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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>
| #Nim | Nim | import sequtils, strutils
const LibXml = "libxml2.so"
type
XmlDocPtr = pointer
XmlXPathContextPtr = pointer
XmlElementKind = enum
xmlElementNode = 1
xmlAttributeNode = 2
xmlTextNode = 3
xmlCdataSectionNode = 4
xmlEntityRefNode = 5
xmlEntityNode = 6
xmlPiNode = 7
xmlCommentNode = 8
xmlDocumentNode = 9
xmlDocumentTypeNode = 10
xmlDocumentFragNode = 11
xmlNotationNode = 12
xmlHtmlDocumentNode = 13
xmlDtdNode = 14
xmlElementDecl = 15
xmlAttributeDecl = 16
xmlEntityDecl = 17
xmlNamespaceDecl = 18
xmlXincludeStart = 19
xmlXincludeEnd = 20
XmlNsKind = XmlElementKind
XmlNsPtr = ptr XmlNs
XmlNs = object
next: XmlNsPtr
kind: XmlNsKind
href: cstring
prefix: cstring
private: pointer
context: XmlDocPtr
XmlAttrPtr = pointer
XmlNodePtr = ptr XmlNode
XmlNode = object
private: pointer
kind: XmlElementKind
name: cstring
children: XmlNodePtr
last: XmlNodePtr
parent: XmlNodePtr
next: XmlNodePtr
prev: XmlNodePtr
doc: XmlDocPtr
ns: XmlNsPtr
content: cstring
properties: XmlAttrPtr
nsDef: XmlNsPtr
psvi: pointer
line: cushort
extra: cushort
XmlNodeSetPtr = ptr XmlNodeSet
XmlNodeSet = object
nodeNr: cint
nodeMax: cint
nodeTab: ptr UncheckedArray[XmlNodePtr]
XmlPathObjectKind = enum
xpathUndefined
xpathNodeset
xpathBoolean
xpathNumber
xpathString
xpathPoint
xpathRange
xpathLocationset
xpathUsers
xpathXsltTree
XmlXPathObjectPtr = ptr XmlXPathObject
XmlXPathObject = object
kind: XmlPathObjectKind
nodeSetVal: XmlNodeSetPtr
boolVal: cint
floatVal: cdouble
stringVal: cstring
user: pointer
index: cint
user2: pointer
index2: cint
XmlSaveCtxtPtr = pointer
XmlBufferPtr = pointer
# Declaration of needed "libxml2" procedures.
proc xmlParseFile(docName: cstring): XmlDocPtr
{.cdecl, dynlib: LibXml, importc: "xmlParseFile".}
proc xmlXPathNewContext(doc: XmlDocPtr): XmlXPathContextPtr
{.cdecl, dynlib: LibXml, importc: "xmlXPathNewContext".}
proc xmlXPathEvalExpression(str: cstring; ctxt: XmlXPathContextPtr): XmlXPathObjectPtr
{.cdecl, dynlib: LibXml, importc: "xmlXPathEvalExpression".}
proc xmlXPathFreeContext(ctxt: XmlXPathContextPtr)
{.cdecl, dynlib: LibXml, importc: "xmlXPathFreeContext".}
proc xmlXPathFreeObject(obj: XmlXPathObjectPtr)
{.cdecl, dynlib: LibXml, importc: "xmlXPathFreeObject".}
proc xmlSaveToBuffer(vuffer: XmlBufferPtr; encoding: cstring; options: cint): XmlSaveCtxtPtr
{.cdecl, dynlib: LibXml, importc: "xmlSaveToBuffer".}
proc xmlBufferCreate(): XmlBufferPtr
{.cdecl, dynlib: LibXml, importc: "xmlBufferCreate".}
proc xmlBufferFree(buf: XmlBufferPtr)
{.cdecl, dynlib: LibXml, importc: "xmlBufferCreate".}
proc xmlBufferContent(buf: XmlBufferPtr): cstring
{.cdecl, dynlib: LibXml, importc: "xmlBufferContent".}
proc xmlSaveTree(ctxt: XmlSaveCtxtPtr; cur: XmlNodePtr): clong
{.cdecl, dynlib: LibXml, importc: "xmlSaveTree".}
proc xmlSaveClose(ctxt: XmlSaveCtxtPtr)
{.cdecl, dynlib: LibXml, importc: "xmlSaveClose".}
proc `$`(node: XmlNodePtr): string =
## Return the representation of a node.
let buffer = xmlBufferCreate()
let saveContext = xmlSaveToBuffer(buffer, nil, 0)
discard saveContext.xmlSaveTree(node)
saveContext.xmlSaveClose()
result = $buffer.xmlBufferContent()
xmlBufferFree(buffer)
iterator nodes(xpath: string; context: XmlXPathContextPtr): XmlNodePtr =
## Yield the nodes which fit the XPath request.
let xpathObj = xmlXPathEvalExpression(xpath, context)
if xpathObj.isNil:
quit "Failed to evaluate XPath: " & xpath, QuitFailure
assert xpathObj.kind == xpathNodeset
let nodeSet = xpathObj.nodeSetVal
if not nodeSet.isNil:
for i in 0..<nodeSet.nodeNr:
yield nodeSet.nodeTab[i]
xmlXPathFreeObject(xpathObj)
# Load and parse XML file.
let doc = xmlParseFile("xpath_test.xml")
if doc.isNil:
quit "Unable to load and parse document", QuitFailure
# Create an XPath context.
let context = xmlXPathNewContext(doc)
if context.isNil:
quit "Failed to create XPath context", QuitFailure
var xpath = "//section[1]/item[1]"
echo "Request $#:".format(xpath)
for node in nodes(xpath, context):
echo node
echo()
xpath = "//price/text()"
echo "Request $#:".format(xpath)
for node in nodes(xpath, context):
echo node.content
echo()
xpath = "//name"
echo "Request $#:".format(xpath)
let names = toSeq(nodes(xpath, context)).mapIt(it.children.content)
echo names
xmlXPathFreeContext(context) |
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>
| #Objeck | Objeck |
use XML;
bundle Default {
class Test {
function : Main(args : String[]) ~ Nil {
in := String->New();
in->Append("<inventory title=\"OmniCorp Store #45x10^3\">");
in->Append("<section name=\"health\">");
in->Append("<item upc=\"123456789\" stock=\"12\">");
in->Append("<name>Invisibility Cream</name>");
in->Append("<price>14.50</price>");
in->Append("<description>Makes you invisible</description>");
in->Append("</item>");
in->Append("<item upc=\"445322344\" stock=\"18\">");
in->Append("<name>Levitation Salve</name>");
in->Append("<price>23.99</price>");
in->Append("<description>Levitate yourself for up to 3 hours per application</description>");
in->Append("</item>");
in->Append("</section>");
in->Append("<section name=\"food\">");
in->Append("<item upc=\"485672034\" stock=\"653\">");
in->Append("<name>Blork and Freen Instameal</name>");
in->Append("<price>4.95</price>");
in->Append("<description>A tasty meal in a tablet; just add water</description>");
in->Append("</item>");
in->Append("<item upc=\"132957764\" stock=\"44\">");
in->Append("<name>Grob winglets</name>");
in->Append("<price>3.56</price>");
in->Append("<description>Tender winglets of Grob. Just add water</description>");
in->Append("</item>");
in->Append("</section>");
in->Append("</inventory>");
parser := XmlParser->New(in);
if(parser->Parse()) {
# get first item
results := parser->FindElements("//inventory/section[1]/item[1]");
if(results <> Nil) {
IO.Console->Instance()->Print("items: ")->PrintLine(results->Size());
};
# get all prices
results := parser->FindElements("//inventory/section/item/price");
if(results <> Nil) {
each(i : results) {
element := results->Get(i)->As(XMLElement);
element->GetContent()->PrintLine();
};
};
# get names
results := parser->FindElements("//inventory/section/item/name");
if(results <> Nil) {
IO.Console->Instance()->Print("names: ")->PrintLine(results->Size());
};
};
}
}
}
|
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.
| #Logo | Logo | to taijitu :r
; Draw a classic Taoist taijitu of the given radius centered on the current
; turtle position. The "eyes" are placed along the turtle's heading, the
; filled one in front, the open one behind.
; don't bother doing anything if the pen is not down
if not pendown? [stop]
; useful derivative values
localmake "r2 (ashift :r -1)
localmake "r4 (ashift :r2 -1)
localmake "r8 (ashift :r4 -1)
; remember where we started
localmake "start pos
; draw outer circle
pendown
arc 360 :r
; draw upper half of S
penup
forward :r2
pendown
arc 180 :r2
; and filled inner eye
arc 360 :r8
fill
; draw lower half of S
penup
back :r
pendown
arc -180 :r2
; other inner eye
arc 360 :r8
; fill this half of the symbol
penup
forward :r4
fill
; put the turtle back where it started
setpos :start
pendown
end
; demo code to produce image at right
clearscreen
pendown
hideturtle
taijitu 100
penup
forward 150
left 90
forward 150
pendown
taijitu 75 |
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.
| #Lua | Lua | function circle(x, y, c, r)
return (r * r) >= (x * x) / 4 + ((y - c) * (y - c))
end
function pixel(x, y, r)
if circle(x, y, -r / 2, r / 6) then
return '#'
end
if circle(x, y, r / 2, r / 6) then
return '.'
end
if circle(x, y, -r / 2, r / 2) then
return '.'
end
if circle(x, y, r / 2, r / 2) then
return '#'
end
if circle(x, y, 0, r) then
if x < 0 then
return '.'
else
return '#'
end
end
return ' '
end
function yinYang(r)
for y=-r,r do
for x=-2*r,2*r do
io.write(pixel(x, y, r))
end
print()
end
end
yinYang(18) |
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
| #Elixir | Elixir |
iex(1)> yc = fn f -> (fn x -> x.(x) end).(fn y -> f.(fn arg -> y.(y).(arg) end) end) end
#Function<6.90072148/1 in :erl_eval.expr/5>
iex(2)> fac = fn f -> fn n -> if n < 2 do 1 else n * f.(n-1) end end end
#Function<6.90072148/1 in :erl_eval.expr/5>
iex(3)> for i <- 0..9, do: yc.(fac).(i)
[1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880]
iex(4)> fib = fn f -> fn n -> if n == 0 do 0 else (if n == 1 do 1 else f.(n-1) + f.(n-2) end) end end end
#Function<6.90072148/1 in :erl_eval.expr/5>
iex(5)> for i <- 0..9, do: yc.(fib).(i)
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34]
|
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
| #Forth | Forth | 0 value diag
: south diag abs + cell+ ;
' cell+ value zig
' south value zag
: init ( n -- )
1- cells negate to diag
['] cell+ to zig
['] south to zag ;
: swap-diag zig zag to zig to zag ;
: put ( n addr -- n+1 addr )
2dup ! swap 1+ swap ;
: turn ( addr -- addr+E/S )
zig execute swap-diag
diag negate to diag ;
: zigzag ( matrix n -- )
{ n } n init
0 swap
n 1 ?do
put turn
i 0 do put diag + loop
loop
swap-diag
n 1 ?do
put turn
n i 1+ ?do put diag + loop
loop
! ;
: .matrix ( n matrix -- )
over 0 do
cr
over 0 do
dup @ 3 .r cell+
loop
loop 2drop ;
: test ( n -- ) here over zigzag here .matrix ;
5 test
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 ok |
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
| #Python | Python | >>> y = str( 5**4**3**2 )
>>> print ("5**4**3**2 = %s...%s and has %i digits" % (y[:20], y[-20:], len(y)))
5**4**3**2 = 62060698786608744707...92256259918212890625 and has 183231 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
| #Quackery | Quackery | > quackery
Welcome to Quackery.
Enter "leave" to leave the shell.
/O> 5 4 3 2 ** ** **
... number$ dup 20 split swap echo$
... say "..." -20 split echo$ drop cr
... size echo say " digits" cr
...
62060698786608744707...92256259918212890625
183231 digits
Stack empty.
/O> |
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
| #REXX | REXX |
/* REXX ***************************************************************
* 11.10.2012 Walter Pachl
**********************************************************************/
fib='13 8 5 3 2 1'
Do i=6 To 1 By -1 /* Prepare Fibonacci Numbers */
Parse Var fib f.i fib /* f.1 ... f.7 */
End
Do n=0 To 20 /* for all numbers in the task */
m=n /* copy of number */
r='' /* result for n */
Do i=6 To 1 By -1 /* loop through numbers */
If m>=f.i Then Do /* f.i must be used */
r=r||1 /* 1 into result */
m=m-f.i /* subtract */
End
Else /* f.i is larger than the rest */
r=r||0 /* 0 into result */
End
r=strip(r,'L','0') /* strip leading zeros */
If r='' Then r='0' /* take care of 0 */
Say right(n,2)': 'right(r,6) /* show result */
End |
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.
| #DUP | DUP | 100[$][0^:1-]# {initialize doors}
%
[s;[$101<][$$;~\:s;+]#%]d: {function d, switch door state function}
1s:[s;101<][d;!s;1+s:]# {increment step width from 1 to 100, execute function d each time}
1[$101<][$$.' ,;['o,'p,'e,'n,10,]['c,'l,'o,'s,'e,'d,10,]?1+]# {loop through doors, print door number and state} |
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)
| #i | i | main
//Fixed-length arrays.
f $= array.integer[1]()
f[0] $= 2
print(f[0])
//Dynamic arrays.
d $= list.integer()
d[+] $= 2
print(d[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
| #smart_BASIC | smart 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
| #SNOBOL4 | SNOBOL4 | OUTPUT = (0 ** 0)
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
| #SQL | SQL |
SQL> SELECT POWER(0,0) FROM dual;
|
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
| #Julia | Julia |
# Julia 1.0
using Combinatorics
function make(str, test )
filter(test, collect( permutations(split(str))) )
end
men = make("danish english german norwegian swedish",
x -> "norwegian" == x[1])
drinks = make("beer coffee milk tea water", x -> "milk" == x[3])
#Julia 1.0 compatible
colors = make("blue green red white yellow",
x -> 1 == findfirst(c -> c == "white", x) - findfirst(c -> c == "green",x))
pets = make("birds cats dog horse zebra")
smokes = make("blend blue-master dunhill pall-mall prince")
function eq(x, xs, y, ys)
findfirst(xs, x) == findfirst(ys, y)
end
function adj(x, xs, y, ys)
1 == abs(findfirst(xs, x) - findfirst(ys, y))
end
function print_houses(n, pet, nationality, colors, drink, smokes)
println("$n, $pet, $nationality $colors $drink $smokes")
end
for m = men, c = colors
if eq("red",c, "english",m) && adj("norwegian",m, "blue",c)
for d = drinks
if eq("danish",m, "tea",d) && eq("coffee",d,"green",c)
for s = smokes
if eq("yellow",c,"dunhill",s) &&
eq("blue-master",s,"beer",d) &&
eq("german",m,"prince",s)
for p = pets
if eq("birds",p,"pall-mall",s) &&
eq("swedish",m,"dog",p) &&
adj("blend",s,"cats",p) &&
adj("horse",p,"dunhill",s)
println("Zebra is owned by ", m[findfirst(c -> c == "zebra", p)])
println("Houses:")
for house_num in 1:5
print_houses(house_num,p[house_num],m[house_num],c[house_num],d[house_num],s[house_num])
end
end
end
end
end
end
end
end
end
|
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>
| #Oz | Oz | declare
[XMLParser] = {Module.link ['x-oz://system/xml/Parser.ozf']}
proc {Main Data}
Parser = {New XMLParser.parser init}
[Doc] = {Parser parseVS(Data $)}
FirstItem = {XPath Doc [inventory section item]}.1
Prices = {XPath Doc [inventory section item price Text]}
Names = {XPath Doc [inventory section item name]}
in
{ForAll Prices System.showInfo}
end
%%
%% Emulation of some XPath functionality:
%%
fun {XPath Doc Path}
P|Pr = Path
in
Doc.name = P %% assert
{FoldL Pr XPathStep [Doc]}
end
fun {XPathStep Elements P}
if {IsProcedure P} then
{Map Elements P}
else
{FilteredChildren Elements P}
end
end
%% A flat list of all Type-children of all Elements.
fun {FilteredChildren Elements Type}
{Flatten
{Map Elements
fun {$ E}
{Filter E.children
fun {$ X}
case X of element(name:!Type ...) then true
else false
end
end}
end}}
end
%% PCDATA of an element as a ByteString
fun {Text Element}
Texts = for Child in Element.children collect:C do
case Child of text(data:BS ...) then {C BS} end
end
in
{FoldR Texts ByteString.append {ByteString.make nil}}
end
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>"
in
{Main Data} |
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.
| #Maple | Maple |
with(plottools):
with(plots):
yingyang := r -> display(
circle([0, 0], r),
disk([0, 1/2*r], 1/10*r, colour = black),
disk([0, -1/2*r], 1/10*r, colour = white),
disk([0, -1/2*r], 1/2*r, colour = black),
inequal({1/4*r^2 <= x^2 + (y - 1/2*r)^2, 1/4*r^2 <= x^2 + (y + 1/2*r)^2, x^2 + y^2 <=
r^2}, x = 0 .. r, y = -r .. r, grid = [100, 100], colour = black),
scaling = constrained, axes = none
);
|
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
| #Elm | Elm | module Main exposing ( main )
import Html exposing ( Html, text )
-- As with most of the strict (non-deferred or non-lazy) languages,
-- this is the Z-combinator with the additional value parameter...
-- wrap type conversion to avoid recursive type definition...
type Mu a b = Roll (Mu a b -> a -> b)
unroll : Mu a b -> (Mu a b -> a -> b) -- unwrap it...
unroll (Roll x) = x
-- note lack of beta reduction using values...
fixz : ((a -> b) -> (a -> b)) -> (a -> b)
fixz f = let g r = f (\ v -> unroll r r v) in g (Roll g)
facz : Int -> Int
-- facz = fixz <| \ f n -> if n < 2 then 1 else n * f (n - 1) -- inefficient recursion
facz = fixz (\ f n i -> if i < 2 then n else f (i * n) (i - 1)) 1 -- efficient tailcall
fibz : Int -> Int
-- fibz = fixz <| \ f n -> if n < 2 then n else f (n - 1) + f (n - 2) -- inefficient recursion
fibz = fixz (\ fn f s i -> if i < 2 then f else fn s (f + s) (i - 1)) 1 1 -- efficient tailcall
-- by injecting laziness, we can get the true Y-combinator...
-- as this includes laziness, there is no need for the type wrapper!
fixy : ((() -> a) -> a) -> a
fixy f = f <| \ () -> fixy f -- direct function recursion
-- the above is not value recursion but function recursion!
-- fixv f = let x = f x in x -- not allowed by task or by Elm!
-- we can make Elm allow it by injecting laziness...
-- fixv f = let x = f () x in x -- but now value recursion not function recursion
facy : Int -> Int
-- facy = fixy <| \ f n -> if n < 2 then 1 else n * f () (n - 1) -- inefficient recursion
facy = fixy (\ f n i -> if i < 2 then n else f () (i * n) (i - 1)) 1 -- efficient tailcall
fiby : Int -> Int
-- fiby = fixy <| \ f n -> if n < 2 then n else f () (n - 1) + f (n - 2) -- inefficient recursion
fiby = fixy (\ fn f s i -> if i < 2 then f else fn () s (f + s) (i - 1)) 1 1 -- efficient tailcall
-- something that can be done with a true Y-Combinator that
-- can't be done with the Z combinator...
-- given an infinite Co-Inductive Stream (CIS) defined as...
type CIS a = CIS a (() -> CIS a) -- infinite lazy stream!
mapCIS : (a -> b) -> CIS a -> CIS b -- uses function to map
mapCIS cf cis =
let mp (CIS head restf) = CIS (cf head) <| \ () -> mp (restf()) in mp cis
-- now we can define a Fibonacci stream as follows...
fibs : () -> CIS Int
fibs() = -- two recursive fix's, second already lazy...
let fibsgen = fixy (\ fn (CIS (f, s) restf) ->
CIS (s, f + s) (\ () -> fn () (restf())))
in fixy (\ cisthnk -> fibsgen (CIS (0, 1) cisthnk))
|> mapCIS (\ (v, _) -> v)
nCISs2String : Int -> CIS a -> String -- convert n CIS's to String
nCISs2String n cis =
let loop i (CIS head restf) rslt =
if i <= 0 then rslt ++ " )" else
loop (i - 1) (restf()) (rslt ++ " " ++ Debug.toString head)
in loop n cis "("
-- unfortunately, if we need CIS memoization so as
-- to make a true lazy list, Elm doesn't support it!!!
main : Html Never
main =
String.fromInt (facz 10) ++ " " ++ String.fromInt (fibz 10)
++ " " ++ String.fromInt (facy 10) ++ " " ++ String.fromInt (fiby 10)
++ " " ++ nCISs2String 20 (fibs())
|> text |
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
| #Fortran | Fortran | PROGRAM ZIGZAG
IMPLICIT NONE
INTEGER, PARAMETER :: size = 5
INTEGER :: zzarray(size,size), x(size*size), y(size*size), i, j
! index arrays
x = (/ ((j, i = 1, size), j = 1, size) /)
y = (/ ((i, i = 1, size), j = 1, size) /)
! Sort indices
DO i = 2, size*size
j = i - 1
DO WHILE (j>=1 .AND. (x(j)+y(j)) > (x(i)+y(i)))
j = j - 1
END DO
x(j+1:i) = cshift(x(j+1:i),-1)
y(j+1:i) = cshift(y(j+1:i),-1)
END DO
! Create zig zag array
DO i = 1, size*size
IF (MOD(x(i)+y(i), 2) == 0) THEN
zzarray(x(i),y(i)) = i - 1
ELSE
zzarray(y(i),x(i)) = i - 1
END IF
END DO
! Print zig zag array
DO j = 1, size
DO i = 1, size
WRITE(*, "(I5)", ADVANCE="NO") zzarray(i,j)
END DO
WRITE(*,*)
END DO
END PROGRAM ZIGZAG |
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
| #R | R | library(gmp)
large <- pow.bigz(5, pow.bigz(4, pow.bigz(3, 2)))
largestr <- as.character(large)
cat("first 20 digits:", substr(largestr, 1, 20), "\n",
"last 20 digits:", substr(largestr, nchar(largestr) - 19, nchar(largestr)), "\n",
"number of digits: ", nchar(largestr), "\n") |
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
| #Racket | Racket | #lang racket
(define answer (number->string (foldr expt 1 '(5 4 3 2))))
(define len (string-length answer))
(printf "Got ~a digits~n" len)
(printf "~a ... ~a~n"
(substring answer 0 20)
(substring answer (- len 20) len))
|
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
| #Ring | Ring |
# Project : Zeckendorf number representation
see "0 0" + nl
for n = 1 to 20
see "" + n + " " + zeckendorf(n) + nl
next
func zeckendorf(n)
fib = list(45)
fib[1] = 1
fib[2] = 1
i = 2
o = ""
while fib[i] <= n
i = i + 1
fib[i] = fib[i-1] + fib[i-2]
end
while i != 2
i = i - 1
if n >= fib[i]
o = o + "1"
n = n - fib[i]
else
o = o + "0"
ok
end
return o
|
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
| #Ruby | Ruby | def zeckendorf
return to_enum(__method__) unless block_given?
x = 0
loop do
bin = x.to_s(2)
yield bin unless bin.include?("11")
x += 1
end
end
zeckendorf.take(21).each_with_index{|x,i| puts "%3d: %8s"% [i, x]} |
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.
| #DWScript | DWScript | var doors : array [1..100] of Boolean;
var i, j : Integer;
for i := 1 to 100 do
for j := i to 100 do
if (j mod i) = 0 then
doors[j] := not doors[j];F
for i := 1 to 100 do
if doors[i] then
PrintLn('Door '+IntToStr(i)+' is 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)
| #Io | Io | foo := list("foo", "bar", "baz")
foo at(1) println // bar
foo append("Foobarbaz")
foo println
foo atPut(2, "barbaz") // baz becomes barbaz |
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
| #Standard_ML | Standard ML | - Math.pow (0.0, 0.0);
val it = 1.0 : real
|
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
| #Stata | Stata | . display 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
| #Swift | Swift | import Darwin
print(pow(0.0,0.0)) |
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
| #Kotlin | Kotlin | // version 1.1.3
fun nextPerm(perm: IntArray): Boolean {
val size = perm.size
var k = -1
for (i in size - 2 downTo 0) {
if (perm[i] < perm[i + 1]) {
k = i
break
}
}
if (k == -1) return false // last permutation
for (l in size - 1 downTo k) {
if (perm[k] < perm[l]) {
val temp = perm[k]
perm[k] = perm[l]
perm[l] = temp
var m = k + 1
var n = size - 1
while (m < n) {
val temp2 = perm[m]
perm[m++] = perm[n]
perm[n--] = temp2
}
break
}
}
return true
}
fun check(a1: Int, a2: Int, v1: Int, v2: Int): Boolean {
for (i in 0..4)
if (p[a1][i] == v1) return p[a2][i] == v2
return false
}
fun checkLeft(a1: Int, a2: Int, v1: Int, v2: Int): Boolean {
for (i in 0..3)
if (p[a1][i] == v1) return p[a2][i + 1] == v2
return false
}
fun checkRight(a1: Int, a2: Int, v1: Int, v2: Int): Boolean {
for (i in 1..4)
if (p[a1][i] == v1) return p[a2][i - 1] == v2
return false
}
fun checkAdjacent(a1: Int, a2: Int, v1: Int, v2: Int): Boolean {
return checkLeft(a1, a2, v1, v2) || checkRight(a1, a2, v1, v2)
}
val colors = listOf("Red", "Green", "White", "Yellow", "Blue")
val nations = listOf("English", "Swede", "Danish", "Norwegian", "German")
val animals = listOf("Dog", "Birds", "Cats", "Horse", "Zebra")
val drinks = listOf("Tea", "Coffee", "Milk", "Beer", "Water")
val smokes = listOf("Pall Mall", "Dunhill", "Blend", "Blue Master", "Prince")
val p = Array(120) { IntArray(5) { -1 } } // stores all permutations of numbers 0..4
fun fillHouses(): Int {
var solutions = 0
for (c in 0..119) {
if (!checkLeft(c, c, 1, 2)) continue // C5 : Green left of white
for (n in 0..119) {
if (p[n][0] != 3) continue // C10: Norwegian in First
if (!check(n, c, 0, 0)) continue // C2 : English in Red
if (!checkAdjacent(n, c, 3, 4)) continue // C15: Norwegian next to Blue
for (a in 0..119) {
if (!check(a, n, 0, 1)) continue // C3 : Swede has Dog
for (d in 0..119) {
if (p[d][2] != 2) continue // C9 : Middle drinks Milk
if (!check(d, n, 0, 2)) continue // C4 : Dane drinks Tea
if (!check(d, c, 1, 1)) continue // C6 : Green drinks Coffee
for (s in 0..119) {
if (!check(s, a, 0, 1)) continue // C7 : Pall Mall has Birds
if (!check(s, c, 1, 3)) continue // C8 : Yellow smokes Dunhill
if (!check(s, d, 3, 3)) continue // C13: Blue Master drinks Beer
if (!check(s, n, 4, 4)) continue // C14: German smokes Prince
if (!checkAdjacent(s, a, 2, 2)) continue // C11: Blend next to Cats
if (!checkAdjacent(s, a, 1, 3)) continue // C12: Dunhill next to Horse
if (!checkAdjacent(s, d, 2, 4)) continue // C16: Blend next to Water
solutions++
printHouses(c, n, a, d, s)
}
}
}
}
}
return solutions
}
fun printHouses(c: Int, n: Int, a: Int, d: Int, s: Int) {
var owner: String = ""
println("House Color Nation Animal Drink Smokes")
println("===== ====== ========= ====== ====== ===========")
for (i in 0..4) {
val f = "%3d %-6s %-9s %-6s %-6s %-11s\n"
System.out.printf(f, i + 1, colors[p[c][i]], nations[p[n][i]], animals[p[a][i]], drinks[p[d][i]], smokes[p[s][i]])
if (animals[p[a][i]] == "Zebra") owner = nations[p[n][i]]
}
println("\nThe $owner owns the Zebra\n")
}
fun main(args: Array<String>) {
val perm = IntArray(5) { it }
for (i in 0..119) {
for (j in 0..4) p[i][j] = perm[j]
nextPerm(perm)
}
val solutions = fillHouses()
val plural = if (solutions == 1) "" else "s"
println("$solutions solution$plural found")
} |
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>
| #Perl | Perl | use XML::XPath qw();
my $x = XML::XPath->new('<inventory ... </inventory>');
[$x->findnodes('//item[1]')->get_nodelist]->[0];
print $x->findnodes_as_string('//price');
$x->findnodes('//name')->get_nodelist; |
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>
| #Phix | Phix | include builtins/xml.e
constant xml_txt = """
<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>
""",
-- or, of course, xml_txt = get_text("input.xml")
xml = xml_parse(xml_txt)
sequence sections = xml_get_nodes(xml[XML_CONTENTS],"section"),
item1 = {},
prices = {},
names = {}
for s=1 to length(sections) do
sequence items = xml_get_nodes(sections[s],"item")
if item1={} then item1 = items[1] end if
for i=1 to length(items) do
prices = append(prices,xml_get_nodes(items[i],"price")[1][XML_CONTENTS])
names = append(names,xml_get_nodes(items[i],"name")[1][XML_CONTENTS])
end for
end for
puts(1,"===item[1]===\n")
sequence tmp = xml_new_doc(item1)
puts(1,xml_sprint(tmp))
puts(1,"===prices===\n")
pp(prices)
puts(1,"===names===\n")
pp(names,{pp_Maxlen,90})
|
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.
| #Mathematica_.2F_Wolfram_Language | Mathematica / Wolfram Language | YinYang[size_] :=
Graphics[{{Circle[{0, 0}, 2]}, {Disk[{0, 0},
2, {90 Degree, -90 Degree}]}, {White, Disk[{0, 1}, 1]}, {Black,
Disk[{0, -1}, 1]}, {Black, Disk[{0, 1}, 1/4]}, {White,
Disk[{0, -1}, 1/4]}}, ImageSize -> 40 size] |
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.
| #Metapost | Metapost | vardef yinyang(expr u) =
picture pic_;
path p_;
p_ := halfcircle scaled 2u rotated -90 --
halfcircle scaled u rotated 90 shifted (0, 1/2u) reflectedabout ((0,1), (0,-1)) --
halfcircle scaled u rotated -270 shifted (0, -1/2u) -- cycle;
pic_ := nullpicture;
addto pic_ contour fullcircle scaled 2u withcolor black;
addto pic_ contour p_ withcolor white;
addto pic_ doublepath p_ withcolor black withpen pencircle scaled 0.5mm;
addto pic_ contour fullcircle scaled 1/3u shifted (0, 1/2u) withcolor white;
addto pic_ contour fullcircle scaled 1/3u shifted (0, -1/2u) withcolor black;
pic_
enddef;
beginfig(1)
% let's create a Yin Yang symbol with a radius of 5cm
draw yinyang(5cm) shifted (5cm, 5cm);
% and another one, radius 2.5cm, rotated 180 degrees and translated
draw yinyang(2.5cm) rotated 180 shifted (11cm, 11cm);
endfig;
end. |
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
| #Erlang | Erlang | Y = fun(M) -> (fun(X) -> X(X) end)(fun (F) -> M(fun(A) -> (F(F))(A) end) end) end.
Fac = fun (F) ->
fun (0) -> 1;
(N) -> N * F(N-1)
end
end.
Fib = fun(F) ->
fun(0) -> 0;
(1) -> 1;
(N) -> F(N-1) + F(N-2)
end
end.
(Y(Fac))(5). %% 120
(Y(Fib))(8). %% 21 |
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
| #FreeBASIC | FreeBASIC | ' FB 1.05.0 Win64
Dim As Integer n
Do
Input "Enter size of matrix "; n
Loop Until n > 0
Dim zigzag(1 To n, 1 To n) As Integer '' all zero by default
' enter the numbers 0 to (n^2 - 1) in the matrix's anti-diagonals
zigzag(1, 1) = 0
If n > 1 Then
Dim As Integer row = 0, col = 3
Dim As Boolean down = true, increment = true
Dim As Integer i = 0, j = 2, k
Do
If down Then
For k = 1 To j
i += 1
row += 1
col -= 1
zigzag(row, col) = i
Next
down = false
Else
For k = 1 To j
i += 1
row -= 1
col += 1
zigzag(row, col) = i
Next
down = true
End If
If increment Then
j += 1
If j > n Then
j = n - 1
increment = false
End If
Else
j -= 1
If j = 0 Then Exit Do
End If
If down AndAlso increment Then
col += 2
row -= 1
ElseIf Not Down AndAlso increment Then
row += 2
col -= 1
ElseIf down AndAlso Not increment Then
col += 1
Else '' Not down AndAlso NotIncrement
row += 1
End If
Loop
End If
' print zigzag matrix if n < 20
Print
If n < 20 Then
For i As Integer = 1 To n
For j As Integer = 1 To n
Print Using "####"; zigzag(i, j);
Next j
Print
Next i
Else
Print "Matrix is too big to display on 80 column console"
End If
Print
Print "Press any key to quit"
Sleep |
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
| #Raku | Raku | given ~[**] 5, 4, 3, 2 {
say "5**4**3**2 = {.substr: 0,20}...{.substr: *-20} and has {.chars} 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
| #REXX | REXX | /*REXX program calculates and demonstrates arbitrary precision numbers (using powers). */
numeric digits 200000 /*two hundred thousand decimal digits. */
# = 5 ** (4 ** (3 ** 2) ) /*calculate multiple exponentiations. */
true=62060698786608744707...92256259918212890625 /*what answer is supposed to look like.*/
rexx= left(#, 20)'...'right(#, 20) /*the left and right 20 decimal digits.*/
say ' true:' true /*show what the "true" answer is. */
say ' REXX:' rexx /* " " " REXX " " */
say 'digits:' length(#) /* " " " length of answer is. */
say
if true == rexx then say 'passed!' /*either it passed, ··· */
else say 'failed!' /* or it didn't. */
/*stick a fork in it, we're all done. */ |
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
| #Scala | Scala | def zNum( n:BigInt ) : String = {
if( n == 0 ) return "0" // Short-circuit this and return zero if we were given zero
val v = n.abs
val fibs : Stream[BigInt] = { def series(i:BigInt,j:BigInt):Stream[BigInt] = i #:: series(j, i+j); series(1,0).tail.tail.tail }
def z( v:BigInt ) : List[BigInt] = if(v == 0) List() else {val m = fibs(fibs.indexWhere(_>v) - 1); m :: z(v-m)}
val zv = z(v)
// Walk the list of fibonacci numbers from the number that matches the most significant down to 1,
// if the zeckendorf matchs then yield '1' otherwise '0'
val s = (for( i <- (fibs.indexWhere(_==zv(0)) to 0 by -1) ) yield {
if( zv.contains(fibs(i))) "1" else "0"
}).mkString
if( n < 0 ) "-" + s // Using a negative-sign instead of twos-complement
else s
}
// A little test...
(0 to 20) foreach( i => print( zNum(i) + "\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.
| #Dyalect | Dyalect | var doors = Array.Empty(100, false)
for p in 0..99 {
for d in 0..99 {
if (d + 1) % (p + 1) == 0 {
doors[d] = !doors[d];
}
}
}
for d in doors.Indices() when doors[d] {
print("Door \(d+1): 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)
| #J | J | 1 NB. a stand-alone scalar value is an array without any axis
1
NB. invoking any array produces that array as the result
{. array=: 1 3, 6#0 NB. create, name, then get head item of the array: 1 3 0 0 0 0 0 0
1
0 { array NB. another way to get the head item
1
aword=: 'there' NB. a literal array
0 1 3 2 2 { aword NB. multiple items can be drawn in a single action
three
]twoD=: 3 5 $ 'abcdefghijklmnopqrstuvwxyz'
abcde
fghij
klmno
1 { twoD NB. item 1 from twoD - a list of three items
fghij
1 {"1 twoD NB. item 1 from each rank-1 item of twoD (i.e. column 1)
bgl
(<2 2){ twoD NB. bracket indexing is not used in J
m
'X' 1} aword NB. amend item 1
tXere
aword=: 'X' 1 4} aword NB. in-place amend of items 1 and 4
tXerX
'X' (0 0;1 1;2 2)} twoD NB. amend specified items
Xbcde
fXhij
klXno |
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
| #Symsyn | Symsyn |
(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
| #Tcl | Tcl | % expr 0**0
1
% expr 0.0**0.0
1.0 |
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
| #Logtalk | Logtalk |
/* Houses logical puzzle: who owns the zebra and who drinks water?
1) Five colored houses in a row, each with an owner, a pet, cigarettes, and a drink.
2) The English lives in the red house.
3) The Spanish has a dog.
4) They drink coffee in the green house.
5) The Ukrainian drinks tea.
6) The green house is next to the white house.
7) The Winston smoker has a serpent.
8) In the yellow house they smoke Kool.
9) In the middle house they drink milk.
10) The Norwegian lives in the first house from the left.
11) The Chesterfield smoker lives near the man with the fox.
12) In the house near the house with the horse they smoke Kool.
13) The Lucky Strike smoker drinks juice.
14) The Japanese smokes Kent.
15) The Norwegian lives near the blue house.
Who owns the zebra and who drinks water?
*/
:- object(houses).
:- public(houses/1).
:- mode(houses(-list), one).
:- info(houses/1, [
comment is 'Solution to the puzzle.',
argnames is ['Solution']
]).
:- public(print/1).
:- mode(print(+list), one).
:- info(print/1, [
comment is 'Pretty print solution to the puzzle.',
argnames is ['Solution']
]).
houses(Solution) :-
template(Solution), % 1
member(h(english, _, _, _, red), Solution), % 2
member(h(spanish, dog, _, _, _), Solution), % 3
member(h(_, _, _, coffee, green), Solution), % 4
member(h(ukrainian, _, _, tea, _), Solution), % 5
next(h(_, _, _, _, green), h(_, _, _, _, white), Solution), % 6
member(h(_, snake, winston, _, _), Solution), % 7
member(h(_, _, kool, _, yellow), Solution), % 8
Solution = [_, _, h(_, _, _, milk, _), _, _], % 9
Solution = [h(norwegian, _, _, _, _)| _], % 10
next(h(_, fox, _, _, _), h(_, _, chesterfield, _, _), Solution), % 11
next(h(_, _, kool, _, _), h(_, horse, _, _, _), Solution), % 12
member(h(_, _, lucky, juice, _), Solution), % 13
member(h(japonese, _, kent, _, _), Solution), % 14
next(h(norwegian, _, _, _, _), h(_, _, _, _, blue), Solution), % 15
member(h(_, _, _, water, _), Solution), % one of them drinks water
member(h(_, zebra, _, _, _), Solution). % one of them owns a zebra
print([]).
print([House| Houses]) :-
write(House), nl,
print(Houses).
% h(Nationality, Pet, Cigarette, Drink, Color)
template([h(_, _, _, _, _), h(_, _, _, _, _), h(_, _, _, _, _), h(_, _, _, _, _), h(_, _, _, _, _)]).
member(A, [A, _, _, _, _]).
member(B, [_, B, _, _, _]).
member(C, [_, _, C, _, _]).
member(D, [_, _, _, D, _]).
member(E, [_, _, _, _, E]).
next(A, B, [A, B, _, _, _]).
next(B, C, [_, B, C, _, _]).
next(C, D, [_, _, C, D, _]).
next(D, E, [_, _, _, D, E]).
:- end_object.
|
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>
| #PHP | PHP | <?php
//PHP5 only example due to changes in XML extensions between version 4 and 5 (Tested on PHP5.2.0)
$doc = DOMDocument::loadXML('<inventory title="OmniCorp Store #45x10^3">...</inventory>');
//Load from file instead with $doc = DOMDocument::load('filename');
$xpath = new DOMXPath($doc);
/*
1st Task: Retrieve the first "item" element
*/
$nodelist = $xpath->query('//item');
$result = $nodelist->item(0);
/*
2nd task: Perform an action on each "price" element (print it out)
*/
$nodelist = $xpath->query('//price');
for($i = 0; $i < $nodelist->length; $i++)
{
//print each price element in the DOMNodeList instance, $nodelist, as text/xml followed by a newline
print $doc->saveXML($nodelist->item($i))."\n";
}
/*
3rd Task: Get an array of all the "name" elements
*/
$nodelist = $xpath->query('//name');
//our array to hold all the name elements, though in practice you'd probably not need to do this and simply use the DOMNodeList
$result = array();
//a different way of iterating through the DOMNodeList
foreach($nodelist as $node)
{
$result[] = $node;
} |
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.
| #Modula-2 | Modula-2 | MODULE Taijitu;
FROM InOut IMPORT Write, WriteLn;
PROCEDURE YinYang(r: INTEGER);
VAR x, y: INTEGER;
PROCEDURE circle(x, y, c, r: INTEGER): BOOLEAN;
BEGIN
RETURN r*r >= (x DIV 2) * (x DIV 2) + (y-c) * (y-c);
END circle;
PROCEDURE pixel(x, y, r: INTEGER): CHAR;
BEGIN
IF circle(x, y, -r DIV 2, r DIV 6) THEN RETURN '#';
ELSIF circle(x, y, r DIV 2, r DIV 6) THEN RETURN '.';
ELSIF circle(x, y, -r DIV 2, r DIV 2) THEN RETURN '.';
ELSIF circle(x, y, r DIV 2, r DIV 2) THEN RETURN '#';
ELSIF circle(x, y, 0, r) THEN
IF x<0 THEN RETURN '.';
ELSE RETURN '#';
END;
ELSE RETURN ' ';
END;
END pixel;
BEGIN
FOR y := -r TO r DO
FOR x := -2*r TO 2*r DO
Write(pixel(x,y,r));
END;
WriteLn();
END;
END YinYang;
BEGIN
YinYang(4);
WriteLn();
YinYang(8);
END Taijitu. |
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
| #F.23 | F# | type 'a mu = Roll of ('a mu -> 'a) // ' fixes ease syntax colouring confusion with
let unroll (Roll x) = x
// val unroll : 'a mu -> ('a mu -> 'a)
// As with most of the strict (non-deferred or non-lazy) languages,
// this is the Z-combinator with the additional 'a' parameter...
let fix f = let g = fun x a -> f (unroll x x) a in g (Roll g)
// val fix : (('a -> 'b) -> 'a -> 'b) -> 'a -> 'b = <fun>
// Although true to the factorial definition, the
// recursive call is not in tail call position, so can't be optimized
// and will overflow the call stack for the recursive calls for large ranges...
//let fac = fix (fun f n -> if n < 2 then 1I else bigint n * f (n - 1))
// val fac : (int -> BigInteger) = <fun>
// much better progressive calculation in tail call position...
let fac = fix (fun f n i -> if i < 2 then n else f (bigint i * n) (i - 1)) <| 1I
// val fac : (int -> BigInteger) = <fun>
// Although true to the definition of Fibonacci numbers,
// this can't be tail call optimized and recursively repeats calculations
// for a horrendously inefficient exponential performance fib function...
// let fib = fix (fun fnc n -> if n < 2 then n else fnc (n - 1) + fnc (n - 2))
// val fib : (int -> BigInteger) = <fun>
// much better progressive calculation in tail call position...
let fib = fix (fun fnc f s i -> if i < 2 then f else fnc s (f + s) (i - 1)) 1I 1I
// val fib : (int -> BigInteger) = <fun>
[<EntryPoint>]
let main argv =
fac 10 |> printfn "%A" // prints 3628800
fib 10 |> printfn "%A" // prints 55
0 // return an integer exit code |
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
| #GAP | GAP | ZigZag := function(n)
local a, i, j, k;
a := NullMat(n, n);
i := 1;
j := 1;
for k in [0 .. n*n - 1] do
a[i][j] := k;
if (i + j) mod 2 = 0 then
if j < n then
j := j + 1;
else
i := i + 2;
fi;
if i > 1 then
i := i - 1;
fi;
else
if i < n then
i := i + 1;
else
j := j + 2;
fi;
if j > 1 then
j := j - 1;
fi;
fi;
od;
return a;
end;
PrintArray(ZigZag(5));
# [ [ 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 ] ] |
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
| #Ruby | Ruby | y = ( 5**4**3**2 ).to_s
puts "5**4**3**2 = #{y[0..19]}...#{y[-20..-1]} and has #{y.length} 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
| #Run_BASIC | Run BASIC | x$ = str$( 5^(4^(3^2)))
print "Length:";len( x$)
print left$( x$, 20); "......"; right$( x$, 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
| #Scheme | Scheme | (import (rnrs))
(define (getFibList maxNum n1 n2 fibs)
(if (> n2 maxNum)
fibs
(getFibList maxNum n2 (+ n1 n2) (cons n2 fibs))))
(define (getZeckendorf num)
(if (<= num 0)
"0"
(let ((fibs (getFibList num 1 2 (list 1))))
(getZeckString "" num fibs))))
(define (getZeckString zeck num fibs)
(let* ((curFib (car fibs))
(placeZeck (>= num curFib))
(outString (string-append zeck (if placeZeck "1" "0")))
(outNum (if placeZeck (- num curFib) num)))
(if (null? (cdr fibs))
outString
(getZeckString outString outNum (cdr fibs)))))
(let loop ((i 0))
(when (<= i 20)
(for-each
(lambda (item)
(display item))
(list "Z(" i "):\t" (getZeckendorf i)))
(newline)
(loop (+ i 1))))
|
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.
| #Dylan | Dylan | define method doors()
let doors = make(<array>, fill: #f, size: 100);
for (x from 0 below 100)
for (y from x below 100 by x + 1)
doors[y] := ~doors[y]
end
end;
for (x from 1 to 100)
if (doors[x - 1])
format-out("door %d open\n", x)
end
end
end |
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)
| #Java | Java | int[] array = new int[10]; //optionally, replace "new int[10]" with a braced list of ints like "{1, 2, 3}"
array[0] = 42;
System.out.println(array[3]); |
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
| #TI-83_BASIC | TI-83_BASIC | 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
| #uBasic.2F4tH | uBasic/4tH | Print 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
| #Mathematica.2FWolfram_Language | Mathematica/Wolfram Language | 1 2 3 4 5
colors Blue Blue Blue Blue Blue
colors Green Green Green Green Green
colors Red Red Red Red Red
colors White White White White White
colors Yellow Yellow Yellow Yellow Yellow
nationality Dane Dane Dane Dane Dane
nationality English English English English English
nationality German German German German German
nationality Norwegian Norwegian Norwegian Norwegian Norwegian
nationality Swede Swede Swede Swede Swede
beverage Beer Beer Beer Beer Beer
beverage Coffee Coffee Coffee Coffee Coffee
beverage Milk Milk Milk Milk Milk
beverage Tea Tea Tea Tea Tea
beverage Water Water Water Water Water
animal Birds Birds Birds Birds Birds
animal Cats Cats Cats Cats Cats
animal Dog Dog Dog Dog Dog
animal Horse Horse Horse Horse Horse
animal Zebra Zebra Zebra Zebra Zebra
smoke Blend Blend Blend Blend Blend
smoke Blue Master Blue Master Blue Master Blue Master Blue Master
smoke Dunhill Dunhill Dunhill Dunhill Dunhill
smoke Pall Mall Pall Mall Pall Mall Pall Mall Pall Mall
smoke Prince Prince Prince Prince Prince |
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>
| #PicoLisp | PicoLisp | (load "@lib/xm.l")
(let Sections (body (in "file.xml" (xml)))
(pretty (car (body (car Sections))))
(prinl)
(for S Sections
(for L (body S)
(prinl (car (body L 'price))) ) )
(make
(for S Sections
(for L (body S)
(link (car (body L 'name))) ) ) ) ) |
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>
| #PowerShell | PowerShell |
$document = [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>
'@
$query = "/inventory/section/item"
$items = $document.SelectNodes($query)
|
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.
| #NetRexx | NetRexx | /* NetRexx */
options replace format comments java crossref savelog symbols binary
say "<?xml version='1.0' encoding='UTF-8' standalone='no'?>"
say "<!DOCTYPE svg PUBLIC '-//W3C//DTD SVG 1.1//EN'"
say " 'http://www.w3.org/Graphics/SVG/1.1/DTD/svg11.dtd'>"
say "<svg xmlns='http://www.w3.org/2000/svg' version='1.1'"
say " xmlns:xlink='http://www.w3.org/1999/xlink'"
say " width='30' height='30'>"
say " <defs><g id='y'>"
say " <circle cx='0' cy='0' r='200' stroke='black'"
say " fill='white' stroke-width='1'/>"
say " <path d='M0 -200 A 200 200 0 0 0 0 200"
say " 100 100 0 0 0 0 0 100 100 0 0 1 0 -200"
say " z' fill='black'/>"
say " <circle cx='0' cy='100' r='33' fill='white'/>"
say " <circle cx='0' cy='-100' r='33' fill='black'/>"
say " </g></defs>"
say draw_yinyang(20, 0.05)
say draw_yinyang(8, 0.02)
say "</svg>"
return
method draw_yinyang(trans = int, scale = double) inheritable static returns String
yy = String.format(" <use xlink:href='#y' transform='translate(%d,%d) scale(%g)'/>", -
[Object Integer(trans), Integer(trans), Double(scale)])
return yy |
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
| #Factor | Factor | USING: fry kernel math ;
IN: rosettacode.Y
: Y ( quot -- quot )
'[ [ dup call call ] curry @ ] dup call ; inline
: almost-fac ( quot -- quot )
'[ dup zero? [ drop 1 ] [ dup 1 - @ * ] if ] ;
: almost-fib ( quot -- quot )
'[ dup 2 >= [ 1 2 [ - @ ] bi-curry@ bi + ] when ] ; |
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
| #Go | Go | package main
import (
"fmt"
"strconv"
)
func zz(n int) []int {
r := make([]int, n*n)
i := 0
n2 := n * 2
for d := 1; d <= n2; d++ {
x := d - n
if x < 0 {
x = 0
}
y := d - 1
if y > n-1 {
y = n - 1
}
j := n2 - d
if j > d {
j = d
}
for k := 0; k < j; k++ {
if d&1 == 0 {
r[(x+k)*n+y-k] = i
} else {
r[(y-k)*n+x+k] = i
}
i++
}
}
return r
}
func main() {
const n = 5
w := len(strconv.Itoa(n*n - 1))
for i, e := range zz(n) {
fmt.Printf("%*d ", w, e)
if i%n == n-1 {
fmt.Println("")
}
}
} |
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
| #Rust | Rust | extern crate num;
use num::bigint::BigUint;
use num::FromPrimitive;
use num::pow::pow;
fn main() {
let big = BigUint::from_u8(5).unwrap();
let answer_as_string = format!("{}", pow(big,pow(4,pow(3,2))));
// The rest is output formatting.
let first_twenty: String = answer_as_string.chars().take(20).collect();
let last_twenty_reversed: Vec<char> = answer_as_string.chars().rev().take(20).collect();
let last_twenty: String = last_twenty_reversed.into_iter().rev().collect();
println!("Number of digits: {}", answer_as_string.len());
println!("First and last digits: {:?}..{:?}", first_twenty, last_twenty);
} |
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
| #Sather | Sather | class MAIN is
main is
r:INTI;
p1 ::= "62060698786608744707";
p2 ::= "92256259918212890625";
-- computing 5^(4^(3^2)), it could be written
-- also e.g. (5.inti)^((4.inti)^((3.inti)^(2.inti)))
r := (3.pow(2)).inti;
r := (4.inti).pow(r);
r := (5.inti).pow(r);
sr ::= r.str; -- string rappr. of the number
if sr.head(p1.size) = p1
and sr.tail(p2.size) = p2 then
#OUT + "result is ok..\n";
else
#OUT + "oops\n";
end;
#OUT + "# of digits: " + sr.size + "\n";
end;
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
| #Sidef | Sidef | func fib(n) is cached {
n < 2 ? 1
: (fib(n-1) + fib(n-2))
}
func zeckendorf(n) {
n == 0 && return '0'
var i = 1
++i while (fib(i) <= n)
gather {
while (--i > 0) {
var f = fib(i)
f > n ? (take '0')
: (take '1'; n -= f)
}
}.join
}
for n (0..20) {
printf("%4d: %8s\n", n, zeckendorf(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.
| #D.C3.A9j.C3.A0_Vu | Déjà Vu | local :open-doors [ rep 101 false ]
for i range 1 100:
local :j i
while <= j 100:
set-to open-doors j not open-doors! j
set :j + j i
!print\ "Open doors: "
for i range 1 100:
if open-doors! i:
!print\( to-str i " " ) |
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)
| #JavaScript | JavaScript | // Create a new array with length 0
var myArray = new Array();
// Create a new array with length 5
var myArray1 = new Array(5);
// Create an array with 2 members (length is 2)
var myArray2 = new Array("Item1","Item2");
// Create an array with 2 members using an array literal
var myArray3 = ["Item1", "Item2"];
// Assign a value to member [2] (length is now 3)
myArray3[2] = 5;
var x = myArray[2] + myArray.length; // 8
// You can also add a member to an array with the push function (length is now 4)
myArray3.push('Test');
// Elisions are supported, but are buggy in some implementations
var y = [0,1,,]; // length 3, or 4 in buggy implementations
|
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
| #Ursa | Ursa | > out (pow 0 0) endl console
1.0 |
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
| #VBA | VBA | Public Sub zero()
x = 0
y = 0
z = 0 ^ 0
Debug.Print "z ="; z
End Sub |
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
| #VBScript | VBScript | WScript.Echo 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
| #Mercury | Mercury | :- module zebra.
:- interface.
:- import_module io.
:- pred main(io, io).
:- mode main(di, uo) is cc_multi. % or det for all-solutions
:- implementation.
:- import_module list.
:- import_module solutions.
% perm
:- pred my_perm(list(T), list(T)).
:- mode my_perm(in, out) is multi.
my_perm([], []).
my_perm([X | Xs], Perm) :-
my_perm(Xs, PermXs),
my_insert(X, PermXs, Perm).
:- pred my_insert(T, list(T), list(T)).
:- mode my_insert(in, in, out) is multi.
my_insert(X, [], [X]).
my_insert(X, [Y | Ys], Zs) :-
(
Zs = [X, Y | Ys]
;
my_insert(X, Ys, Zs0),
Zs = [Y | Zs0]
).
% The puzzle
:- type person
---> english
; spanish
; ukrainian
; norwegian
; japanese.
:- pred left_of(list(T), T, T).
:- mode left_of(in, in, in) is semidet.
left_of([A, B | _], A, B).
left_of([_ | List], A, B) :- left_of(List, A, B).
:- pred next_to(list(T), T, T).
:- mode next_to(in, in, in) is semidet.
next_to(List, A, B) :-
( left_of(List, A, B)
; left_of(List, B, A)
).
:- pred puzzle({list(person), list(person), list(person), list(person),
list(person)}).
:- mode puzzle(out) is nondet.
puzzle({Houses, Colours, Pets, Drinks, Smokes}) :-
% 10. The Norwegian lives in the first house.
First = norwegian,
perm([english, spanish, ukrainian, japanese],
[Second, Third, Fourth, Fifth]),
% 2. The Englishman lives in the red house.
Red = english,
perm([spanish, ukrainian, norwegian, japanese],
[Green, Ivory, Yellow, Blue]),
% 10. The Norwegian lives in the first house.
% 15. The Norwegian lives next to the blue house.
Second = Blue,
% 6. The green house is immediately to the right of the ivory house.
left_of(Houses, Ivory, Green),
% 3. The Spaniard owns the dog.
Dog = spanish,
perm([english, ukrainian, norwegian, japanese],
[Snails, Fox, Horse, Zebra]),
% 4. Coffee is drunk in the green house.
Green = Coffee,
% 5. The Ukrainian drinks tea.
Tea = ukrainian,
% 9. Milk is drunk in the middle house.
Milk = Third,
perm([english, spanish, norwegian, japanese],
[Coffee, Milk, Juice, Water]),
% 7. The Old Gold smoker owns snails.
Snails = OldGold,
% 8. Kools are smoked in the yellow house.
Kools = Yellow,
% 13. The Lucky Strike smoker drinks orange juice.
LuckyStrike = Juice,
% 14. The Japanese smokes Parliaments.
Parliament = japanese,
perm([english, spanish, ukrainian, norwegian],
[OldGold, Kools, Chesterfield, LuckyStrike]),
% 11. The man who smokes Chesterfields lives in the house
% next to the man with the fox.
next_to(Houses, Chesterfield, Fox),
% 12. Kools are smoked in the house next to the house
% where the horse is kept.
next_to(Houses, Kools, Horse),
Houses = [First, Second, Third, Fourth, Fifth],
Colours = [Red, Green, Ivory, Yellow, Blue],
Pets = [Dog, Snails, Fox, Horse, Zebra],
Drinks = [Coffee, Tea, Milk, Juice, Water],
Smokes = [OldGold, Kools, Chesterfield, LuckyStrike, Parliament].
% Printing a solution
:- pred write_solution({list(person), list(person), list(person),
list(person), list(person)}::in, io::di, io::uo) is det.
write_solution({Houses, Colours, Pets, Drinks, Smokes}, !IO) :-
write_string("--------\n", !IO),
write_assignments(["1st", "2nd", "3rd", "4th", "5th"],
Houses, !IO),
write_assignments(["red", "green", "ivory", "yellow", "blue"],
Colours, !IO),
write_assignments(["dog", "snails", "fox", "horse", "zebra"],
Pets, !IO),
write_assignments(["coffee", "tea", "milk", "juice", "water"],
Drinks, !IO),
write_assignments(["oldgold", "kools", "chesterfield",
"luckystrike", "parliament"], Smokes, !IO).
:- pred write_assignments(list(string)::in, list(person)::in,
io::di, io::uo) is det.
write_assignments(Labels, Persons, !IO) :-
foldl_corresponding(write_assignment, Labels, Persons, !IO),
nl(!IO).
:- pred write_assignment(string::in, person::in, io::di, io::uo) is det.
write_assignment(Label, Person, !IO) :-
write_string(Label, !IO),
write_string(" - ", !IO),
write(Person, !IO),
write_string("\n", !IO).
% main
main(!IO) :-
% Print all solutions.
/*
solutions(puzzle, Solutions),
foldl(write_solution, Solutions, !IO).
*/
% Print solutions as they are found.
/*
unsorted_aggregate(puzzle, write_solution, !IO).
*/
% Print one solution.
( if puzzle(Solution) then
write_solution(Solution, !IO)
else
write_string("No solution found.\n", !IO)
).
|
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>
| #Python | Python | # Python has basic xml parsing built in
from xml.dom import minidom
xmlfile = file("test3.xml") # load xml document from file
xmldoc = minidom.parse(xmlfile).documentElement # parse from file stream or...
xmldoc = minidom.parseString("<inventory title="OmniCorp Store #45x10^3">...</inventory>").documentElement # alternatively, parse a string
# 1st Task: Retrieve the first "item" element
i = xmldoc.getElementsByTagName("item") # get a list of all "item" tags
firstItemElement = i[0] # get the first element
# 2nd task: Perform an action on each "price" element (print it out)
for j in xmldoc.getElementsByTagName("price"): # get a list of all "price" tags
print j.childNodes[0].data # XML Element . TextNode . data of textnode
# 3rd Task: Get an array of all the "name" elements
namesArray = xmldoc.getElementsByTagName("name") |
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>
| #R | R |
library("XML")
doc <- xmlInternalTreeParse("test3.xml")
# Retrieve the first "item" element
getNodeSet(doc, "//item")[[1]]
# Perform an action on each "price" element
sapply(getNodeSet(doc, "//price"), xmlValue)
# Get an array of all the "name" elements
sapply(getNodeSet(doc, "//name"), xmlValue)
|
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.
| #Nim | Nim | import gintro/cairo
proc draw(ctx: Context; x, y, r: float) =
ctx.arc(x, y, r + 1, 1.571, 7.854)
ctx.setSource(0.0, 0.0, 0.0)
ctx.fill()
ctx.arcNegative(x, y - r / 2, r / 2, 1.571, 4.712)
ctx.arc(x, y + r / 2, r / 2, 1.571, 4.712)
ctx.arcNegative(x, y, r, 4.712, 1.571)
ctx.setSource(1.0, 1.0, 1.0)
ctx.fill()
ctx.arc(x, y - r / 2, r / 5, 1.571, 7.854)
ctx.setSource(0.0, 0.0, 0.0)
ctx.fill()
ctx.arc(x, y + r / 2, r / 5, 1.571, 7.854)
ctx.setSource(1.0, 1.0, 1.0)
ctx.fill()
let surface = imageSurfaceCreate(argb32, 200, 200)
let context = newContext(surface)
context.draw(120, 120, 75)
context.draw(35, 35, 30)
let status = surface.writeToPng("yin-yang.png")
assert status == Status.success |
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
| #Falcon | Falcon |
Y = { f => {x=> {n => f(x(x))(n)}} ({x=> {n => f(x(x))(n)}}) }
facStep = { f => {x => x < 1 ? 1 : x*f(x-1) }}
fibStep = { f => {x => x == 0 ? 0 : (x == 1 ? 1 : f(x-1) + f(x-2))}}
YFac = Y(facStep)
YFib = Y(fibStep)
> "Factorial 10: ", YFac(10)
> "Fibonacci 10: ", YFib(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
| #Groovy | Groovy | def zz = { n ->
grid = new int[n][n]
i = 0
for (d in 1..n*2) {
(x, y) = [Math.max(0, d - n), Math.min(n - 1, d - 1)]
Math.min(d, n*2 - d).times {
grid[d%2?y-it:x+it][d%2?x+it:y-it] = i++;
}
}
grid
} |
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
| #Scala | Scala | scala> BigInt(5) modPow (BigInt(4) pow (BigInt(3) pow 2).toInt, BigInt(10) pow 20)
res21: scala.math.BigInt = 92256259918212890625
scala> (BigInt(5) pow (BigInt(4) pow (BigInt(3) pow 2).toInt).toInt).toString
res22: String = 6206069878660874470748320557284679309194219265199117173177383244
78446890420544620839553285931321349485035253770303663683982841794590287939217907
89641300156281305613064874236198955114921296922487632406742326659692228562195387
46210423235340883954495598715281862895110697243759768434501295076608139350684049
01191160699929926568099301259938271975526587719565309995276438998093283175080241
55833224724855977970015112594128926594587205662421861723789001208275184293399910
13912158886504596553858675842231519094813553261073608575593794241686443569888058
92732524316323249492420512640962691673104618378381545202638771401061171968052873
21414945463925055899307933774904078819911387324217976311238875802878310483037255
33789567769926391314746986316354035923183981697660495275234703657750678459919...
scala> res22 take 20
res23: String = 62060698786608744707
scala> res22 length
res24: Int = 183231
scala> |
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
| #Scheme | Scheme | (define x (expt 5 (expt 4 (expt 3 2))))
(define y (number->string x))
(define l (string-length y))
(display (string-append "5**4**3**2 = " (substring y 0 20) "..." (substring y (- l 20) l) " and has " (number->string l) " digits"))
(newline) |
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
| #Simula | Simula | BEGIN
INTEGER N, F0, F1, F2, D;
N := 20;
COMMENT CALCULATE D FROM ANY GIVEN N ;
F1 := 1; F2 := 2; F0 := F1 + F2; D := 2;
WHILE F0 < N DO BEGIN
F1 := F2; F2 := F0; F0 := F1 + F2; D := D + 1;
END;
BEGIN
COMMENT Sinclair ZX81 BASIC Solution ;
TEXT Z1, S1;
INTEGER I, J, Z;
INTEGER ARRAY F(1:D); ! 10 dim f(6) ;
F(1) := 1; ! 20 let f(1)=1 ;
F(2) := 2; ! 30 let f(2)=2 ;
FOR I := 3 STEP 1 UNTIL D DO BEGIN ! 40 for i=3 to 6 ;
F(I) := F(I-2) + F(I-1); ! 50 let f(i)=f(i-2)+f(i-1) ;
END; ! 60 next i ;
FOR I := 0 STEP 1 UNTIL N DO BEGIN ! 70 for i=0 to 20 ;
Z1 :- ""; ! 80 let z$="" ;
S1 :- " "; ! 90 let s$=" " ;
Z := I; ! 100 let z=i ;
FOR J := D STEP -1 UNTIL 1 DO BEGIN ! 110 for j=6 to 1 step -1 ;
IF J=1 THEN S1 :- "0"; ! 120 if j=1 then let s$="0" ;
IF NOT (Z<F(J)) THEN BEGIN ! 130 if z<f(j) then goto 180 ;
Z1 :- Z1 & "1"; ! 140 let z$=z$+"1" ;
Z := Z-F(J); ! 150 let z=z-f(j) ;
S1 :- "0"; ! 160 let s$="0" ;
END ELSE ! 170 goto 190 ;
Z1 :- Z1 & S1; ! 180 let z$=z$+s$ ;
END; ! 190 next j ;
OUTINT(I, 0); OUTCHAR(' '); ! 200 print i ; !" "; !;
IF I<10 THEN OUTCHAR(' '); ! 210 if i<10 then print " "; !;
OUTTEXT(Z1); OUTIMAGE; ! 220 print z$ ;
END; ! 230 next i ;
END;
END |
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.
| #E | E | #!/usr/bin/env rune
var toggles := []
var gets := []
# Set up GUI (and data model)
def frame := <swing:makeJFrame>("100 doors")
frame.getContentPane().setLayout(<awt:makeGridLayout>(10, 10))
for i in 1..100 {
def component := <import:javax.swing.makeJCheckBox>(E.toString(i))
toggles with= fn { component.setSelected(!component.isSelected()) }
gets with= fn { component.isSelected() }
frame.getContentPane().add(component)
}
# Set up termination condition
def done
frame.addWindowListener(def _ {
to windowClosing(event) {
bind done := true
}
match _ {}
})
# Open and close doors
def loop(step, i) {
toggles[i] <- ()
def next := i + step
timer.whenPast(timer.now() + 10, fn {
if (next >= 100) {
if (step >= 100) {
# Done.
} else {
loop <- (step + 1, step)
}
} else {
loop <- (step, i + step)
}
})
}
loop(1, 0)
frame.pack()
frame.show()
interp.waitAtTop(done) |
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)
| #jq | jq | # Create a new array with length 0
[]
# Create a new array of 5 nulls
[][4] = null # setting the element at offset 4 expands the array
# Create an array having the elements 1 and 2 in that order
[1,2]
# Create an array of integers from 0 to 10 inclusive
[ range(0; 11) ]
# If a is an array (of any length), update it so that a[2] is 5
a[2] = 5;
# Append arrays a and b
a + b
# Append an element, e, to an array a
a + [e]
##################################################
# In the following, a is assumed to be [0,1,2,3,4]
# It is not an error to use an out-of-range index:
a[10] # => null
# Negative indices count backwards from after the last element:
a[-1] # => 4
# jq supports simple slice operations but
# only in the forward direction:
a[:1] # => [0]
a[1:] # => [1,2,3,4]
a[2:4] # => [2,3]
a[4:2] # null |
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
| #Verilog | Verilog | module main;
initial begin
$display("0 ^ 0 = ", 0**0);
$finish ;
end
endmodule |
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
| #Visual_Basic_.NET | Visual Basic .NET | Module Program
Sub Main()
Console.Write(0^0)
End Sub
End Module |
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
| #MiniZinc | MiniZinc |
%Solve Zebra Puzzle. Nigel Galloway, August 27th., 2019
include "alldifferent.mzn";
enum N={English,Swedish,Danish,German,Norwegian};
enum I={Tea,Coffee,Milk,Beer,Water};
enum G={Dog,Birds,Cats,Horse,Zebra};
enum E={Red,Green,White,Blue,Yellow};
enum L={PallMall,Dunhill,BlueMaster,Prince,Blend};
array[1..5] of var N: Nz; constraint alldifferent(Nz); constraint Nz[1]=Norwegian; %The Norwegian lives in the first house.
array[1..5] of var I: Iz; constraint alldifferent(Iz); constraint Iz[3]=Milk; %In the middle house they drink milk.
array[1..5] of var G: Gz; constraint alldifferent(Gz);
array[1..5] of var E: Ez; constraint alldifferent(Ez);
array[1..5] of var L: Lz; constraint alldifferent(Lz);
constraint exists(n in 1..5)(Nz[n]=English /\ Ez[n]=Red); %The English man lives in the red house
constraint exists(n in 1..5)(Nz[n]=Swedish /\ Gz[n]=Dog); %The Swede has a dog.
constraint exists(n in 1..5)(Nz[n]=Danish /\ Iz[n]=Tea); %The Dane drinks tea.
constraint exists(n in 1..4)(Ez[n]=Green /\ Ez[n+1]=White); %The green house is immediately to the left of the white house.
constraint exists(n in 1..5)(Ez[n]=Green /\ Iz[n]=Coffee); %They drink coffee in the green house.
constraint exists(n in 1..5)(Lz[n]=PallMall /\ Gz[n]=Birds); %The man who smokes Pall Mall has birds
constraint exists(n in 1..5)(Ez[n]=Yellow /\ Lz[n]=Dunhill); %In the yellow house they smoke Dunhill.
constraint exists(n in 1..4)((Lz[n]=Blend /\ Gz[n+1]=Cats) \/ (Lz[n+1]=Blend /\ Gz[n]=Cats)); %The man who smokes Blend lives in the house next to the house with cats.
constraint exists(n in 1..4)((Gz[n]=Horse /\ Lz[n+1]=Dunhill) \/ (Gz[n+1]=Horse /\ Lz[n]=Dunhill)); %In a house next to the house where they have a horse, they smoke Dunhill.
constraint exists(n in 1..5)(Lz[n]=BlueMaster /\ Iz[n]=Beer); %The man who smokes Blue Master drinks beer.
constraint exists(n in 1..5)(Nz[n]=German /\ Lz[n]=Prince); %The German smokes Prince.
constraint exists(n in 1..4)((Nz[n]=Norwegian /\ Ez[n+1]=Blue) \/ (Nz[n+1]=Norwegian /\ Ez[n]=Blue));%The Norwegian lives next to the blue house.
constraint exists(n in 1..4)((Lz[n]=Blend /\ Iz[n+1]=Water) \/ (Lz[n+1]=Blend /\ Iz[n]=Water)); %They drink water in a house next to the house where they smoke Blend.
var 1..5: n;
constraint Gz[n]=Zebra;
solve satisfy;
output ["The "++show(Nz[n])++" owns the zebra"++"\n\n"++show(Nz)++"\n"++show(Iz)++"\n"++show(Gz)++"\n"++show(Ez)++"\n"++show(Lz)++"\n"];
|
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>
| #Racket | Racket |
#lang at-exp racket
(define input @~a{
<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>})
(require xml xml/path)
(define data (xml->xexpr
((eliminate-whitespace '(inventory section item))
(read-xml/element (open-input-string input)))))
;; Retrieve the first "item" element
(displayln (xexpr->string (se-path* '(item) data)))
;; => <name>Invisibility Cream</name>
;; Perform an action on each "price" element (print it out)
(printf "Prices: ~a\n" (string-join (se-path*/list '(item price) data) ", "))
;; => Prices: 14.50, 23.99, 4.95, 3.56
;; Get an array of all the "name" elements
(se-path*/list '(item name) data)
;; => '("Invisibility Cream" "Levitation Salve" "Blork and Freen Instameal" "Grob winglets")
|
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>
| #Raku | Raku | use XML::XPath;
my $XML = XML::XPath.new(xml => q:to/END/);
<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>
END
put "First item:\n", $XML.find('//item[1]')[0];
put "\nPrice elements:";
.contents.put for $XML.find('//price').List;
put "\nName elements:\n", $XML.find('//name')».contents.join: ', '; |
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.
| #OCaml | OCaml | open Graphics
let draw_yinyang x y radius black white =
let hr = radius / 2 in
let sr = radius / 6 in
set_color black;
set_line_width 6;
draw_circle x y radius;
set_line_width 0;
set_color black;
fill_arc x y radius radius 270 450;
set_color white;
fill_arc x y radius radius 90 270;
fill_arc x (y + hr) hr hr 270 450;
set_color black;
fill_arc x (y - hr) hr hr 90 270;
fill_circle x (y + hr) sr;
set_color white;
fill_circle x (y - hr) sr
let () =
open_graph "";
let width = size_x()
and height = size_y() in
set_color (rgb 200 200 200);
fill_rect 0 0 width height;
let w = width / 3
and h = height / 3 in
let r = (min w h) / 3 in
draw_yinyang w (h*2) (r*2) black white;
draw_yinyang (w*2) h r blue magenta;
ignore(read_key()) |
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
| #Forth | Forth | \ Address of an xt.
variable 'xt
\ Make room for an xt.
: xt, ( -- ) here 'xt ! 1 cells allot ;
\ Store xt.
: !xt ( xt -- ) 'xt @ ! ;
\ Compile fetching the xt.
: @xt, ( -- ) 'xt @ postpone literal postpone @ ;
\ Compile the Y combinator.
: y, ( xt1 -- xt2 ) >r :noname @xt, r> compile, postpone ; ;
\ Make a new instance of the Y combinator.
: y ( xt1 -- xt2 ) xt, y, dup !xt ; |
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
| #Haskell | Haskell | import Data.Array (Array, array, bounds, range, (!))
import Text.Printf (printf)
import Data.List (sortBy)
compZig :: (Int, Int) -> (Int, Int) -> Ordering
compZig (x, y) (x_, y_) = compare (x + y) (x_ + y_) <> go x y
where
go x y
| even (x + y) = compare x x_
| otherwise = compare y y_
zigZag :: (Int, Int) -> Array (Int, Int) Int
zigZag upper = array b $ zip (sortBy compZig (range b)) [0 ..]
where
b = ((0, 0), upper) |
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
| #Seed7 | Seed7 | $ include "seed7_05.s7i";
include "bigint.s7i";
const proc: main is func
local
var bigInteger: fiveToThePowerOf262144 is 5_ ** 4 ** 3 ** 2;
var string: numberAsString is str(fiveToThePowerOf262144);
begin
writeln("5**4**3**2 = " <& numberAsString[..20] <&
"..." <& numberAsString[length(numberAsString) - 19 ..]);
writeln("decimal digits: " <& length(numberAsString));
end func; |
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
| #Sidef | Sidef | var x = 5**(4**(3**2));
var y = x.to_s;
printf("5**4**3**2 = %s...%s and has %i digits\n", y.ft(0,19), y.ft(-20), y.len); |
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
| #Sinclair_ZX81_BASIC | Sinclair ZX81 BASIC | 10 DIM F(6)
20 LET F(1)=1
30 LET F(2)=2
40 FOR I=3 TO 6
50 LET F(I)=F(I-2)+F(I-1)
60 NEXT I
70 FOR I=0 TO 20
80 LET Z$=""
90 LET S$=" "
100 LET Z=I
110 FOR J=6 TO 1 STEP -1
120 IF J=1 THEN LET S$="0"
130 IF Z<F(J) THEN GOTO 180
140 LET Z$=Z$+"1"
150 LET Z=Z-F(J)
160 LET S$="0"
170 GOTO 190
180 LET Z$=Z$+S$
190 NEXT J
200 PRINT I;" ";
210 IF I<10 THEN PRINT " ";
220 PRINT Z$
230 NEXT 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.
| #EasyLang | EasyLang | len d[] 101
for p = 1 to 100
i = p
while i <= 100
d[i] = 1 - d[i]
i += p
.
.
for i = 1 to 100
if d[i] = 1
print i
.
. |
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)
| #Jsish | Jsish | /* Arrays in Jsi */
// Create a new array with length 0
var myArray = new Array();
;myArray;
// In Jsi, typeof [] is "array". In ECMAScript, typeof [] is "object"
;typeof [];
// Create a new array with length 5
var myArray1 = new Array(5);
;myArray1;
// Create an array with 2 members (length is 2)
var myArray2 = new Array("Item1","Item2");
;myArray2;
;myArray2.length;
// Create an array with 2 members using an array literal
var myArray3 = ["Item1", "Item2"];
;myArray3;
// Assign a value to member [2] (length is now 3)
myArray3[2] = 5;
;myArray3;
;myArray3.length;
var x = myArray3[2] + myArray3.length; // 8
;x;
// You can also add a member to an array with the push function (length is now 4)
myArray3.push('Test');
;myArray3;
;myArray3.length;
// Empty array entries in a literal is a syntax error, elisions not allowed
//var badSyntax = [1,2,,4];
/*
=!EXPECTSTART!=
myArray ==> []
typeof [] ==> array
myArray1 ==> [ undefined, undefined, undefined, undefined, undefined ]
myArray2 ==> [ "Item1", "Item2" ]
myArray2.length ==> 2
myArray3 ==> [ "Item1", "Item2" ]
myArray3 ==> [ "Item1", "Item2", 5 ]
myArray3.length ==> 3
x ==> 8
myArray3 ==> [ "Item1", "Item2", 5, "Test" ]
myArray3.length ==> 4
=!EXPECTEND!=
*/ |
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
| #Vlang | Vlang | // Zero to the zero power, in V
// Tectonics: v run zero-to-the-zero-power.v
module main
import math
// starts here
// V does not include an exponentiation operator, but uses a math module
pub fn main() {
println(math.pow(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
| #Wren | Wren | System.print(0.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
| #Nial | Nial |
remove is op x xs {filter (not (x =)) xs}
append_map is transformer func op seq { \
reduce (op x xs { (func x) link xs}) (seq append []) }
permutations is op seq { \
if empty seq then [[]] else \
(append_map \
(op head {each (op tail {head hitch tail}) \
(permutations (remove head seq))}) \
seq) \
endif}
f is find
tokenize is op str{string_split ' ' str}
mk is tr pred op str {filter pred permutations tokenize str}
eq is op x xs y ys{f x xs = f y ys}
adj is op x xs y ys{1 = abs(f x xs - f y ys)}
run is { \
men := mk (op xs {0 = f 'norwegian' xs}) \
'danish english german norwegian swedish'; \
colors := mk (op xs {1 = ((f 'white' xs) - (f 'green' xs))}) \
'blue green red white yellow'; \
drinks := mk (op xs {2 = f 'milk' xs}) 'beer coffee milk tea water'; \
pets := mk (op xs {l}) 'birds cats dog horse zebra'; \
smokes := mk (op xs {l}) 'blend blue-master dunhill pall-mall prince'; \
for m with men do \
for c with colors do \
if (eq 'english' m 'red' c) and \
(adj 'norwegian' m 'blue' c) then \
for d with drinks do \
if (eq 'danish' m 'tea' d) and \
(eq 'coffee' d 'green' c) then \
for s with smokes do \
if (eq 'yellow' c 'dunhill' s) and \
(eq 'blue-master' s 'beer' d) and \
(eq 'german' m 'prince' s) then \
for p with pets do \
if (eq 'birds' p 'pall-mall' s) and \
(eq 'swedish' m 'dog' p) and \
(adj 'blend' s 'cats' p) and \
(adj 'horse' p 'dunhill' s) then \
write (0 blend (p m c d s)) \
endif \
endfor \
endif \
endfor \
endif \
endfor \
endif \
endfor \
endfor }
abs(time - (run; time))
|
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>
| #Rascal | Rascal | import lang::xml::DOM;
import Prelude;
public void get_first_item(loc a){
D = parseXMLDOM(readFile(a));
top-down-break visit(D){
case E:element(_,"item",_): return println(xmlPretty(E));
};
}
public void print_prices(loc a){
D = parseXMLDOM(readFile(a));
for(/element(_,"price",[charData(/str p)]) := D)
println(p);
}
public list[str] get_names(loc a){
D = parseXMLDOM(readFile(a));
L = [];
for(/element(_,"name",[charData(/str n)]) := D)
L += n;
return L;
} |
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>
| #REXX | REXX | /*REXX program to parse various queries on an XML document (from a file). */
iFID='XPATH.XML' /*name of the input XML file (doc). */
$= /*string will contain the file's text. */
do j=1 while lines(iFID)\==0 /*read the entire file into a string. */
$=$ linein(iFID) /*append the line to the $ string. */
end /*j*/
/* [↓] show 1st ITEM in the document*/
parse var $ '<item ' item "</item>"
say center('first item:',length(space(item)),'─') /*display a nice header.*/
say space(item)
/* [↓] show all PRICES in the document*/
prices= /*nullify the list and add/append to it*/
$$=$ /*start with a fresh copy of document. */
do until $$='' /* [↓] keep parsing string until done.*/
parse var $$ '<price>' price '</price>' $$
prices=prices price /*add/append the price to the list. */
end /*until*/
say
say center('prices:',length(space(prices)),'─') /*display a nice header.*/
say space(prices)
/* [↓] show all NAMES in the document*/
names.= /*nullify the list and add/append to it*/
L=length(' names: ') /*maximum length of any one list name. */
$$=$ /*start with a fresh copy of document. */
do #=1 until $$='' /* [↓] keep parsing string until done.*/
parse var $$ '<name>' names.# '</name>' $$
L=max(L,length(names.#)) /*L: is used to find the widest name. */
end /*#*/
names.0=#-1; say /*adjust the number of names (DO loop).*/
say center('names:',L,'─') /*display a nicely formatted header. */
do k=1 for names.0 /*display all the names in the list. */
say names.k /*display a name from the NAMES list.*/
end /*k*/
/*stick a fork in it, we're all done. */ |
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.
| #PARI.2FGP | PARI/GP | YinYang(r)={ for(y=-r,r, print(concat(apply( x->
if( x^2+y^2>r^2, " ",
[y<0,y>0,x>0][logint((x^2+(abs(y)-r/2)^2)<<8\r^2+1,2)\3+1], "#", "."
), [-r..r]
))))
} |
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.
| #Pascal | Pascal | //Written for TU Berlin
//Compiled with fpc
Program yingyang;
Uses Math;
const
scale_x=2;
scale_y=1;
black='#';
white='.';
clear=' ';
function inCircle(centre_x:Integer;centre_y:Integer;radius:Integer;x:Integer;y:Integer):Boolean ;
begin
inCircle:=power(x-centre_x,2)+power(y-centre_y,2)<=power(radius,2);
end;
function bigCircle(radius:Integer;x:Integer;y:Integer):Boolean ;
begin
bigCircle:=inCircle(0,0,radius,x,y);
end;
function whiteSemiCircle(radius:Integer;x:Integer;y:Integer):Boolean ;
begin
whiteSemiCircle:=inCircle(0,radius div 2 ,radius div 2,x,y);
end;
function smallBlackCircle(radius:Integer;x:Integer;y:Integer):Boolean ;
begin
smallBlackCircle:=inCircle(0,radius div 2 ,radius div 6,x,y);
end;
function blackSemiCircle(radius:Integer;x:Integer;y:Integer):Boolean ;
begin
blackSemiCircle:=inCircle(0,-radius div 2 ,radius div 2,x,y);
end;
function smallWhiteCircle(radius:Integer;x:Integer;y:Integer):Boolean ;
begin
smallWhiteCircle:=inCircle(0,-radius div 2 ,radius div 6,x,y);
end;
var
radius,sy,sx,x,y:Integer;
begin
writeln('Please type a radius:');
readln(radius);
if radius<3 then begin writeln('A radius bigger than 3');halt end;
sy:=round(radius*scale_y);
while(sy>=-round(radius*scale_y)) do begin
sx:=-round(radius*scale_x);
while(sx<=round(radius*scale_x)) do begin
x:=sx div scale_x;
y:=sy div scale_y;
if bigCircle(radius,x,y) then begin
if (whiteSemiCircle(radius,x,y)) then if smallblackCircle(radius,x,y) then write(black) else write(white) else if blackSemiCircle(radius,x,y) then if smallWhiteCircle(radius,x,y) then write(white) else write(black) else if x>0 then write(white) else write(black);
end
else write(clear);
sx:=sx+1
end;
writeln;
sy:=sy-1;
end;
end.
|
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
| #FreeBASIC | FreeBASIC | Function Y(f As String) As String
Y = f
End Function
Function fib(n As Long) As Long
Dim As Long n1 = 0, n2 = 1, k, sum
For k = 1 To Abs(n)
sum = n1 + n2
n1 = n2
n2 = sum
Next k
Return Iif(n < 0, (n1 * ((-1) ^ ((-n)+1))), n1)
End Function
Function fac(n As Long) As Long
Dim As Long r = 1, i
For i = 2 To n
r *= i
Next i
Return r
End Function
Function execute(s As String, n As Integer) As Long
Return Iif (s = "fac", fac(n), fib(n))
End Function
Sub test(nombre As String)
Dim f As String: f = Y(nombre)
Print !"\n"; f; ":";
For i As Integer = 1 To 10
Print execute(f, i);
Next i
End Sub
test("fac")
test("fib")
Sleep |
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