInstance().newXPath()
-- Extract attribute from 1st item element
say expr1
say " "(Node xpath.evaluate(expr1, doc, XPathConstants.NODE)).getAttributes().getNamedItem(attr1)
say
-- Extract and display all price elments
nodes = NodeList xpath.evaluate(expr2, doc, XPathConstants.NODESET)
say expr2
loop i_ = 0 to nodes.getLength() - 1
say Rexx(nodes.item(i_).getTextContent()).format(10, 2)
end i_
say
-- Extract elements and store in an ArrayList
nameList = java.util.List
nameList = ArrayList()
nodes = NodeList xpath.evaluate(expr3, doc, XPathConstants.NODESET)
loop i_ = 0 to nodes.getLength() - 1
nameList.add(nodes.item(i_).getTextContent())
end i_
-- display contents of ArrayList
say expr3
loop n_ = 0 to nameList.size() - 1
say " "nameList.get(n_)
end n_
say
catch ex = Exception
ex.printStackTrace()
end
return
|
http://rosettacode.org/wiki/Yin_and_yang | Yin and yang | One well-known symbol of the philosophy of duality known as yin and yang is the taijitu.
Task
Create a function that, given a parameter representing size, generates such a symbol scaled to the requested size.
Generate and display the symbol for two different (small) sizes.
| #Lambdatalk | Lambdatalk |
{{SVG 580 580}
{YY 145 145 300}
{YY 270 195 50}
{YY 270 345 50}
}
{def YY
{lambda {:x :y :s}
{{G :x :y :s}
{CIRCLE 0.5 0.5 0.5 black 0 0}
{{G 0.5 0 1} {HALF_CIRCLE}}
{CIRCLE 0.5 0.25 0.25 black 0 0}
{CIRCLE 0.5 0.75 0.25 white 0 0}
{CIRCLE 0.5 0.25 0.1 white 0 0}
{CIRCLE 0.5 0.75 0.1 black 0 0}
{CIRCLE 0.5 0.5 0.5 none gray 0.01} }}}
{def CIRCLE
{lambda {:x :y :r :f :s :w}
{circle {@ cx=":x" cy=":y" r=":r"
fill=":f" stroke=":s" stroke-width=":w"}}}}
{def HALF_CIRCLE
{path {@ d="M 0 0 A 0.5 0.5 0 0 0 0 1" fill="white"}}}
{def SVG
{lambda {:w :h}
svg {@ width=":w" height=":h"
style="box-shadow:0 0 8px #888;"}}}
{def G
{lambda {:x :y :s}
g {@ transform="translate(:x,:y) scale(:s,:s)"}}}
Output: Sorry, I was unable to upload the following PNG picture (45kb). Need help.
http://lambdaway.free.fr/lambdawalks/data/lambdatalk_yinyang.png
|
http://rosettacode.org/wiki/Y_combinator | Y combinator | In strict functional programming and the lambda calculus, functions (lambda expressions) don't have state and are only allowed to refer to arguments of enclosing functions.
This rules out the usual definition of a recursive function wherein a function is associated with the state of a variable and this variable's state is used in the body of the function.
The Y combinator is itself a stateless function that, when applied to another stateless function, returns a recursive version of the function.
The Y combinator is the simplest of the class of such functions, called fixed-point combinators.
Task
Define the stateless Y combinator and use it to compute factorials and Fibonacci numbers from other stateless functions or lambda expressions.
Cf
Jim Weirich: Adventures in Functional Programming
| #Elena | Elena | import extensions;
singleton YCombinator
{
fix(func)
= (f){(x){ x(x) }((g){ f((x){ (g(g))(x) })})}(func);
}
public program()
{
var fib := YCombinator.fix:(f => (i => (i <= 1) ? i : (f(i-1) + f(i-2)) ));
var fact := YCombinator.fix:(f => (i => (i == 0) ? 1 : (f(i-1) * i) ));
console.printLine("fib(10)=",fib(10));
console.printLine("fact(10)=",fact(10));
} |
http://rosettacode.org/wiki/Zig-zag_matrix | Zig-zag matrix | Task
Produce a zig-zag array.
A zig-zag array is a square arrangement of the first N2 natural numbers, where the
numbers increase sequentially as you zig-zag along the array's anti-diagonals.
For a graphical representation, see JPG zigzag (JPG uses such arrays to encode images).
For example, given 5, produce this array:
0 1 5 6 14
2 4 7 13 15
3 8 12 16 21
9 11 17 20 22
10 18 19 23 24
Related tasks
Spiral matrix
Identity matrix
Ulam spiral (for primes)
See also
Wiktionary entry: anti-diagonals
| #Fan | Fan | using gfx // for Point; convenient x/y wrapper
**
** A couple methods for generating a 'zigzag' array like
**
** 0 1 5 6
** 2 4 7 12
** 3 8 11 13
** 9 10 14 15
**
class ZigZag
{
** return an n x n array of uninitialized Int
static Int[][] makeSquareArray(Int n)
{
Int[][] grid := Int[][,] {it.size=n}
n.times |i| { grid[i] = Int[,] {it.size=n} }
return grid
}
Int[][] zig(Int n)
{
grid := makeSquareArray(n)
move := |Int i, Int j->Point|
{ return j < n - 1 ? Point(i <= 0 ? 0 : i-1, j+1) : Point(i+1, j) }
pt := Point(0,0)
(n*n).times |i| {
grid[pt.y][pt.x] = i
if ((pt.x+pt.y)%2 != 0) pt = move(pt.x,pt.y)
else {tmp:= move(pt.y,pt.x); pt = Point(tmp.y, tmp.x) }
}
return grid
}
public static Int[][] zag(Int size)
{
data := makeSquareArray(size)
Int i := 1
Int j := 1
for (element:=0; element < size * size; element++)
{
data[i - 1][j - 1] = element
if((i + j) % 2 == 0) {
// Even stripes
if (j < size) {
j++
} else {
i += 2
}
if (i > 1) {
i--
}
} else {
// Odd stripes
if (i < size) {
i++;
} else {
j += 2
}
if (j > 1) {
j--
}
}
}
return data;
}
Void print(Int[][] data)
{
data.each |row|
{
buf := StrBuf()
row.each |num|
{
buf.add(num.toStr.justr(3))
}
echo(buf)
}
}
Void main()
{
echo("zig method:")
print(zig(8))
echo("\nzag method:")
print(zag(8))
}
} |
http://rosettacode.org/wiki/Arbitrary-precision_integers_(included) | Arbitrary-precision integers (included) | Using the in-built capabilities of your language, calculate the integer value of:
5
4
3
2
{\displaystyle 5^{4^{3^{2}}}}
Confirm that the first and last twenty digits of the answer are:
62060698786608744707...92256259918212890625
Find and show the number of decimal digits in the answer.
Note: Do not submit an implementation of arbitrary precision arithmetic. The intention is to show the capabilities of the language as supplied. If a language has a single, overwhelming, library of varied modules that is endorsed by its home site – such as CPAN for Perl or Boost for C++ – then that may be used instead.
Strictly speaking, this should not be solved by fixed-precision numeric libraries where the precision has to be manually set to a large value; although if this is the only recourse then it may be used with a note explaining that the precision must be set manually to a large enough value.
Related tasks
Long multiplication
Exponentiation order
exponentiation operator
Exponentiation with infix operators in (or operating on) the base
| #Prolog | Prolog |
task(Length) :-
N is 5^4^3^2,
number_codes(N, Codes),
append(`62060698786608744707`, _, Codes),
append(_, `92256259918212890625`, Codes),
length(Codes, Length).
|
http://rosettacode.org/wiki/Arbitrary-precision_integers_(included) | Arbitrary-precision integers (included) | Using the in-built capabilities of your language, calculate the integer value of:
5
4
3
2
{\displaystyle 5^{4^{3^{2}}}}
Confirm that the first and last twenty digits of the answer are:
62060698786608744707...92256259918212890625
Find and show the number of decimal digits in the answer.
Note: Do not submit an implementation of arbitrary precision arithmetic. The intention is to show the capabilities of the language as supplied. If a language has a single, overwhelming, library of varied modules that is endorsed by its home site – such as CPAN for Perl or Boost for C++ – then that may be used instead.
Strictly speaking, this should not be solved by fixed-precision numeric libraries where the precision has to be manually set to a large value; although if this is the only recourse then it may be used with a note explaining that the precision must be set manually to a large enough value.
Related tasks
Long multiplication
Exponentiation order
exponentiation operator
Exponentiation with infix operators in (or operating on) the base
| #PureBasic | PureBasic | IncludeFile "Decimal.pbi"
;- Declare the variables that will be used
Define.Decimal *a
Define n, L$, R$, out$, digits.s
;- 4^3^2 is withing 32 bit range, so normal procedures can be used
n=Pow(4,Pow(3,2))
;- 5^n is larger then 31^2, so the same library call as in the "Long multiplication" task is used
*a=PowerDecimal(IntegerToDecimal(5),IntegerToDecimal(n))
;- Convert the large number into a string & present the results
out$=DecimalToString(*a)
L$ = Left(out$,20)
R$ = Right(out$,20)
digits=Str(Len(out$))
out$="First 20 & last 20 chars of 5^4^3^2 are;"+#CRLF$+L$+#CRLF$+R$+#CRLF$
out$+"and the result is "+digits+" digits long."
MessageRequester("Arbitrary-precision integers, PureBasic",out$) |
http://rosettacode.org/wiki/Zeckendorf_number_representation | Zeckendorf number representation | Just as numbers can be represented in a positional notation as sums of multiples of the powers of ten (decimal) or two (binary); all the positive integers can be represented as the sum of one or zero times the distinct members of the Fibonacci series.
Recall that the first six distinct Fibonacci numbers are: 1, 2, 3, 5, 8, 13.
The decimal number eleven can be written as 0*13 + 1*8 + 0*5 + 1*3 + 0*2 + 0*1 or 010100 in positional notation where the columns represent multiplication by a particular member of the sequence. Leading zeroes are dropped so that 11 decimal becomes 10100.
10100 is not the only way to make 11 from the Fibonacci numbers however; 0*13 + 1*8 + 0*5 + 0*3 + 1*2 + 1*1 or 010011 would also represent decimal 11. For a true Zeckendorf number there is the added restriction that no two consecutive Fibonacci numbers can be used which leads to the former unique solution.
Task
Generate and show here a table of the Zeckendorf number representations of the decimal numbers zero to twenty, in order.
The intention in this task to find the Zeckendorf form of an arbitrary integer. The Zeckendorf form can be iterated by some bit twiddling rather than calculating each value separately but leave that to another separate task.
Also see
OEIS A014417 for the the sequence of required results.
Brown's Criterion - Numberphile
Related task
Fibonacci sequence
| #Raku | Raku | printf "%2d: %8s\n", $_, zeckendorf($_) for 0 .. 20;
multi zeckendorf(0) { '0' }
multi zeckendorf($n is copy) {
constant FIBS = (1,2, *+* ... *).cache;
[~] map {
$n -= $_ if my $digit = $n >= $_;
+$digit;
}, reverse FIBS ...^ * > $n;
} |
http://rosettacode.org/wiki/100_doors | 100 doors | There are 100 doors in a row that are all initially closed.
You make 100 passes by the doors.
The first time through, visit every door and toggle the door (if the door is closed, open it; if it is open, close it).
The second time, only visit every 2nd door (door #2, #4, #6, ...), and toggle it.
The third time, visit every 3rd door (door #3, #6, #9, ...), etc, until you only visit the 100th door.
Task
Answer the question: what state are the doors in after the last pass? Which are open, which are closed?
Alternate:
As noted in this page's discussion page, the only doors that remain open are those whose numbers are perfect squares.
Opening only those doors is an optimization that may also be expressed;
however, as should be obvious, this defeats the intent of comparing implementations across programming languages.
| #Draco | Draco | proc nonrec main() void:
byte DOORS = 100;
[DOORS+1] bool door_open;
unsigned DOORS i, j;
/* make sure all doors are closed */
for i from 1 upto DOORS do door_open[i] := false od;
/* pass through the doors */
for i from 1 upto DOORS do
for j from i by i upto DOORS do
door_open[j] := not door_open[j]
od
od;
/* show the open doors */
for i from 1 upto DOORS do
if door_open[i] then
writeln("Door ", i, " is open.")
fi
od
corp |
http://rosettacode.org/wiki/Arrays | Arrays | This task is about arrays.
For hashes or associative arrays, please see Creating an Associative Array.
For a definition and in-depth discussion of what an array is, see Array.
Task
Show basic array syntax in your language.
Basically, create an array, assign a value to it, and retrieve an element (if available, show both fixed-length arrays and
dynamic arrays, pushing a value into it).
Please discuss at Village Pump: Arrays.
Please merge code in from these obsolete tasks:
Creating an Array
Assigning Values to an Array
Retrieving an Element of an Array
Related tasks
Collections
Creating an Associative Array
Two-dimensional array (runtime)
| #Icon_and_Unicon | Icon and Unicon | record aThing(a, b, c) # arbitrary object (record or class) for illustration
procedure main()
A0 := [] # empty list
A0 := list() # empty list (default size 0)
A0 := list(0) # empty list (literal size 0)
A1 := list(10) # 10 elements, default initializer &null
A2 := list(10, 1) # 10 elements, initialized to 1
# literal array construction - arbitrary dynamically typed members
A3 := [1, 2, 3, ["foo", "bar", "baz"], aThing(1, 2, 3), "the end"]
# left-end workers
# NOTE: get() is a synonym for pop() which allows nicely-worded use of put() and get() to implement queues
#
Q := [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
x := pop(A0) # x is 1
x := get(A0) # x is 2
push(Q,0)
# Q is now [0,3, 4, 5, 6, 7, 8, 9, 10]
# right-end workers
x := pull(Q) # x is 10
put(Q, 100) # Q is now [0, 3, 4, 5, 6, 7, 8, 9, 100]
# push and put return the list they are building
# they also can have multiple arguments which work like repeated calls
Q2 := put([],1,2,3) # Q2 is [1,2,3]
Q3 := push([],1,2,3) # Q3 is [3,2,1]
Q4 := push(put(Q2),4),0] # Q4 is [0,1,2,3,4] and so is Q2
# array access follows with A as the sample array
A := [10, 20, 30, 40, 50, 60, 70, 80, 90, 100]
# get element indexed from left
x := A[1] # x is 10
x := A[2] # x is 20
x := A[10] # x is 100
# get element indexed from right
x := A[-1] # x is 100
x := A[-2] # x is 90
x := A[-10] # x is 10
# copy array to show assignment to elements
B := copy(A)
# assign element indexed from left
B[1] := 11
B[2] := 21
B[10] := 101
# B is now [11, 21, 30, 50, 60, 60, 70, 80, 90, 101]
# assign element indexed from right - see below
B[-1] := 102
B[-2] := 92
B[-10] := 12
# B is now [12, 21, 30, 50, 60, 60, 70, 80, 92, 102]
# list slicing
# the unusual nature of the slice - returning 1 less element than might be expected
# in many languages - is best understood if you imagine indexes as pointing to BEFORE
# the item of interest. When a slice is made, the elements between the two points are
# collected. eg in the A[3 : 6] sample, it will get the elements between the [ ] marks
#
# sample list: 10 20 [30 40 50] 60 70 80 90 100
# positive indexes: 1 2 3 4 5 6 7 8 9 10 11
# non-positive indexes: -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0
#
# I have deliberately drawn the indexes between the positions of the values.
# The nature of this indexing brings simplicity to string operations
#
# list slicing can also use non-positive indexes to access values from the right.
# The final index of 0 shown above shows how the end of the list can be nominated
# without having to know it's length
#
# NOTE: list slices are distinct lists, so assigning to the slice
# or a member of the slice does not change the values in A
#
# Another key fact to understand: once the non-positive indexes and length-offsets are
# resolved to a simple positive index, the index pair (if two are given) are swapped
# if necessary to yield the elements between the two.
#
S := A[3 : 6] # S is [30, 40, 50]
S := A[6 : 3] # S is [30, 40, 50] not illegal or erroneous
S := A[-5 : -8] # S is [30, 40, 50]
S := A[-8 : -5] # S is [30, 40, 50] also legal and meaningful
# list slicing with length request
S := A[3 +: 3] # S is [30, 40, 50]
S := A[6 -: 3] # S is [30, 40, 50]
S := A[-8 +: 3] # S is [30, 40, 50]
S := A[-5 -: 3] # S is [30, 40, 50]
S := A[-8 -: -3] # S is [30, 40, 50]
S := A[-5 +: -3] # S is [30, 40, 50]
end |
http://rosettacode.org/wiki/Zero_to_the_zero_power | Zero to the zero power | Some computer programming languages are not exactly consistent (with other computer programming languages)
when raising zero to the zeroth power: 00
Task
Show the results of raising zero to the zeroth power.
If your computer language objects to 0**0 or 0^0 at compile time, you may also try something like:
x = 0
y = 0
z = x**y
say 'z=' z
Show the result here.
And of course use any symbols or notation that is supported in your computer programming language for exponentiation.
See also
The Wiki entry: Zero to the power of zero.
The Wiki entry: History of differing points of view.
The MathWorld™ entry: exponent laws.
Also, in the above MathWorld™ entry, see formula (9):
x
0
=
1
{\displaystyle x^{0}=1}
.
The OEIS entry: The special case of zero to the zeroth power
| #Sidef | Sidef | [0, Complex(0, 0)].each {|n|
say n**n
} |
http://rosettacode.org/wiki/Zero_to_the_zero_power | Zero to the zero power | Some computer programming languages are not exactly consistent (with other computer programming languages)
when raising zero to the zeroth power: 00
Task
Show the results of raising zero to the zeroth power.
If your computer language objects to 0**0 or 0^0 at compile time, you may also try something like:
x = 0
y = 0
z = x**y
say 'z=' z
Show the result here.
And of course use any symbols or notation that is supported in your computer programming language for exponentiation.
See also
The Wiki entry: Zero to the power of zero.
The Wiki entry: History of differing points of view.
The MathWorld™ entry: exponent laws.
Also, in the above MathWorld™ entry, see formula (9):
x
0
=
1
{\displaystyle x^{0}=1}
.
The OEIS entry: The special case of zero to the zeroth power
| #Sinclair_ZX81_BASIC | Sinclair ZX81 BASIC | PRINT 0**0 |
http://rosettacode.org/wiki/Zero_to_the_zero_power | Zero to the zero power | Some computer programming languages are not exactly consistent (with other computer programming languages)
when raising zero to the zeroth power: 00
Task
Show the results of raising zero to the zeroth power.
If your computer language objects to 0**0 or 0^0 at compile time, you may also try something like:
x = 0
y = 0
z = x**y
say 'z=' z
Show the result here.
And of course use any symbols or notation that is supported in your computer programming language for exponentiation.
See also
The Wiki entry: Zero to the power of zero.
The Wiki entry: History of differing points of view.
The MathWorld™ entry: exponent laws.
Also, in the above MathWorld™ entry, see formula (9):
x
0
=
1
{\displaystyle x^{0}=1}
.
The OEIS entry: The special case of zero to the zeroth power
| #Smalltalk | Smalltalk |
0 raisedTo: 0
0.0 raisedTo: 0.0
|
http://rosettacode.org/wiki/Zebra_puzzle | Zebra puzzle | Zebra puzzle
You are encouraged to solve this task according to the task description, using any language you may know.
The Zebra puzzle, a.k.a. Einstein's Riddle,
is a logic puzzle which is to be solved programmatically.
It has several variants, one of them this:
There are five houses.
The English man lives in the red house.
The Swede has a dog.
The Dane drinks tea.
The green house is immediately to the left of the white house.
They drink coffee in the green house.
The man who smokes Pall Mall has birds.
In the yellow house they smoke Dunhill.
In the middle house they drink milk.
The Norwegian lives in the first house.
The man who smokes Blend lives in the house next to the house with cats.
In a house next to the house where they have a horse, they smoke Dunhill.
The man who smokes Blue Master drinks beer.
The German smokes Prince.
The Norwegian lives next to the blue house.
They drink water in a house next to the house where they smoke Blend.
The question is, who owns the zebra?
Additionally, list the solution for all the houses.
Optionally, show the solution is unique.
Related tasks
Dinesman's multiple-dwelling problem
Twelve statements
| #jq | jq |
# Attempt to unify the input object with the specified object
def unify( object ):
# Attempt to unify the input object with the specified tag:value
def unify2(tag; value):
if . == null then null
elif .[tag] == value then .
elif .[tag] == null then .[tag] = value
else null
end;
reduce (object|keys[]) as $key
(.; unify2($key; object[$key]) );
# Input: an array
# Output: if the i-th element can be made to satisfy the condition,
# then the updated array, otherwise empty.
def enforce(i; cond):
if 0 <= i and i < length
then
(.[i] | cond) as $ans
| if $ans then .[i] = $ans else empty end
else empty
end ; |
Subsets and Splits
No saved queries yet
Save your SQL queries to embed, download, and access them later. Queries will appear here once saved.