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http://rosettacode.org/wiki/Ulam_spiral_(for_primes) | Ulam spiral (for primes) | An Ulam spiral (of primes) is a method of visualizing primes when expressed in a (normally counter-clockwise) outward spiral (usually starting at 1), constructed on a square grid, starting at the "center".
An Ulam spiral is also known as a prime spiral.
The first grid (green) is shown with sequential integers, starting at 1.
In an Ulam spiral of primes, only the primes are shown (usually indicated by some glyph such as a dot or asterisk), and all non-primes as shown as a blank (or some other whitespace).
Of course, the grid and border are not to be displayed (but they are displayed here when using these Wiki HTML tables).
Normally, the spiral starts in the "center", and the 2nd number is to the viewer's right and the number spiral starts from there in a counter-clockwise direction.
There are other geometric shapes that are used as well, including clock-wise spirals.
Also, some spirals (for the 2nd number) is viewed upwards from the 1st number instead of to the right, but that is just a matter of orientation.
Sometimes, the starting number can be specified to show more visual striking patterns (of prime densities).
[A larger than necessary grid (numbers wise) is shown here to illustrate the pattern of numbers on the diagonals (which may be used by the method to orientate the direction of spiral-construction algorithm within the example computer programs)].
Then, in the next phase in the transformation of the Ulam prime spiral, the non-primes are translated to blanks.
In the orange grid below, the primes are left intact, and all non-primes are changed to blanks.
Then, in the final transformation of the Ulam spiral (the yellow grid), translate the primes to a glyph such as a • or some other suitable glyph.
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The Ulam spiral becomes more visually obvious as the grid increases in size.
Task
For any sized N × N grid, construct and show an Ulam spiral (counter-clockwise) of primes starting at some specified initial number (the default would be 1), with some suitably dotty (glyph) representation to indicate primes, and the absence of dots to indicate non-primes.
You should demonstrate the generator by showing at Ulam prime spiral large enough to (almost) fill your terminal screen.
Related tasks
Spiral matrix
Zig-zag matrix
Identity matrix
Sequence of primes by Trial Division
See also
Wikipedia entry: Ulam spiral
MathWorld™ entry: Prime Spiral
| #Haskell | Haskell | import Data.List
import Data.Numbers.Primes
ulam n representation = swirl n . map representation |
http://rosettacode.org/wiki/Truncatable_primes | Truncatable primes | A truncatable prime is a prime number that when you successively remove digits from one end of the prime, you are left with a new prime number.
Examples
The number 997 is called a left-truncatable prime as the numbers 997, 97, and 7 are all prime.
The number 7393 is a right-truncatable prime as the numbers 7393, 739, 73, and 7 formed by removing digits from its right are also prime.
No zeroes are allowed in truncatable primes.
Task
The task is to find the largest left-truncatable and right-truncatable primes less than one million (base 10 is implied).
Related tasks
Find largest left truncatable prime in a given base
Sieve of Eratosthenes
See also
Truncatable Prime from MathWorld.]
| #C.2B.2B | C++ | #include <iostream>
#include "prime_sieve.hpp"
bool is_left_truncatable(const prime_sieve& sieve, int p) {
for (int n = 10, q = p; p > n; n *= 10) {
if (!sieve.is_prime(p % n) || q == p % n)
return false;
q = p % n;
}
return true;
}
bool is_right_truncatable(const prime_sieve& sieve, int p) {
for (int q = p/10; q > 0; q /= 10) {
if (!sieve.is_prime(q))
return false;
}
return true;
}
int main() {
const int limit = 1000000;
// find the prime numbers up to the limit
prime_sieve sieve(limit + 1);
int largest_left = 0;
int largest_right = 0;
// find largest left truncatable prime
for (int p = limit; p >= 2; --p) {
if (sieve.is_prime(p) && is_left_truncatable(sieve, p)) {
largest_left = p;
break;
}
}
// find largest right truncatable prime
for (int p = limit; p >= 2; --p) {
if (sieve.is_prime(p) && is_right_truncatable(sieve, p)) {
largest_right = p;
break;
}
}
// write results to standard output
std::cout << "Largest left truncatable prime is " << largest_left << '\n';
std::cout << "Largest right truncatable prime is " << largest_right << '\n';
return 0;
} |
http://rosettacode.org/wiki/Unbias_a_random_generator | Unbias a random generator |
P
1
{\displaystyle P_{1}}
P
0
{\displaystyle P_{0}}
P
1
{\displaystyle P_{1}}
P
0
{\displaystyle P_{0}}
P
0
{\displaystyle P_{0}}
P
1
{\displaystyle P_{1}}
Task details
Use your language's random number generator to create a function/method/subroutine/... randN that returns a one or a zero, but with one occurring, on average, 1 out of N times, where N is an integer from the range 3 to 6 inclusive.
Create a function unbiased that uses only randN as its source of randomness to become an unbiased generator of random ones and zeroes.
For N over its range, generate and show counts of the outputs of randN and unbiased(randN).
The actual unbiasing should be done by generating two numbers at a time from randN and only returning a 1 or 0 if they are different. As long as you always return the first number or always return the second number, the probabilities discussed above should take over the biased probability of randN.
This task is an implementation of Von Neumann debiasing, first described in a 1951 paper.
| #Visual_Basic_.NET | Visual Basic .NET | Module Module1
Dim random As New Random()
Function RandN(n As Integer) As Boolean
Return random.Next(0, n) = 0
End Function
Function Unbiased(n As Integer) As Boolean
Dim flip1 As Boolean
Dim flip2 As Boolean
Do
flip1 = RandN(n)
flip2 = RandN(n)
Loop While flip1 = flip2
Return flip1
End Function
Sub Main()
For n = 3 To 6
Dim biasedZero = 0
Dim biasedOne = 0
Dim unbiasedZero = 0
Dim unbiasedOne = 0
For i = 1 To 100000
If RandN(n) Then
biasedOne += 1
Else
biasedZero += 1
End If
If Unbiased(n) Then
unbiasedOne += 1
Else
unbiasedZero += 1
End If
Next
Console.WriteLine("(N = {0}):".PadRight(17) + "# of 0" + vbTab + "# of 1" + vbTab + "% of 0" + vbTab + "% of 1", n)
Console.WriteLine("(Biased: {0}):".PadRight(15) + "{0}" + vbTab + "{1}" + vbTab + "{2}" + vbTab + "{3}", biasedZero, biasedOne, biasedZero / 1000, biasedOne / 1000)
Console.WriteLine("(UnBiased: {0}):".PadRight(15) + "{0}" + vbTab + "{1}" + vbTab + "{2}" + vbTab + "{3}", unbiasedZero, unbiasedOne, unbiasedZero / 1000, unbiasedOne / 1000)
Next
End Sub
End Module |
http://rosettacode.org/wiki/Truncate_a_file | Truncate a file | Task
Truncate a file to a specific length. This should be implemented as a routine that takes two parameters: the filename and the required file length (in bytes).
Truncation can be achieved using system or library calls intended for such a task, if such methods exist, or by creating a temporary file of a reduced size and renaming it, after first deleting the original file, if no other method is available. The file may contain non human readable binary data in an unspecified format, so the routine should be "binary safe", leaving the contents of the untruncated part of the file unchanged.
If the specified filename does not exist, or the provided length is not less than the current file length, then the routine should raise an appropriate error condition.
On some systems, the provided file truncation facilities might not change the file or may extend the file, if the specified length is greater than the current length of the file.
This task permits the use of such facilities. However, such behaviour should be noted, or optionally a warning message relating to an non change or increase in file size may be implemented.
| #PicoLisp | PicoLisp | (de truncate (File Len)
(native "@" "truncate" 'I File Len) ) |
http://rosettacode.org/wiki/Truncate_a_file | Truncate a file | Task
Truncate a file to a specific length. This should be implemented as a routine that takes two parameters: the filename and the required file length (in bytes).
Truncation can be achieved using system or library calls intended for such a task, if such methods exist, or by creating a temporary file of a reduced size and renaming it, after first deleting the original file, if no other method is available. The file may contain non human readable binary data in an unspecified format, so the routine should be "binary safe", leaving the contents of the untruncated part of the file unchanged.
If the specified filename does not exist, or the provided length is not less than the current file length, then the routine should raise an appropriate error condition.
On some systems, the provided file truncation facilities might not change the file or may extend the file, if the specified length is greater than the current length of the file.
This task permits the use of such facilities. However, such behaviour should be noted, or optionally a warning message relating to an non change or increase in file size may be implemented.
| #PL.2FI | PL/I |
/* Parameters to be read in by the program are: */
/* 1. the name of the file to be truncated, and */
/* 2. the size of file after truncation. */
truncate_file: procedure options (main); /* 12 July 2012 */
declare (i, n) fixed binary (31);
declare filename character(50) varying;
declare in file record input, out file record output;
put ('What is the name of the file to be truncated?');
get edit (filename) (L);
put ('What is the length of file to be retained?');
get (n);
begin;
declare c(n) character (1), ch character (1);
open file (in) title ('/' || filename || ',type(fixed),recsize(1)' )
input;
do i = 1 to n; read file (in) into (ch); c(i) = ch; end;
close file (in);
open file (out) title ('/' || filename || ',append(n),type(fixed),
recsize(' || trim(n) || ')' );
write file (out) from (c);
close file (out);
end;
end truncate_file;
|
http://rosettacode.org/wiki/Universal_Turing_machine | Universal Turing machine | One of the foundational mathematical constructs behind computer science
is the universal Turing Machine.
(Alan Turing introduced the idea of such a machine in 1936–1937.)
Indeed one way to definitively prove that a language
is turing-complete
is to implement a universal Turing machine in it.
Task
Simulate such a machine capable
of taking the definition of any other Turing machine and executing it.
Of course, you will not have an infinite tape,
but you should emulate this as much as is possible.
The three permissible actions on the tape are "left", "right" and "stay".
To test your universal Turing machine (and prove your programming language
is Turing complete!), you should execute the following two Turing machines
based on the following definitions.
Simple incrementer
States: q0, qf
Initial state: q0
Terminating states: qf
Permissible symbols: B, 1
Blank symbol: B
Rules:
(q0, 1, 1, right, q0)
(q0, B, 1, stay, qf)
The input for this machine should be a tape of 1 1 1
Three-state busy beaver
States: a, b, c, halt
Initial state: a
Terminating states: halt
Permissible symbols: 0, 1
Blank symbol: 0
Rules:
(a, 0, 1, right, b)
(a, 1, 1, left, c)
(b, 0, 1, left, a)
(b, 1, 1, right, b)
(c, 0, 1, left, b)
(c, 1, 1, stay, halt)
The input for this machine should be an empty tape.
Bonus:
5-state, 2-symbol probable Busy Beaver machine from Wikipedia
States: A, B, C, D, E, H
Initial state: A
Terminating states: H
Permissible symbols: 0, 1
Blank symbol: 0
Rules:
(A, 0, 1, right, B)
(A, 1, 1, left, C)
(B, 0, 1, right, C)
(B, 1, 1, right, B)
(C, 0, 1, right, D)
(C, 1, 0, left, E)
(D, 0, 1, left, A)
(D, 1, 1, left, D)
(E, 0, 1, stay, H)
(E, 1, 0, left, A)
The input for this machine should be an empty tape.
This machine runs for more than 47 millions steps.
| #APL | APL | :Namespace Turing
⍝ Run Turing machine until it halts
∇r←RunTuring (rules init halts blank itape);state;rt;lt;next
state←init
lt←⍬
rt←,blank
:If 0≠≢itape ⋄ rt←itape ⋄ :EndIf
:While ~(⊂state)∊halts
next←((⊂state(⊃rt))≡¨↓rules[;⍳2])⌿rules
'No rule applies!'⎕SIGNAL(0=≢next)/11
(⊃rt)←⊃next[1;3]
state←⊃next[1;5]
:Select ⊃next[1;4]
:Case 'stay' ⋄ ⍝nothing
:Case 'right'
lt,⍨←⊃rt
rt←1↓rt
:If 0=≢rt ⋄ rt←,blank ⋄ :EndIf
:Case 'left'
:If 0=≢lt ⋄ lt←,blank ⋄ :EndIf
rt,⍨←⊃lt
lt←1↓lt
:Else
'Invalid action'⎕SIGNAL 11
:EndSelect
:EndWhile
r←(⌽lt),rt
∇
⍝ Display the resulting tape neatly
∇r←len Display t
r←(len⌊≢t)↑t
→(len≥≢t)/0
r,←'... (total length: ',(⍕≢t),')'
∇
⍝ Simple incrementer
∇t←∆1_SimpleIncrementer
t ←⊂'q0' '1' '1' 'right' 'q0'
t,←⊂'q0' 'B' '1' 'stay' 'qf'
t←(↑t) 'q0' (,⊂'qf') 'B' '111'
∇
⍝ Three state beaver
∇t←∆2_ThreeStateBeaver
t ←⊂'a' '0' '1' 'right' 'b'
t,←⊂'a' '1' '1' 'left' 'c'
t,←⊂'b' '0' '1' 'left' 'a'
t,←⊂'b' '1' '1' 'right' 'b'
t,←⊂'c' '0' '1' 'left' 'b'
t,←⊂'c' '1' '1' 'stay' 'halt'
t←(↑t) 'a' (,⊂'halt') '0' ''
∇
⍝ Five state beaver
∇t←∆3_FiveStateBeaver
t ←⊂'A' '0' '1' 'right' 'B'
t,←⊂'A' '1' '1' 'left' 'C'
t,←⊂'B' '0' '1' 'right' 'C'
t,←⊂'B' '1' '1' 'right' 'B'
t,←⊂'C' '0' '1' 'right' 'D'
t,←⊂'C' '1' '0' 'left' 'E'
t,←⊂'D' '0' '1' 'left' 'A'
t,←⊂'D' '1' '1' 'left' 'D'
t,←⊂'E' '0' '1' 'stay' 'H'
t,←⊂'E' '1' '0' 'left' 'A'
t←(↑t) 'A' (,⊂'H') '0' ''
∇
⍝ Run all of them and display the results
∇RunAll;m;ms
ms←('∆'=⊃¨ms)/ms←⎕NL¯3
:For m :In ms
⎕←(1↓m),': ',(32 Display RunTuring ⍎m)
:EndFor
∇
:EndNamespace |
http://rosettacode.org/wiki/Trigonometric_functions | Trigonometric functions | Task
If your language has a library or built-in functions for trigonometry, show examples of:
sine
cosine
tangent
inverses (of the above)
using the same angle in radians and degrees.
For the non-inverse functions, each radian/degree pair should use arguments that evaluate to the same angle (that is, it's not necessary to use the same angle for all three regular functions as long as the two sine calls use the same angle).
For the inverse functions, use the same number and convert its answer to radians and degrees.
If your language does not have trigonometric functions available or only has some available, write functions to calculate the functions based on any known approximation or identity.
| #ActionScript | ActionScript | trace("Radians:");
trace("sin(Pi/4) = ", Math.sin(Math.PI/4));
trace("cos(Pi/4) = ", Math.cos(Math.PI/4));
trace("tan(Pi/4) = ", Math.tan(Math.PI/4));
trace("arcsin(0.5) = ", Math.asin(0.5));
trace("arccos(0.5) = ", Math.acos(0.5));
trace("arctan(0.5) = ", Math.atan(0.5));
trace("arctan2(-1,-2) = ", Math.atan2(-1,-2));
trace("\nDegrees")
trace("sin(45) = ", Math.sin(45 * Math.PI/180));
trace("cos(45) = ", Math.cos(45 * Math.PI/180));
trace("tan(45) = ", Math.tan(45 * Math.PI/180));
trace("arcsin(0.5) = ", Math.asin(0.5)*180/Math.PI);
trace("arccos(0.5) = ", Math.acos(0.5)*180/Math.PI);
trace("arctan(0.5) = ", Math.atan(0.5)*180/Math.PI);
trace("arctan2(-1,-2) = ", Math.atan2(-1,-2)*180/Math.PI);
|
http://rosettacode.org/wiki/Trabb_Pardo%E2%80%93Knuth_algorithm | Trabb Pardo–Knuth algorithm | The TPK algorithm is an early example of a programming chrestomathy.
It was used in Donald Knuth and Luis Trabb Pardo's Stanford tech report The Early Development of Programming Languages.
The report traces the early history of work in developing computer languages in the 1940s and 1950s, giving several translations of the algorithm.
From the wikipedia entry:
ask for 11 numbers to be read into a sequence S
reverse sequence S
for each item in sequence S
result := call a function to do an operation
if result overflows
alert user
else
print result
The task is to implement the algorithm:
Use the function:
f
(
x
)
=
|
x
|
0.5
+
5
x
3
{\displaystyle f(x)=|x|^{0.5}+5x^{3}}
The overflow condition is an answer of greater than 400.
The 'user alert' should not stop processing of other items of the sequence.
Print a prompt before accepting eleven, textual, numeric inputs.
You may optionally print the item as well as its associated result, but the results must be in reverse order of input.
The sequence S may be 'implied' and so not shown explicitly.
Print and show the program in action from a typical run here. (If the output is graphical rather than text then either add a screendump or describe textually what is displayed).
| #Agena | Agena | scope # TPK algorithm in Agena
local y;
local a := [];
local f := proc( t :: number ) is return sqrt(abs(t))+5*t*t*t end;
for i from 0 to 10 do a[i] := tonumber( io.read() ) od;
for i from 10 to 0 by - 1 do
y:=f(a[i]);
if y > 400
then print( "TOO LARGE" )
else printf( "%10.4f\n", y )
fi
od
epocs |
http://rosettacode.org/wiki/Trabb_Pardo%E2%80%93Knuth_algorithm | Trabb Pardo–Knuth algorithm | The TPK algorithm is an early example of a programming chrestomathy.
It was used in Donald Knuth and Luis Trabb Pardo's Stanford tech report The Early Development of Programming Languages.
The report traces the early history of work in developing computer languages in the 1940s and 1950s, giving several translations of the algorithm.
From the wikipedia entry:
ask for 11 numbers to be read into a sequence S
reverse sequence S
for each item in sequence S
result := call a function to do an operation
if result overflows
alert user
else
print result
The task is to implement the algorithm:
Use the function:
f
(
x
)
=
|
x
|
0.5
+
5
x
3
{\displaystyle f(x)=|x|^{0.5}+5x^{3}}
The overflow condition is an answer of greater than 400.
The 'user alert' should not stop processing of other items of the sequence.
Print a prompt before accepting eleven, textual, numeric inputs.
You may optionally print the item as well as its associated result, but the results must be in reverse order of input.
The sequence S may be 'implied' and so not shown explicitly.
Print and show the program in action from a typical run here. (If the output is graphical rather than text then either add a screendump or describe textually what is displayed).
| #ALGOL_60 | ALGOL 60 | begin
integer i; real y; real array a[0:10];
real procedure f(t); value t; real t;
f:=sqrt(abs(t))+5*t^3;
for i:=0 step 1 until 10 do inreal(0, a[i]);
for i:=10 step -1 until 0 do
begin
y:=f(a[i]);
if y > 400 then outstring(1, "TOO LARGE")
else outreal(1,y);
outchar(1, "\n", 1)
end
end |
http://rosettacode.org/wiki/Tree_from_nesting_levels | Tree from nesting levels | Given a flat list of integers greater than zero, representing object nesting
levels, e.g. [1, 2, 4],
generate a tree formed from nested lists of those nesting level integers where:
Every int appears, in order, at its depth of nesting.
If the next level int is greater than the previous then it appears in a sub-list of the list containing the previous item
The generated tree data structure should ideally be in a languages nested list format that can
be used for further calculations rather than something just calculated for printing.
An input of [1, 2, 4] should produce the equivalent of: [1, [2, [[4]]]]
where 1 is at depth1, 2 is two deep and 4 is nested 4 deep.
[1, 2, 4, 2, 2, 1] should produce [1, [2, [[4]], 2, 2], 1].
All the nesting integers are in the same order but at the correct nesting
levels.
Similarly [3, 1, 3, 1] should generate [[[3]], 1, [[3]], 1]
Task
Generate and show here the results for the following inputs:
[]
[1, 2, 4]
[3, 1, 3, 1]
[1, 2, 3, 1]
[3, 2, 1, 3]
[3, 3, 3, 1, 1, 3, 3, 3]
| #Phix | Phix | function test(sequence s, integer level=1, idx=1)
sequence res = {}, part
while idx<=length(s) do
switch compare(s[idx],level) do
case +1: {idx,part} = test(s,level+1,idx)
res = append(res,part)
case 0: res &= s[idx]
case -1: idx -= 1 exit
end switch
idx += 1
end while
return iff(level=1?res:{idx,res})
end function
constant tests = {{},
{1, 2, 4},
-- {1, 2, 4, 2, 2, 1}, -- (fine too)
{3, 1, 3, 1},
{1, 2, 3, 1},
{3, 2, 1, 3},
{3, 3, 3, 1, 1, 3, 3, 3}}
for i=1 to length(tests) do
sequence ti = tests[i],
res = test(ti),
rpp = ppf(res,{pp_Nest,3,pp_Indent,4})
printf(1,"%v nests to %v\n or %s\n",{ti,res,rpp})
end for
|
http://rosettacode.org/wiki/Twelve_statements | Twelve statements | This puzzle is borrowed from math-frolic.blogspot.
Given the following twelve statements, which of them are true?
1. This is a numbered list of twelve statements.
2. Exactly 3 of the last 6 statements are true.
3. Exactly 2 of the even-numbered statements are true.
4. If statement 5 is true, then statements 6 and 7 are both true.
5. The 3 preceding statements are all false.
6. Exactly 4 of the odd-numbered statements are true.
7. Either statement 2 or 3 is true, but not both.
8. If statement 7 is true, then 5 and 6 are both true.
9. Exactly 3 of the first 6 statements are true.
10. The next two statements are both true.
11. Exactly 1 of statements 7, 8 and 9 are true.
12. Exactly 4 of the preceding statements are true.
Task
When you get tired of trying to figure it out in your head,
write a program to solve it, and print the correct answer or answers.
Extra credit
Print out a table of near misses, that is, solutions that are contradicted by only a single statement.
| #Go | Go | package main
import "fmt"
// its' not too much more work to check all the permutations concurrently
var solution = make(chan int)
var nearMiss = make(chan int)
var done = make(chan bool)
func main() {
// iterate and use the bits as the permutation
for i := 0; i < 4096; i++ {
go checkPerm(i)
}
// collect the misses and list them after the complete solution(s)
var ms []int
for i := 0; i < 4096; {
select {
case <-done:
i++
case s := <-solution:
print12("solution", s)
case m := <-nearMiss:
ms = append(ms, m)
}
}
for _, m := range ms {
print12("near miss", m)
}
}
func print12(label string, bits int) {
fmt.Print(label, ":")
for i := 1; i <= 12; i++ {
if bits&1 == 1 {
fmt.Print(" ", i)
}
bits >>= 1
}
fmt.Println()
}
func checkPerm(tz int) {
// closure returns true if tz bit corresponding to
// 1-based statement number is 1.
ts := func(n uint) bool {
return tz>>(n-1)&1 == 1
}
// variadic closure returns number of statements listed as arguments
// which have corresponding tz bit == 1.
ntrue := func(xs ...uint) int {
nt := 0
for _, x := range xs {
if ts(x) {
nt++
}
}
return nt
}
// a flag used on repeated calls to test.
// set to true when first contradiction is found.
// if another is found, this function (checkPerm) can "short circuit"
// and return immediately without checking additional statements.
var con bool
// closure called to test each statement
test := func(statement uint, b bool) {
switch {
case ts(statement) == b:
case con:
panic("bail")
default:
con = true
}
}
// short circuit mechanism
defer func() {
if x := recover(); x != nil {
if msg, ok := x.(string); !ok && msg != "bail" {
panic(x)
}
}
done <- true
}()
// 1. This is a numbered list of twelve statements.
test(1, true)
// 2. Exactly 3 of the last 6 statements are true.
test(2, ntrue(7, 8, 9, 10, 11, 12) == 3)
// 3. Exactly 2 of the even-numbered statements are true.
test(3, ntrue(2, 4, 6, 8, 10, 12) == 2)
// 4. If statement 5 is true, then statements 6 and 7 are both true.
test(4, !ts(5) || ts(6) && ts(7))
// 5. The 3 preceding statements are all false.
test(5, !ts(4) && !ts(3) && !ts(2))
// 6. Exactly 4 of the odd-numbered statements are true.
test(6, ntrue(1, 3, 5, 7, 9, 11) == 4)
// 7. Either statement 2 or 3 is true, but not both.
test(7, ts(2) != ts(3))
// 8. If statement 7 is true, then 5 and 6 are both true.
test(8, !ts(7) || ts(5) && ts(6))
// 9. Exactly 3 of the first 6 statements are true.
test(9, ntrue(1, 2, 3, 4, 5, 6) == 3)
// 10. The next two statements are both true.
test(10, ts(11) && ts(12))
// 11. Exactly 1 of statements 7, 8 and 9 are true.
test(11, ntrue(7, 8, 9) == 1)
// 12. Exactly 4 of the preceding statements are true.
test(12, ntrue(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11) == 4)
// no short circuit? send permutation as either near miss or solution
if con {
nearMiss <- tz
} else {
solution <- tz
}
} |
http://rosettacode.org/wiki/Truth_table | Truth table | A truth table is a display of the inputs to, and the output of a Boolean function organized as a table where each row gives one combination of input values and the corresponding value of the function.
Task
Input a Boolean function from the user as a string then calculate and print a formatted truth table for the given function.
(One can assume that the user input is correct).
Print and show output for Boolean functions of two and three input variables, but any program should not be limited to that many variables in the function.
Either reverse-polish or infix notation expressions are allowed.
Related tasks
Boolean values
Ternary logic
See also
Wolfram MathWorld entry on truth tables.
some "truth table" examples from Google.
| #Go | Go | package main
import (
"bufio"
"errors"
"fmt"
"go/ast"
"go/parser"
"go/token"
"os"
"reflect"
)
func main() {
in := bufio.NewScanner(os.Stdin)
for {
fmt.Print("Expr: ")
in.Scan()
if err := in.Err(); err != nil {
fmt.Println(err)
return
}
if !tt(in.Text()) {
return
}
}
}
func tt(expr string) bool {
// call library parser
tree, err := parser.ParseExpr(expr)
if err != nil {
fmt.Println(err)
return false
}
// create handy object to pass around
e := &evaluator{nil, map[string]bool{}, tree}
// library tree traversal function calls e.Visit for each node.
// use this to collect variables of the expression.
ast.Walk(e, tree)
// print headings for truth table
for _, n := range e.names {
fmt.Printf("%-6s", n)
}
fmt.Println(" ", expr)
// start recursive table generation function on first variable
e.evalVar(0)
return true
}
type evaluator struct {
names []string // variables, in order of appearance
val map[string]bool // map variables to boolean values
tree ast.Expr // parsed expression as ast
}
// visitor function called by library Walk function.
// builds a list of unique variable names.
func (e *evaluator) Visit(n ast.Node) ast.Visitor {
if id, ok := n.(*ast.Ident); ok {
if !e.val[id.Name] {
e.names = append(e.names, id.Name)
e.val[id.Name] = true
}
}
return e
}
// method recurses for each variable of the truth table, assigning it to
// false, then true. At bottom of recursion, when all variables are
// assigned, it evaluates the expression and outputs one line of the
// truth table
func (e *evaluator) evalVar(nx int) bool {
if nx == len(e.names) {
// base case
v, err := evalNode(e.tree, e.val)
if err != nil {
fmt.Println(" ", err)
return false
}
// print variable values
for _, n := range e.names {
fmt.Printf("%-6t", e.val[n])
}
// print expression value
fmt.Println(" ", v)
return true
}
// recursive case
for _, v := range []bool{false, true} {
e.val[e.names[nx]] = v
if !e.evalVar(nx + 1) {
return false
}
}
return true
}
// recursively evaluate ast
func evalNode(nd ast.Node, val map[string]bool) (bool, error) {
switch n := nd.(type) {
case *ast.Ident:
return val[n.Name], nil
case *ast.BinaryExpr:
x, err := evalNode(n.X, val)
if err != nil {
return false, err
}
y, err := evalNode(n.Y, val)
if err != nil {
return false, err
}
switch n.Op {
case token.AND:
return x && y, nil
case token.OR:
return x || y, nil
case token.XOR:
return x != y, nil
default:
return unsup(n.Op)
}
case *ast.UnaryExpr:
x, err := evalNode(n.X, val)
if err != nil {
return false, err
}
switch n.Op {
case token.XOR:
return !x, nil
default:
return unsup(n.Op)
}
case *ast.ParenExpr:
return evalNode(n.X, val)
}
return unsup(reflect.TypeOf(nd))
}
func unsup(i interface{}) (bool, error) {
return false, errors.New(fmt.Sprintf("%v unsupported", i))
}
|
http://rosettacode.org/wiki/Ulam_spiral_(for_primes) | Ulam spiral (for primes) | An Ulam spiral (of primes) is a method of visualizing primes when expressed in a (normally counter-clockwise) outward spiral (usually starting at 1), constructed on a square grid, starting at the "center".
An Ulam spiral is also known as a prime spiral.
The first grid (green) is shown with sequential integers, starting at 1.
In an Ulam spiral of primes, only the primes are shown (usually indicated by some glyph such as a dot or asterisk), and all non-primes as shown as a blank (or some other whitespace).
Of course, the grid and border are not to be displayed (but they are displayed here when using these Wiki HTML tables).
Normally, the spiral starts in the "center", and the 2nd number is to the viewer's right and the number spiral starts from there in a counter-clockwise direction.
There are other geometric shapes that are used as well, including clock-wise spirals.
Also, some spirals (for the 2nd number) is viewed upwards from the 1st number instead of to the right, but that is just a matter of orientation.
Sometimes, the starting number can be specified to show more visual striking patterns (of prime densities).
[A larger than necessary grid (numbers wise) is shown here to illustrate the pattern of numbers on the diagonals (which may be used by the method to orientate the direction of spiral-construction algorithm within the example computer programs)].
Then, in the next phase in the transformation of the Ulam prime spiral, the non-primes are translated to blanks.
In the orange grid below, the primes are left intact, and all non-primes are changed to blanks.
Then, in the final transformation of the Ulam spiral (the yellow grid), translate the primes to a glyph such as a • or some other suitable glyph.
65
64
63
62
61
60
59
58
57
66
37
36
35
34
33
32
31
56
67
38
17
16
15
14
13
30
55
68
39
18
5
4
3
12
29
54
69
40
19
6
1
2
11
28
53
70
41
20
7
8
9
10
27
52
71
42
21
22
23
24
25
26
51
72
43
44
45
46
47
48
49
50
73
74
75
76
77
78
79
80
81
61
59
37
31
67
17
13
5
3
29
19
2
11
53
41
7
71
23
43
47
73
79
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
The Ulam spiral becomes more visually obvious as the grid increases in size.
Task
For any sized N × N grid, construct and show an Ulam spiral (counter-clockwise) of primes starting at some specified initial number (the default would be 1), with some suitably dotty (glyph) representation to indicate primes, and the absence of dots to indicate non-primes.
You should demonstrate the generator by showing at Ulam prime spiral large enough to (almost) fill your terminal screen.
Related tasks
Spiral matrix
Zig-zag matrix
Identity matrix
Sequence of primes by Trial Division
See also
Wikipedia entry: Ulam spiral
MathWorld™ entry: Prime Spiral
| #J | J | spiral =: ,~ $ [: /: }.@(2 # >:@i.@-) +/\@# <:@+: $ (, -)@(1&,) |
http://rosettacode.org/wiki/Truncatable_primes | Truncatable primes | A truncatable prime is a prime number that when you successively remove digits from one end of the prime, you are left with a new prime number.
Examples
The number 997 is called a left-truncatable prime as the numbers 997, 97, and 7 are all prime.
The number 7393 is a right-truncatable prime as the numbers 7393, 739, 73, and 7 formed by removing digits from its right are also prime.
No zeroes are allowed in truncatable primes.
Task
The task is to find the largest left-truncatable and right-truncatable primes less than one million (base 10 is implied).
Related tasks
Find largest left truncatable prime in a given base
Sieve of Eratosthenes
See also
Truncatable Prime from MathWorld.]
| #Clojure | Clojure | (use '[clojure.contrib.lazy-seqs :only [primes]])
(def prime?
(let [mem (ref #{})
primes (ref primes)]
(fn [n]
(dosync
(if (< n (first @primes))
(@mem n)
(let [[mems ss] (split-with #(<= % n) @primes)]
(ref-set primes ss)
((commute mem into mems) n)))))))
(defn drop-lefts [n]
(let [dropl #(if (< % 10) 0 (Integer. (subs (str %) 1)))]
(->> (iterate dropl n)
(take-while pos? ,)
next)))
(defn drop-rights [n]
(->> (iterate #(quot % 10) n)
next
(take-while pos? ,)))
(defn truncatable-left? [n]
(every? prime? (drop-lefts n)))
(defn truncatable-right? [n]
(every? prime? (drop-rights n)))
user> (->> (for [p primes
:while (< p 1000000)
:when (not-any? #{\0} (str p))
:let [l? (if (truncatable-left? p) p 0)
r? (if (truncatable-right? p) p 0)]
:when (or l? r?)]
[l? r?])
((juxt #(apply max-key first %) #(apply max-key second %)) ,)
((juxt ffirst (comp second second)) ,)
(map vector ["left truncatable: " "right truncatable: "] ,))
(["left truncatable: " 998443] ["right truncatable: " 739399]) |
http://rosettacode.org/wiki/Unbias_a_random_generator | Unbias a random generator |
P
1
{\displaystyle P_{1}}
P
0
{\displaystyle P_{0}}
P
1
{\displaystyle P_{1}}
P
0
{\displaystyle P_{0}}
P
0
{\displaystyle P_{0}}
P
1
{\displaystyle P_{1}}
Task details
Use your language's random number generator to create a function/method/subroutine/... randN that returns a one or a zero, but with one occurring, on average, 1 out of N times, where N is an integer from the range 3 to 6 inclusive.
Create a function unbiased that uses only randN as its source of randomness to become an unbiased generator of random ones and zeroes.
For N over its range, generate and show counts of the outputs of randN and unbiased(randN).
The actual unbiasing should be done by generating two numbers at a time from randN and only returning a 1 or 0 if they are different. As long as you always return the first number or always return the second number, the probabilities discussed above should take over the biased probability of randN.
This task is an implementation of Von Neumann debiasing, first described in a 1951 paper.
| #Wren | Wren | import "random" for Random
import "/fmt" for Fmt
var rand = Random.new()
var biased = Fn.new { |n| rand.float() < 1 / n }
var unbiased = Fn.new { |n|
while (true) {
var a = biased.call(n)
var b = biased.call(n)
if (a != b) return a
}
}
var m = 50000
var f = "$d: $2.2f\% $2.2f\%"
for (n in 3..6) {
var c1 = 0
var c2 = 0
for (i in 0...m) {
if (biased.call(n)) c1 = c1 + 1
if (unbiased.call(n)) c2 = c2 + 1
}
Fmt.print(f, n, 100 * c1 / m, 100 * c2 / m)
} |
http://rosettacode.org/wiki/Truncate_a_file | Truncate a file | Task
Truncate a file to a specific length. This should be implemented as a routine that takes two parameters: the filename and the required file length (in bytes).
Truncation can be achieved using system or library calls intended for such a task, if such methods exist, or by creating a temporary file of a reduced size and renaming it, after first deleting the original file, if no other method is available. The file may contain non human readable binary data in an unspecified format, so the routine should be "binary safe", leaving the contents of the untruncated part of the file unchanged.
If the specified filename does not exist, or the provided length is not less than the current file length, then the routine should raise an appropriate error condition.
On some systems, the provided file truncation facilities might not change the file or may extend the file, if the specified length is greater than the current length of the file.
This task permits the use of such facilities. However, such behaviour should be noted, or optionally a warning message relating to an non change or increase in file size may be implemented.
| #PowerBASIC | PowerBASIC | SUB truncateFile (file AS STRING, length AS DWORD)
IF LEN(DIR$(file)) THEN
DIM f AS LONG
f = FREEFILE
OPEN file FOR BINARY AS f
IF length > LOF(f) THEN
CLOSE f
ERROR 62 'Input past end
ELSE
SEEK f, length + 1
SETEOF f
CLOSE f
END IF
ELSE
ERROR 53 'File not found
END IF
END SUB |
http://rosettacode.org/wiki/Truncate_a_file | Truncate a file | Task
Truncate a file to a specific length. This should be implemented as a routine that takes two parameters: the filename and the required file length (in bytes).
Truncation can be achieved using system or library calls intended for such a task, if such methods exist, or by creating a temporary file of a reduced size and renaming it, after first deleting the original file, if no other method is available. The file may contain non human readable binary data in an unspecified format, so the routine should be "binary safe", leaving the contents of the untruncated part of the file unchanged.
If the specified filename does not exist, or the provided length is not less than the current file length, then the routine should raise an appropriate error condition.
On some systems, the provided file truncation facilities might not change the file or may extend the file, if the specified length is greater than the current length of the file.
This task permits the use of such facilities. However, such behaviour should be noted, or optionally a warning message relating to an non change or increase in file size may be implemented.
| #Powershell | Powershell | Function Truncate-File(fname) {
$null | Set-Content -Path "$fname"
}
|
http://rosettacode.org/wiki/Truncate_a_file | Truncate a file | Task
Truncate a file to a specific length. This should be implemented as a routine that takes two parameters: the filename and the required file length (in bytes).
Truncation can be achieved using system or library calls intended for such a task, if such methods exist, or by creating a temporary file of a reduced size and renaming it, after first deleting the original file, if no other method is available. The file may contain non human readable binary data in an unspecified format, so the routine should be "binary safe", leaving the contents of the untruncated part of the file unchanged.
If the specified filename does not exist, or the provided length is not less than the current file length, then the routine should raise an appropriate error condition.
On some systems, the provided file truncation facilities might not change the file or may extend the file, if the specified length is greater than the current length of the file.
This task permits the use of such facilities. However, such behaviour should be noted, or optionally a warning message relating to an non change or increase in file size may be implemented.
| #PureBasic | PureBasic | Procedure SetFileSize(File$, length.q)
Protected fh, pos, i
If FileSize(File$) < length
Debug "File to small, is a directory or does not exist."
ProcedureReturn #False
Else
fh = OpenFile(#PB_Any, File$)
FileSeek(fh, length)
TruncateFile(fh)
CloseFile(fh)
EndIf
ProcedureReturn #True
EndProcedure |
http://rosettacode.org/wiki/Universal_Turing_machine | Universal Turing machine | One of the foundational mathematical constructs behind computer science
is the universal Turing Machine.
(Alan Turing introduced the idea of such a machine in 1936–1937.)
Indeed one way to definitively prove that a language
is turing-complete
is to implement a universal Turing machine in it.
Task
Simulate such a machine capable
of taking the definition of any other Turing machine and executing it.
Of course, you will not have an infinite tape,
but you should emulate this as much as is possible.
The three permissible actions on the tape are "left", "right" and "stay".
To test your universal Turing machine (and prove your programming language
is Turing complete!), you should execute the following two Turing machines
based on the following definitions.
Simple incrementer
States: q0, qf
Initial state: q0
Terminating states: qf
Permissible symbols: B, 1
Blank symbol: B
Rules:
(q0, 1, 1, right, q0)
(q0, B, 1, stay, qf)
The input for this machine should be a tape of 1 1 1
Three-state busy beaver
States: a, b, c, halt
Initial state: a
Terminating states: halt
Permissible symbols: 0, 1
Blank symbol: 0
Rules:
(a, 0, 1, right, b)
(a, 1, 1, left, c)
(b, 0, 1, left, a)
(b, 1, 1, right, b)
(c, 0, 1, left, b)
(c, 1, 1, stay, halt)
The input for this machine should be an empty tape.
Bonus:
5-state, 2-symbol probable Busy Beaver machine from Wikipedia
States: A, B, C, D, E, H
Initial state: A
Terminating states: H
Permissible symbols: 0, 1
Blank symbol: 0
Rules:
(A, 0, 1, right, B)
(A, 1, 1, left, C)
(B, 0, 1, right, C)
(B, 1, 1, right, B)
(C, 0, 1, right, D)
(C, 1, 0, left, E)
(D, 0, 1, left, A)
(D, 1, 1, left, D)
(E, 0, 1, stay, H)
(E, 1, 0, left, A)
The input for this machine should be an empty tape.
This machine runs for more than 47 millions steps.
| #AutoHotkey | AutoHotkey | ; By Uberi, http://www.autohotkey.com/board/topic/58599-turing-machine/
SetBatchLines, -1
OnExit, Exit
SaveFilePath := A_ScriptFullPath ".ini"
; Defaults are for a 2-state_3-symbol turning machine. Format:
; machine state symbol on tape, symbol on tape | tape shift (- is left, + is right, 0 is halt) | machine state
, Rule1 := "A0,1|1|B"
, Rule2 := "A1,2|-1|A"
, Rule3 := "A2,1|-1|A"
, Rule4 := "B0,2|-1|A"
, Rule5 := "B1,2|1|B"
, Rule6 := "B2,0|1|A"
; no error check is run on this input, so be sure states and symbols align with actions
IniRead, UseSaveFile, %SaveFilePath%, Global, UseSaveFile, 1 ; on exit, save state to text file so I can resume on next run
IniRead, MaxIterations, %SaveFilePath%, Global, MaxIterations, 100000 ; set as %A_Space% to run indefinitely
IniRead, Section, %SaveFilePath%, Global, Section, 2-state_3-symbol ; The name of the machine to run. Options defined:
; 2-state_3-symbol
; Simple_incrementer
; Three-state_busy_beaver
; Probable_busy_beaver_Wikipedia
IniRead, States, %SaveFilePath%, %Section%, States, A|B ; valid states
IniRead, InitialState, %SaveFilePath%, %Section%, InitialState, A ; start state
IniRead, TerminalState, %SaveFilePath%, %Section%, TerminalState, C ; end state
IniRead, Symbols, %SaveFilePath%, %Section%, Symbols, 0,1,2 ; valid symbols
IniRead, DefaultCell, %SaveFilePath%, %Section%, DefaultCell, 0 ; the default symbol of any cell not defined on input tape
IniRead, ProgramCode, %SaveFilePath%, %Section%, ProgramCode, 10101|01010 ; start tape
Iniread, RuleCount, %SaveFilePath%, %Section%, RuleCount, 6 ; number of actions to read
Loop, %RuleCount%
{
IniRead, Temp1, %SaveFilePath%, %Section%, Rule%A_Index%, % Rule%A_Index%
StringSplit, Temp, Temp1, `,
Action%Temp1% := Temp2
}
IniRead, Index, %SaveFilePath%, SavedState, Index, 0
IniRead, IterationCount, %SaveFilePath%, SavedState, IterationCount, 0
IniRead, State, %SaveFilePath%, SavedState, State, %InitialState%
If IterationCount > 0
IniRead, ProgramCode, %SaveFilePath%, SavedState, ProgramCode, %ProgramCode%
IfNotInString, ProgramCode, |
ProgramCode := "|" ProgramCode
StringSplit, Temp, ProgramCode, |
NegativeCells := Temp1, PositiveCells := Temp2
Loop, Parse, Symbols, |
Color%A_LoopField% := hex(mod((A_Index+1/(2**((A_Index-1)//7))-1)/7,1)*16777215) ; unlimited number of unique colors
Color%DefaultCell% := "White"
Gui, Color, Black
Gui, +ToolWindow +AlwaysOnTop +LastFound -Caption
WindowID := WinExist()
OnMessage(0x201, "WM_LBUTTONDOWN")
Gui, Font, s6 cWhite, Arial
Loop, 61 ; display 30 cell symbols on each side of current index
{
Temp1 := ((A_Index - 1) * 15) + 1
Gui, Add, Progress, x%Temp1% y1 w14 h40 vCell%A_Index% BackgroundWhite
Gui, Add, Text, x%Temp1% y42 w15 h10 vLabel%A_Index% Center
}
Gui, Add, Text, x2 y54 w26 h10 vState
Gui, Add, Text, x35 y54 w50 h10 vCurrentCell
Gui, Add, Text, x350 y54 w158 h10 vActions
Gui, Add, Text, x844 y54 w33 h10, Iterations:
Gui, Add, Text, x884 y54 w29 h10 vIterations Right
Gui, Font, s4 cWhite Bold, Arial
Gui, Add, Text, x450 y1 w15 h10 Center, V
GuiControl, Move, Cell31, x451 y8 w14 h33
Gui, Show, y20 w916 h64, Wolfram's 2-State 3-Symbol Turing Machine ;'
;MaxIndex := ProgramOffset + StrLen(ProgramCode), MinIndex := ProgramOffset ; not implemented
While, ((MaxIterations = "") || IterationCount <= MaxIterations) ; process until limit is reached, if any
{
Loop, 61 ; color each cell per its current symbol
{ ; must run for all displayed cells because they are not directly mapped to shifting tape
TempIndex := (Index + A_Index) - 31
GuiControl, , Label%A_Index%, %TempIndex%
CellColor := CellGet(TempIndex)
, CellColor := Color%CellColor%
GuiControl, +Background%CellColor%, Cell%A_Index%
}
CurrentCell := CellGet(Index)
GuiControl, , State, State: %State%
GuiControl, , CurrentCell, Current Cell: %CurrentCell%
GuiControl, , Iterations, %IterationCount%
If (State = TerminalState)
Break
StringSplit, Temp, Action%State%%CurrentCell%, |
GuiControl, , Actions, % "Actions: Print " . Temp1 . ", Move " . ((Temp2 = -1) ? "left" : "right") . ", " . ((State <> Temp3) ? "Switch to state " . Temp3 : "Do not switch state")
IterationCount++
, CellPut(Index,Temp1)
, Index += Temp2
, State := Temp3
;, (Index > MaxIndex) ? MaxIndex := Index : ""
;, (Index < MinIndex) ? MinIndex := Index : ""
Sleep, 0.1*1000
}
MsgBox, 64, Complete, Completed %IterationCount% iterations of the Turing machine.
Return
; Hotkeys and functions:
~Pause::Pause
GuiEscape:
GuiClose:
ExitApp
Exit:
If UseSaveFile
{
IniWrite, %Index%, %SaveFilePath%, %Section%, Index
IniWrite, %IterationCount%, %SaveFilePath%, %Section%, IterationCount
IniWrite, %State%, %SaveFilePath%, %Section%, State
IniWrite, %NegativeCells%|%PositiveCells%, %SaveFilePath%, %Section%, ProgramCode
}
ExitApp
CellGet(Index)
{
global NegativeCells, PositiveCells, DefaultCell
Temp1 := (Index < 0) ? SubStr(NegativeCells,Abs(Index),1) : SubStr(PositiveCells,Index + 1,1)
Return, (Temp1 = "") ? DefaultCell : Temp1
}
CellPut(Index,Char)
{
global NegativeCells, PositiveCells, DefaultCell
static StrGetFunc := "StrGet" ; workaround to hide function from AHK Basic (which does not have or require it)
CharType := A_IsUnicode ? "UShort" : "UChar"
, (Index < 0)
? (Index := 0 - Index
, Temp1 := Index - StrLen(NegativeCells)
, (Temp1 > 0)
? (VarSetCapacity(Pad,64) ; these three functions are quirks in AHK's memory management (not required)
, VarSetCapacity(Pad,0)
, VarSetCapacity(Pad,Temp1,Asc(DefaultCell))
, NegativeCells .= A_IsUnicode ? %StrGetFunc%(&Pad,Temp1,"CP0") : Pad)
: ""
, NumPut(Asc(Char),NegativeCells,(Index - 1) << !!A_IsUnicode,CharType) )
: (Temp1 := Index - StrLen(PositiveCells) + 1
, (Temp1 > 0)
? (VarSetCapacity(Pad,64) ; these three functions are quirks in AHK's memory management (not required)
, VarSetCapacity(Pad,0)
, VarSetCapacity(Pad,Temp1,Asc(DefaultCell))
, PositiveCells .= A_IsUnicode ? %StrGetFunc%(&Pad,Temp1,"CP0") : Pad)
: ""
, NumPut(Asc(Char),PositiveCells,Index << !!A_IsUnicode,CharType) )
}
Hex(p_Integer)
{
PtrType:=(A_PtrSize=8) ? "Ptr":"UInt"
l_Format:="`%0" . 6 . "I64X"
VarSetCapacity(l_Argument,8)
NumPut(p_Integer,l_Argument,0,"Int64")
VarSetCapacity(l_Buffer,A_IsUnicode ? 12:6,0)
DllCall(A_IsUnicode ? "msvcrt\_vsnwprintf":"msvcrt\_vsnprintf"
,"Str",l_Buffer ;-- Storage location for output
,"UInt",6 ;-- Maximum number of characters to write
,"Str",l_Format ;-- Format specification
,PtrType,&l_Argument) ;-- Argument
Return l_Buffer
}
WM_LBUTTONDOWN()
{
If (A_Gui = 1)
PostMessage, 0xA1, 2
} |
http://rosettacode.org/wiki/Trigonometric_functions | Trigonometric functions | Task
If your language has a library or built-in functions for trigonometry, show examples of:
sine
cosine
tangent
inverses (of the above)
using the same angle in radians and degrees.
For the non-inverse functions, each radian/degree pair should use arguments that evaluate to the same angle (that is, it's not necessary to use the same angle for all three regular functions as long as the two sine calls use the same angle).
For the inverse functions, use the same number and convert its answer to radians and degrees.
If your language does not have trigonometric functions available or only has some available, write functions to calculate the functions based on any known approximation or identity.
| #Ada | Ada | with Ada.Numerics.Elementary_Functions;
use Ada.Numerics.Elementary_Functions;
with Ada.Float_Text_Io; use Ada.Float_Text_Io;
with Ada.Text_IO; use Ada.Text_IO;
procedure Trig is
Degrees_Cycle : constant Float := 360.0;
Radians_Cycle : constant Float := 2.0 * Ada.Numerics.Pi;
Angle_Degrees : constant Float := 45.0;
Angle_Radians : constant Float := Ada.Numerics.Pi / 4.0;
procedure Put (V1, V2 : Float) is
begin
Put (V1, Aft => 5, Exp => 0);
Put (" ");
Put (V2, Aft => 5, Exp => 0);
New_Line;
end Put;
begin
Put (Sin (Angle_Degrees, Degrees_Cycle),
Sin (Angle_Radians, Radians_Cycle));
Put (Cos (Angle_Degrees, Degrees_Cycle),
Cos (Angle_Radians, Radians_Cycle));
Put (Tan (Angle_Degrees, Degrees_Cycle),
Tan (Angle_Radians, Radians_Cycle));
Put (Cot (Angle_Degrees, Degrees_Cycle),
Cot (Angle_Radians, Radians_Cycle));
Put (ArcSin (Sin (Angle_Degrees, Degrees_Cycle), Degrees_Cycle),
ArcSin (Sin (Angle_Radians, Radians_Cycle), Radians_Cycle));
Put (Arccos (Cos (Angle_Degrees, Degrees_Cycle), Degrees_Cycle),
Arccos (Cos (Angle_Radians, Radians_Cycle), Radians_Cycle));
Put (Arctan (Y => Tan (Angle_Degrees, Degrees_Cycle)),
Arctan (Y => Tan (Angle_Radians, Radians_Cycle)));
Put (Arccot (X => Cot (Angle_Degrees, Degrees_Cycle)),
Arccot (X => Cot (Angle_Degrees, Degrees_Cycle)));
end Trig; |
http://rosettacode.org/wiki/Trabb_Pardo%E2%80%93Knuth_algorithm | Trabb Pardo–Knuth algorithm | The TPK algorithm is an early example of a programming chrestomathy.
It was used in Donald Knuth and Luis Trabb Pardo's Stanford tech report The Early Development of Programming Languages.
The report traces the early history of work in developing computer languages in the 1940s and 1950s, giving several translations of the algorithm.
From the wikipedia entry:
ask for 11 numbers to be read into a sequence S
reverse sequence S
for each item in sequence S
result := call a function to do an operation
if result overflows
alert user
else
print result
The task is to implement the algorithm:
Use the function:
f
(
x
)
=
|
x
|
0.5
+
5
x
3
{\displaystyle f(x)=|x|^{0.5}+5x^{3}}
The overflow condition is an answer of greater than 400.
The 'user alert' should not stop processing of other items of the sequence.
Print a prompt before accepting eleven, textual, numeric inputs.
You may optionally print the item as well as its associated result, but the results must be in reverse order of input.
The sequence S may be 'implied' and so not shown explicitly.
Print and show the program in action from a typical run here. (If the output is graphical rather than text then either add a screendump or describe textually what is displayed).
| #ALGOL_68 | ALGOL 68 | [ 0 : 10 ]REAL a;
PROC f = ( REAL t )REAL:
sqrt(ABS t)+5*t*t*t;
FOR i FROM LWB a TO UPB a DO read( ( a[ i ] ) ) OD;
FOR i FROM UPB a BY -1 TO LWB a DO
REAL y=f(a[i]);
IF y > 400 THEN print( ( "TOO LARGE", newline ) )
ELSE print( ( fixed( y, -9, 4 ), newline ) )
FI
OD |
http://rosettacode.org/wiki/Trabb_Pardo%E2%80%93Knuth_algorithm | Trabb Pardo–Knuth algorithm | The TPK algorithm is an early example of a programming chrestomathy.
It was used in Donald Knuth and Luis Trabb Pardo's Stanford tech report The Early Development of Programming Languages.
The report traces the early history of work in developing computer languages in the 1940s and 1950s, giving several translations of the algorithm.
From the wikipedia entry:
ask for 11 numbers to be read into a sequence S
reverse sequence S
for each item in sequence S
result := call a function to do an operation
if result overflows
alert user
else
print result
The task is to implement the algorithm:
Use the function:
f
(
x
)
=
|
x
|
0.5
+
5
x
3
{\displaystyle f(x)=|x|^{0.5}+5x^{3}}
The overflow condition is an answer of greater than 400.
The 'user alert' should not stop processing of other items of the sequence.
Print a prompt before accepting eleven, textual, numeric inputs.
You may optionally print the item as well as its associated result, but the results must be in reverse order of input.
The sequence S may be 'implied' and so not shown explicitly.
Print and show the program in action from a typical run here. (If the output is graphical rather than text then either add a screendump or describe textually what is displayed).
| #ALGOL_W | ALGOL W | begin
real y; real array a( 0 :: 10 );
real procedure f( real value t );
sqrt(abs(t))+5*t*t*t;
for i:=0 until 10 do read( a(i) );
r_format := "A"; r_w := 9; r_d := 4;
for i:=10 step -1 until 0 do
begin
y:=f(a(i));
if y > 400 then write( "TOO LARGE" )
else write( y );
end
end. |
http://rosettacode.org/wiki/Tree_from_nesting_levels | Tree from nesting levels | Given a flat list of integers greater than zero, representing object nesting
levels, e.g. [1, 2, 4],
generate a tree formed from nested lists of those nesting level integers where:
Every int appears, in order, at its depth of nesting.
If the next level int is greater than the previous then it appears in a sub-list of the list containing the previous item
The generated tree data structure should ideally be in a languages nested list format that can
be used for further calculations rather than something just calculated for printing.
An input of [1, 2, 4] should produce the equivalent of: [1, [2, [[4]]]]
where 1 is at depth1, 2 is two deep and 4 is nested 4 deep.
[1, 2, 4, 2, 2, 1] should produce [1, [2, [[4]], 2, 2], 1].
All the nesting integers are in the same order but at the correct nesting
levels.
Similarly [3, 1, 3, 1] should generate [[[3]], 1, [[3]], 1]
Task
Generate and show here the results for the following inputs:
[]
[1, 2, 4]
[3, 1, 3, 1]
[1, 2, 3, 1]
[3, 2, 1, 3]
[3, 3, 3, 1, 1, 3, 3, 3]
| #Python | Python | def to_tree(x, index=0, depth=1):
so_far = []
while index < len(x):
this = x[index]
if this == depth:
so_far.append(this)
elif this > depth:
index, deeper = to_tree(x, index, depth + 1)
so_far.append(deeper)
else: # this < depth:
index -=1
break
index += 1
return (index, so_far) if depth > 1 else so_far
if __name__ == "__main__":
from pprint import pformat
def pnest(nest:list, width: int=9) -> str:
text = pformat(nest, width=width).replace('\n', '\n ')
print(f" OR {text}\n")
exercises = [
[],
[1, 2, 4],
[3, 1, 3, 1],
[1, 2, 3, 1],
[3, 2, 1, 3],
[3, 3, 3, 1, 1, 3, 3, 3],
]
for flat in exercises:
nest = to_tree(flat)
print(f"{flat} NESTS TO: {nest}")
pnest(nest) |
http://rosettacode.org/wiki/Twelve_statements | Twelve statements | This puzzle is borrowed from math-frolic.blogspot.
Given the following twelve statements, which of them are true?
1. This is a numbered list of twelve statements.
2. Exactly 3 of the last 6 statements are true.
3. Exactly 2 of the even-numbered statements are true.
4. If statement 5 is true, then statements 6 and 7 are both true.
5. The 3 preceding statements are all false.
6. Exactly 4 of the odd-numbered statements are true.
7. Either statement 2 or 3 is true, but not both.
8. If statement 7 is true, then 5 and 6 are both true.
9. Exactly 3 of the first 6 statements are true.
10. The next two statements are both true.
11. Exactly 1 of statements 7, 8 and 9 are true.
12. Exactly 4 of the preceding statements are true.
Task
When you get tired of trying to figure it out in your head,
write a program to solve it, and print the correct answer or answers.
Extra credit
Print out a table of near misses, that is, solutions that are contradicted by only a single statement.
| #Groovy | Groovy | enum Rule {
r01( 1, { r()*.num == (1..12) }),
r02( 2, { r(7..12).count { it.truth } == 3 }),
r03( 3, { r(2..12, 2).count { it.truth } == 2 }),
r04( 4, { r(5).truth ? r(6).truth && r(7).truth : true }),
r05( 5, { r(2..4).count { it.truth } == 0 }),
r06( 6, { r(1..11, 2).count { it.truth } == 4 }),
r07( 7, { r(2).truth != r(3).truth }),
r08( 8, { r(7).truth ? r(5).truth && r(6).truth : true }),
r09( 9, { r(1..6).count { it.truth } == 3 }),
r10(10, { r(11).truth && r(12).truth }),
r11(11, { r(7..9).count { it.truth } == 1 }),
r12(12, { r(1..11).count { it.truth } == 4 });
final int num
final Closure statement
boolean truth
static final List<Rule> rules = [ null, r01, r02, r03, r04, r05, r06, r07, r08, r09, r10, r11, r12]
private Rule(num, statement) {
this.num = num
this.statement = statement
}
public static Rule r(int index) { rules[index] }
public static List<Rule> r() { rules[1..12] }
public static List<Rule> r(List<Integer> indices) { rules[indices] }
public static List<Rule> r(IntRange indices) { rules[indices] }
public static List<Rule> r(IntRange indices, int step) { r(indices.step(step)) }
public static void setAllTruth(int bits) {
(1..12).each { r(it).truth = !(bits & (1 << (12 - it))) }
}
public static void evaluate() {
def nearMisses = [:]
(0..<(2**12)).each { i ->
setAllTruth(i)
def truthCandidates = r().findAll { it.truth }
def truthMatchCount = r().count { it.statement() == it.truth }
if (truthMatchCount == 12) {
println ">Solution< ${truthCandidates*.num}"
} else if (truthMatchCount == 11) {
def miss = (1..12).find { r(it).statement() != r(it).truth }
nearMisses << [(truthCandidates): miss]
}
}
nearMisses.each { truths, miss ->
printf ("Near Miss: %-21s (failed %2d)\n", "${truths*.num}", miss)
}
}
}
Rule.evaluate() |
http://rosettacode.org/wiki/Truth_table | Truth table | A truth table is a display of the inputs to, and the output of a Boolean function organized as a table where each row gives one combination of input values and the corresponding value of the function.
Task
Input a Boolean function from the user as a string then calculate and print a formatted truth table for the given function.
(One can assume that the user input is correct).
Print and show output for Boolean functions of two and three input variables, but any program should not be limited to that many variables in the function.
Either reverse-polish or infix notation expressions are allowed.
Related tasks
Boolean values
Ternary logic
See also
Wolfram MathWorld entry on truth tables.
some "truth table" examples from Google.
| #Haskell | Haskell | import Control.Monad (mapM, foldM, forever)
import Data.List (unwords, unlines, nub)
import Data.Maybe (fromJust)
truthTable expr = let
tokens = words expr
operators = ["&", "|", "!", "^", "=>"]
variables = nub $ filter (not . (`elem` operators)) tokens
table = zip variables <$> mapM (const [True,False]) variables
results = map (\r -> (map snd r) ++ (calculate tokens) r) table
header = variables ++ ["result"]
in
showTable $ header : map (map show) results
-- Performs evaluation of token sequence in a given context.
-- The context is an assoc-list, which binds variable and it's value.
-- Here the monad is simple ((->) r).
calculate :: [String] -> [(String, Bool)] -> [Bool]
calculate = foldM interprete []
where
interprete (x:y:s) "&" = (: s) <$> pure (x && y)
interprete (x:y:s) "|" = (: s) <$> pure (x || y)
interprete (x:y:s) "^" = (: s) <$> pure (x /= y)
interprete (x:y:s) "=>" = (: s) <$> pure (not y || x)
interprete (x:s) "!" = (: s) <$> pure (not x)
interprete s var = (: s) <$> fromJust . lookup var
-- pretty printing
showTable tbl = unlines $ map (unwords . map align) tbl
where
align txt = take colWidth $ txt ++ repeat ' '
colWidth = max 6 $ maximum $ map length (head tbl)
main = forever $ getLine >>= putStrLn . truthTable |
http://rosettacode.org/wiki/Ulam_spiral_(for_primes) | Ulam spiral (for primes) | An Ulam spiral (of primes) is a method of visualizing primes when expressed in a (normally counter-clockwise) outward spiral (usually starting at 1), constructed on a square grid, starting at the "center".
An Ulam spiral is also known as a prime spiral.
The first grid (green) is shown with sequential integers, starting at 1.
In an Ulam spiral of primes, only the primes are shown (usually indicated by some glyph such as a dot or asterisk), and all non-primes as shown as a blank (or some other whitespace).
Of course, the grid and border are not to be displayed (but they are displayed here when using these Wiki HTML tables).
Normally, the spiral starts in the "center", and the 2nd number is to the viewer's right and the number spiral starts from there in a counter-clockwise direction.
There are other geometric shapes that are used as well, including clock-wise spirals.
Also, some spirals (for the 2nd number) is viewed upwards from the 1st number instead of to the right, but that is just a matter of orientation.
Sometimes, the starting number can be specified to show more visual striking patterns (of prime densities).
[A larger than necessary grid (numbers wise) is shown here to illustrate the pattern of numbers on the diagonals (which may be used by the method to orientate the direction of spiral-construction algorithm within the example computer programs)].
Then, in the next phase in the transformation of the Ulam prime spiral, the non-primes are translated to blanks.
In the orange grid below, the primes are left intact, and all non-primes are changed to blanks.
Then, in the final transformation of the Ulam spiral (the yellow grid), translate the primes to a glyph such as a • or some other suitable glyph.
65
64
63
62
61
60
59
58
57
66
37
36
35
34
33
32
31
56
67
38
17
16
15
14
13
30
55
68
39
18
5
4
3
12
29
54
69
40
19
6
1
2
11
28
53
70
41
20
7
8
9
10
27
52
71
42
21
22
23
24
25
26
51
72
43
44
45
46
47
48
49
50
73
74
75
76
77
78
79
80
81
61
59
37
31
67
17
13
5
3
29
19
2
11
53
41
7
71
23
43
47
73
79
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
The Ulam spiral becomes more visually obvious as the grid increases in size.
Task
For any sized N × N grid, construct and show an Ulam spiral (counter-clockwise) of primes starting at some specified initial number (the default would be 1), with some suitably dotty (glyph) representation to indicate primes, and the absence of dots to indicate non-primes.
You should demonstrate the generator by showing at Ulam prime spiral large enough to (almost) fill your terminal screen.
Related tasks
Spiral matrix
Zig-zag matrix
Identity matrix
Sequence of primes by Trial Division
See also
Wikipedia entry: Ulam spiral
MathWorld™ entry: Prime Spiral
| #Java | Java | import java.util.Arrays;
public class Ulam{
enum Direction{
RIGHT, UP, LEFT, DOWN;
}
private static String[][] genUlam(int n){
return genUlam(n, 1);
}
private static String[][] genUlam(int n, int i){
String[][] spiral = new String[n][n];
Direction dir = Direction.RIGHT;
int j = i;
int y = n / 2;
int x = (n % 2 == 0) ? y - 1 : y; //shift left for even n's
while(j <= ((n * n) - 1 + i)){
spiral[y][x] = isPrime(j) ? String.format("%4d", j) : " ---";
switch(dir){
case RIGHT:
if(x <= (n - 1) && spiral[y - 1][x] == null && j > i) dir = Direction.UP; break;
case UP:
if(spiral[y][x - 1] == null) dir = Direction.LEFT; break;
case LEFT:
if(x == 0 || spiral[y + 1][x] == null) dir = Direction.DOWN; break;
case DOWN:
if(spiral[y][x + 1] == null) dir = Direction.RIGHT; break;
}
switch(dir){
case RIGHT: x++; break;
case UP: y--; break;
case LEFT: x--; break;
case DOWN: y++; break;
}
j++;
}
return spiral;
}
public static boolean isPrime(int a){
if(a == 2) return true;
if(a <= 1 || a % 2 == 0) return false;
long max = (long)Math.sqrt(a);
for(long n = 3; n <= max; n += 2){
if(a % n == 0) return false;
}
return true;
}
public static void main(String[] args){
String[][] ulam = genUlam(9);
for(String[] row : ulam){
System.out.println(Arrays.toString(row).replaceAll(",", ""));
}
System.out.println();
for(String[] row : ulam){
System.out.println(Arrays.toString(row).replaceAll("\\[\\s+\\d+", "[ * ").replaceAll("\\s+\\d+", " * ").replaceAll(",", ""));
}
}
} |
http://rosettacode.org/wiki/Truncatable_primes | Truncatable primes | A truncatable prime is a prime number that when you successively remove digits from one end of the prime, you are left with a new prime number.
Examples
The number 997 is called a left-truncatable prime as the numbers 997, 97, and 7 are all prime.
The number 7393 is a right-truncatable prime as the numbers 7393, 739, 73, and 7 formed by removing digits from its right are also prime.
No zeroes are allowed in truncatable primes.
Task
The task is to find the largest left-truncatable and right-truncatable primes less than one million (base 10 is implied).
Related tasks
Find largest left truncatable prime in a given base
Sieve of Eratosthenes
See also
Truncatable Prime from MathWorld.]
| #CoffeeScript | CoffeeScript | # You could have symmetric algorithms for max right and left
# truncatable numbers, but they lend themselves to slightly
# different optimizations.
max_right_truncatable_number = (n, f) ->
# This algorithm only evaluates 37 numbers for primeness to
# get the max right truncatable prime < 1000000. Its
# optimization is that it prunes candidates for
# the first n-1 digits before having to iterate through
# the 10 possibilities for the last digit.
if n < 10
candidate = n
while candidate > 0
return candidate if f(candidate)
candidate -= 1
else
left = Math.floor n / 10
while left > 0
left = max_right_truncatable_number left, f
right = 9
while right > 0
candidate = left * 10 + right
return candidate if candidate <= n and f(candidate)
right -= 1
left -= 1
throw Error "none found"
max_left_truncatable_number = (max, f) ->
# This is a pretty straightforward countdown. The first
# optimization here would probably be to cache results of
# calling f on small numbers.
is_left_truncatable = (n) ->
candidate = 0
power_of_ten = 1
while n > 0
r = n % 10
return false if r == 0
n = Math.floor n / 10
candidate = r * power_of_ten + candidate
power_of_ten *= 10
return false unless f(candidate)
true
do ->
n = max
while n > 0
return n if is_left_truncatable n, f
n -= 1
throw Error "none found"
is_prime = (n) ->
return false if n == 1
return true if n == 2
for d in [2..n]
return false if n % d == 0
return true if d * d >= n
console.log "right", max_right_truncatable_number(999999, is_prime)
console.log "left", max_left_truncatable_number(999999, is_prime)
|
http://rosettacode.org/wiki/Unbias_a_random_generator | Unbias a random generator |
P
1
{\displaystyle P_{1}}
P
0
{\displaystyle P_{0}}
P
1
{\displaystyle P_{1}}
P
0
{\displaystyle P_{0}}
P
0
{\displaystyle P_{0}}
P
1
{\displaystyle P_{1}}
Task details
Use your language's random number generator to create a function/method/subroutine/... randN that returns a one or a zero, but with one occurring, on average, 1 out of N times, where N is an integer from the range 3 to 6 inclusive.
Create a function unbiased that uses only randN as its source of randomness to become an unbiased generator of random ones and zeroes.
For N over its range, generate and show counts of the outputs of randN and unbiased(randN).
The actual unbiasing should be done by generating two numbers at a time from randN and only returning a 1 or 0 if they are different. As long as you always return the first number or always return the second number, the probabilities discussed above should take over the biased probability of randN.
This task is an implementation of Von Neumann debiasing, first described in a 1951 paper.
| #zkl | zkl | fcn randN(N){ (not (0).random(N)).toInt() }
fcn unbiased(randN){ while((a:=randN())==randN()){} a } |
http://rosettacode.org/wiki/Truncate_a_file | Truncate a file | Task
Truncate a file to a specific length. This should be implemented as a routine that takes two parameters: the filename and the required file length (in bytes).
Truncation can be achieved using system or library calls intended for such a task, if such methods exist, or by creating a temporary file of a reduced size and renaming it, after first deleting the original file, if no other method is available. The file may contain non human readable binary data in an unspecified format, so the routine should be "binary safe", leaving the contents of the untruncated part of the file unchanged.
If the specified filename does not exist, or the provided length is not less than the current file length, then the routine should raise an appropriate error condition.
On some systems, the provided file truncation facilities might not change the file or may extend the file, if the specified length is greater than the current length of the file.
This task permits the use of such facilities. However, such behaviour should be noted, or optionally a warning message relating to an non change or increase in file size may be implemented.
| #Python | Python |
def truncate_file(name, length):
if not os.path.isfile(name):
return False
if length >= os.path.getsize(name):
return False
with open(name, 'ab') as f:
f.truncate(length)
return True
|
http://rosettacode.org/wiki/Truncate_a_file | Truncate a file | Task
Truncate a file to a specific length. This should be implemented as a routine that takes two parameters: the filename and the required file length (in bytes).
Truncation can be achieved using system or library calls intended for such a task, if such methods exist, or by creating a temporary file of a reduced size and renaming it, after first deleting the original file, if no other method is available. The file may contain non human readable binary data in an unspecified format, so the routine should be "binary safe", leaving the contents of the untruncated part of the file unchanged.
If the specified filename does not exist, or the provided length is not less than the current file length, then the routine should raise an appropriate error condition.
On some systems, the provided file truncation facilities might not change the file or may extend the file, if the specified length is greater than the current length of the file.
This task permits the use of such facilities. However, such behaviour should be noted, or optionally a warning message relating to an non change or increase in file size may be implemented.
| #R | R |
truncate_file <- function(filename, n_bytes) {
# check file exists and size is greater than n_bytes
stopifnot(
"file does not exist"= file.exists(filename),
"not enough bytes in file"= file.size(filename) >= n_bytes
)
# read bytes from input file
input.con <- file(filename, "rb")
bindata <- readBin(input.con, integer(), n=n_bytes/4)
close(input.con)
# write bytes to temporary file
tmp.filename <- tempfile()
output.con <- file(tmp.filename, "wb")
writeBin(bindata, output.con)
close(output.con)
# double check that everything worked before overwriting original file
stopifnot(
"temp file is not expected size"= file.size(tmp.filename) == n_bytes
)
# overwrite input file with temporary file
file.rename(tmp.filename, filename)
}
|
http://rosettacode.org/wiki/Truncate_a_file | Truncate a file | Task
Truncate a file to a specific length. This should be implemented as a routine that takes two parameters: the filename and the required file length (in bytes).
Truncation can be achieved using system or library calls intended for such a task, if such methods exist, or by creating a temporary file of a reduced size and renaming it, after first deleting the original file, if no other method is available. The file may contain non human readable binary data in an unspecified format, so the routine should be "binary safe", leaving the contents of the untruncated part of the file unchanged.
If the specified filename does not exist, or the provided length is not less than the current file length, then the routine should raise an appropriate error condition.
On some systems, the provided file truncation facilities might not change the file or may extend the file, if the specified length is greater than the current length of the file.
This task permits the use of such facilities. However, such behaviour should be noted, or optionally a warning message relating to an non change or increase in file size may be implemented.
| #Racket | Racket |
#lang racket
(define (truncate file size)
(unless (file-exists? file) (error 'truncat "missing file: ~a" file))
(when (> size (file-size file)) (printf "Warning: extending file size.\n"))
(call-with-output-file* file #:exists 'update
(λ(o) (file-truncate o size))))
|
http://rosettacode.org/wiki/Universal_Turing_machine | Universal Turing machine | One of the foundational mathematical constructs behind computer science
is the universal Turing Machine.
(Alan Turing introduced the idea of such a machine in 1936–1937.)
Indeed one way to definitively prove that a language
is turing-complete
is to implement a universal Turing machine in it.
Task
Simulate such a machine capable
of taking the definition of any other Turing machine and executing it.
Of course, you will not have an infinite tape,
but you should emulate this as much as is possible.
The three permissible actions on the tape are "left", "right" and "stay".
To test your universal Turing machine (and prove your programming language
is Turing complete!), you should execute the following two Turing machines
based on the following definitions.
Simple incrementer
States: q0, qf
Initial state: q0
Terminating states: qf
Permissible symbols: B, 1
Blank symbol: B
Rules:
(q0, 1, 1, right, q0)
(q0, B, 1, stay, qf)
The input for this machine should be a tape of 1 1 1
Three-state busy beaver
States: a, b, c, halt
Initial state: a
Terminating states: halt
Permissible symbols: 0, 1
Blank symbol: 0
Rules:
(a, 0, 1, right, b)
(a, 1, 1, left, c)
(b, 0, 1, left, a)
(b, 1, 1, right, b)
(c, 0, 1, left, b)
(c, 1, 1, stay, halt)
The input for this machine should be an empty tape.
Bonus:
5-state, 2-symbol probable Busy Beaver machine from Wikipedia
States: A, B, C, D, E, H
Initial state: A
Terminating states: H
Permissible symbols: 0, 1
Blank symbol: 0
Rules:
(A, 0, 1, right, B)
(A, 1, 1, left, C)
(B, 0, 1, right, C)
(B, 1, 1, right, B)
(C, 0, 1, right, D)
(C, 1, 0, left, E)
(D, 0, 1, left, A)
(D, 1, 1, left, D)
(E, 0, 1, stay, H)
(E, 1, 0, left, A)
The input for this machine should be an empty tape.
This machine runs for more than 47 millions steps.
| #BASIC | BASIC | • R$(), an array of rules;
• T$, an input tape (where an empty string stands for a blank tape);
• B$, a character to use as a blank;
• S$, an initial state;
• H$, a halting state.
|
http://rosettacode.org/wiki/Totient_function | Totient function | The totient function is also known as:
Euler's totient function
Euler's phi totient function
phi totient function
Φ function (uppercase Greek phi)
φ function (lowercase Greek phi)
Definitions (as per number theory)
The totient function:
counts the integers up to a given positive integer n that are relatively prime to n
counts the integers k in the range 1 ≤ k ≤ n for which the greatest common divisor gcd(n,k) is equal to 1
counts numbers ≤ n and prime to n
If the totient number (for N) is one less than N, then N is prime.
Task
Create a totient function and:
Find and display (1 per line) for the 1st 25 integers:
the integer (the index)
the totient number for that integer
indicate if that integer is prime
Find and display the count of the primes up to 100
Find and display the count of the primes up to 1,000
Find and display the count of the primes up to 10,000
Find and display the count of the primes up to 100,000 (optional)
Show all output here.
Related task
Perfect totient numbers
Also see
Wikipedia: Euler's totient function.
MathWorld: totient function.
OEIS: Euler totient function phi(n).
| #11l | 11l | F f(n)
R sum((1..n).filter(k -> gcd(@n, k) == 1).map(k -> 1))
F is_prime(n)
R f(n) == n - 1
L(n) 1..25
print(‘ f(#.) == #.’.format(n, f(n))‘’(I is_prime(n) {‘, is prime’} E ‘’))
V count = 0
L(n) 1..10'000
count += is_prime(n)
I n C (100, 1000, 10'000)
print(‘Primes up to #.: #.’.format(n, count)) |
http://rosettacode.org/wiki/Topswops | Topswops | Topswops is a card game created by John Conway in the 1970's.
Assume you have a particular permutation of a set of n cards numbered 1..n on both of their faces, for example the arrangement of four cards given by [2, 4, 1, 3] where the leftmost card is on top.
A round is composed of reversing the first m cards where m is the value of the topmost card.
Rounds are repeated until the topmost card is the number 1 and the number of swaps is recorded.
For our example the swaps produce:
[2, 4, 1, 3] # Initial shuffle
[4, 2, 1, 3]
[3, 1, 2, 4]
[2, 1, 3, 4]
[1, 2, 3, 4]
For a total of four swaps from the initial ordering to produce the terminating case where 1 is on top.
For a particular number n of cards, topswops(n) is the maximum swaps needed for any starting permutation of the n cards.
Task
The task is to generate and show here a table of n vs topswops(n) for n in the range 1..10 inclusive.
Note
Topswops is also known as Fannkuch from the German word Pfannkuchen meaning pancake.
Related tasks
Number reversal game
Sorting algorithms/Pancake sort
| #11l | 11l | V best = [0] * 16
F try_swaps(&deck, f, =s, d, n)
I d > :best[n]
:best[n] = d
V i = 0
V k = 1 << s
L s != 0
k >>= 1
s--
I deck[s] == -1 | deck[s] == s
L.break
i [|]= k
I (i [&] f) == i & d + :best[s] <= :best[n]
R d
s++
V deck2 = copy(deck)
k = 1
L(i2) 1 .< s
k <<= 1
I deck2[i2] == -1
I (f [&] k) != 0
L.continue
E I deck2[i2] != i2
L.continue
deck[i2] = i2
L(j) 0 .. i2
deck2[j] = deck[i2 - j]
try_swaps(&deck2, f [|] k, s, 1 + d, n)
F topswops(n)
:best[n] = 0
V deck0 = [-1] * 16
deck0[0] = 0
try_swaps(&deck0, 1, n, 0, n)
R :best[n]
L(i) 1..12
print(‘#2: #.’.format(i, topswops(i))) |
http://rosettacode.org/wiki/Trigonometric_functions | Trigonometric functions | Task
If your language has a library or built-in functions for trigonometry, show examples of:
sine
cosine
tangent
inverses (of the above)
using the same angle in radians and degrees.
For the non-inverse functions, each radian/degree pair should use arguments that evaluate to the same angle (that is, it's not necessary to use the same angle for all three regular functions as long as the two sine calls use the same angle).
For the inverse functions, use the same number and convert its answer to radians and degrees.
If your language does not have trigonometric functions available or only has some available, write functions to calculate the functions based on any known approximation or identity.
| #ALGOL_68 | ALGOL 68 | main:(
REAL pi = 4 * arc tan(1);
# Pi / 4 is 45 degrees. All answers should be the same. #
REAL radians = pi / 4;
REAL degrees = 45.0;
REAL temp;
# sine #
print((sin(radians), " ", sin(degrees * pi / 180), new line));
# cosine #
print((cos(radians), " ", cos(degrees * pi / 180), new line));
# tangent #
print((tan(radians), " ", tan(degrees * pi / 180), new line));
# arcsine #
temp := arc sin(sin(radians));
print((temp, " ", temp * 180 / pi, new line));
# arccosine #
temp := arc cos(cos(radians));
print((temp, " ", temp * 180 / pi, new line));
# arctangent #
temp := arc tan(tan(radians));
print((temp, " ", temp * 180 / pi, new line))
) |
http://rosettacode.org/wiki/Trabb_Pardo%E2%80%93Knuth_algorithm | Trabb Pardo–Knuth algorithm | The TPK algorithm is an early example of a programming chrestomathy.
It was used in Donald Knuth and Luis Trabb Pardo's Stanford tech report The Early Development of Programming Languages.
The report traces the early history of work in developing computer languages in the 1940s and 1950s, giving several translations of the algorithm.
From the wikipedia entry:
ask for 11 numbers to be read into a sequence S
reverse sequence S
for each item in sequence S
result := call a function to do an operation
if result overflows
alert user
else
print result
The task is to implement the algorithm:
Use the function:
f
(
x
)
=
|
x
|
0.5
+
5
x
3
{\displaystyle f(x)=|x|^{0.5}+5x^{3}}
The overflow condition is an answer of greater than 400.
The 'user alert' should not stop processing of other items of the sequence.
Print a prompt before accepting eleven, textual, numeric inputs.
You may optionally print the item as well as its associated result, but the results must be in reverse order of input.
The sequence S may be 'implied' and so not shown explicitly.
Print and show the program in action from a typical run here. (If the output is graphical rather than text then either add a screendump or describe textually what is displayed).
| #APL | APL | ∇ {res}←Trabb;f;S;i;a;y ⍝ define a function Trabb
f←{(0.5*⍨|⍵)+5×⍵*3} ⍝ define a function f
S←,⍎{⍞←⍵ ⋄ (≢⍵)↓⍞}'Please, enter 11 numbers: '
:For i a :InEach (⌽⍳≢S)(⌽S) ⍝ loop through N..1 and reversed S
:If 400<y←f(a)
⎕←'Too large: ',⍕i
:Else
⎕←i,y
:EndIf
:EndFor
∇ |
http://rosettacode.org/wiki/Trabb_Pardo%E2%80%93Knuth_algorithm | Trabb Pardo–Knuth algorithm | The TPK algorithm is an early example of a programming chrestomathy.
It was used in Donald Knuth and Luis Trabb Pardo's Stanford tech report The Early Development of Programming Languages.
The report traces the early history of work in developing computer languages in the 1940s and 1950s, giving several translations of the algorithm.
From the wikipedia entry:
ask for 11 numbers to be read into a sequence S
reverse sequence S
for each item in sequence S
result := call a function to do an operation
if result overflows
alert user
else
print result
The task is to implement the algorithm:
Use the function:
f
(
x
)
=
|
x
|
0.5
+
5
x
3
{\displaystyle f(x)=|x|^{0.5}+5x^{3}}
The overflow condition is an answer of greater than 400.
The 'user alert' should not stop processing of other items of the sequence.
Print a prompt before accepting eleven, textual, numeric inputs.
You may optionally print the item as well as its associated result, but the results must be in reverse order of input.
The sequence S may be 'implied' and so not shown explicitly.
Print and show the program in action from a typical run here. (If the output is graphical rather than text then either add a screendump or describe textually what is displayed).
| #Arturo | Arturo | proc: function [x]->
((abs x) ^ 0.5) + 5 * x ^ 3
ask: function [msg][
to [:floating] first.n: 11 split.words strip input msg
]
loop reverse ask "11 numbers: " 'n [
result: proc n
print [n ":" (result > 400)? -> "TOO LARGE!" -> result]
] |
http://rosettacode.org/wiki/Tree_from_nesting_levels | Tree from nesting levels | Given a flat list of integers greater than zero, representing object nesting
levels, e.g. [1, 2, 4],
generate a tree formed from nested lists of those nesting level integers where:
Every int appears, in order, at its depth of nesting.
If the next level int is greater than the previous then it appears in a sub-list of the list containing the previous item
The generated tree data structure should ideally be in a languages nested list format that can
be used for further calculations rather than something just calculated for printing.
An input of [1, 2, 4] should produce the equivalent of: [1, [2, [[4]]]]
where 1 is at depth1, 2 is two deep and 4 is nested 4 deep.
[1, 2, 4, 2, 2, 1] should produce [1, [2, [[4]], 2, 2], 1].
All the nesting integers are in the same order but at the correct nesting
levels.
Similarly [3, 1, 3, 1] should generate [[[3]], 1, [[3]], 1]
Task
Generate and show here the results for the following inputs:
[]
[1, 2, 4]
[3, 1, 3, 1]
[1, 2, 3, 1]
[3, 2, 1, 3]
[3, 3, 3, 1, 1, 3, 3, 3]
| #Quackery | Quackery | [ stack ] is prev ( --> s )
[ temp take
swap join
temp put ] is add$ ( x --> )
[ dup [] = if done
0 prev put
$ "' " temp put
witheach
[ dup prev take -
over prev put
dup 0 > iff
[ times
[ $ "[ " add$ ] ]
else
[ abs times
[ $ "] " add$ ] ]
number$ space join add$ ]
prev take times
[ $ "] " add$ ]
temp take quackery ] is nesttree ( [ --> [ )
' [ [ ]
[ 1 2 4 ]
[ 3 1 3 1 ]
[ 1 2 3 1 ]
[ 3 2 1 3 ]
[ 3 3 3 1 1 3 3 3 ] ]
witheach
[ dup echo say " --> "
nesttree echo cr cr ] |
http://rosettacode.org/wiki/Tree_from_nesting_levels | Tree from nesting levels | Given a flat list of integers greater than zero, representing object nesting
levels, e.g. [1, 2, 4],
generate a tree formed from nested lists of those nesting level integers where:
Every int appears, in order, at its depth of nesting.
If the next level int is greater than the previous then it appears in a sub-list of the list containing the previous item
The generated tree data structure should ideally be in a languages nested list format that can
be used for further calculations rather than something just calculated for printing.
An input of [1, 2, 4] should produce the equivalent of: [1, [2, [[4]]]]
where 1 is at depth1, 2 is two deep and 4 is nested 4 deep.
[1, 2, 4, 2, 2, 1] should produce [1, [2, [[4]], 2, 2], 1].
All the nesting integers are in the same order but at the correct nesting
levels.
Similarly [3, 1, 3, 1] should generate [[[3]], 1, [[3]], 1]
Task
Generate and show here the results for the following inputs:
[]
[1, 2, 4]
[3, 1, 3, 1]
[1, 2, 3, 1]
[3, 2, 1, 3]
[3, 3, 3, 1, 1, 3, 3, 3]
| #Raku | Raku | sub new_level ( @stack --> Nil ) {
my $e = [];
push @stack.tail, $e;
push @stack, $e;
}
sub to_tree_iterative ( @xs --> List ) {
my $nested = [];
my @stack = $nested;
for @xs -> Int $x {
new_level(@stack) while $x > @stack;
pop @stack while $x < @stack;
push @stack.tail, $x;
}
return $nested;
}
my @tests = (), (1, 2, 4), (3, 1, 3, 1), (1, 2, 3, 1), (3, 2, 1, 3), (3, 3, 3, 1, 1, 3, 3, 3);
say .Str.fmt( '%15s => ' ), .&to_tree_iterative for @tests; |
http://rosettacode.org/wiki/Twelve_statements | Twelve statements | This puzzle is borrowed from math-frolic.blogspot.
Given the following twelve statements, which of them are true?
1. This is a numbered list of twelve statements.
2. Exactly 3 of the last 6 statements are true.
3. Exactly 2 of the even-numbered statements are true.
4. If statement 5 is true, then statements 6 and 7 are both true.
5. The 3 preceding statements are all false.
6. Exactly 4 of the odd-numbered statements are true.
7. Either statement 2 or 3 is true, but not both.
8. If statement 7 is true, then 5 and 6 are both true.
9. Exactly 3 of the first 6 statements are true.
10. The next two statements are both true.
11. Exactly 1 of statements 7, 8 and 9 are true.
12. Exactly 4 of the preceding statements are true.
Task
When you get tired of trying to figure it out in your head,
write a program to solve it, and print the correct answer or answers.
Extra credit
Print out a table of near misses, that is, solutions that are contradicted by only a single statement.
| #Haskell | Haskell | import Data.List (findIndices)
tf :: [[Int] -> Bool] -> [[Int]]
tf = traverse (const [1, 0])
wrongness :: [Int] -> [[Int] -> Bool] -> [Int]
wrongness ns ps = findIndices id (zipWith (/=) ns (map (fromEnum . ($ ns)) ps))
statements :: [[Int] -> Bool]
statements =
[ (== 12) . length
, 3 ⊂ [length statements - 6 ..]
, 2 ⊂ [1,3 ..]
, 4 → [4 .. 6]
, 0 ⊂ [1 .. 3]
, 4 ⊂ [0,2 ..]
, 1 ⊂ [1, 2]
, 6 → [4 .. 6]
, 3 ⊂ [0 .. 5]
, 2 ⊂ [10, 11]
, 1 ⊂ [6, 7, 8]
, 4 ⊂ [0 .. 10]
]
where
(⊂), (→) :: Int -> [Int] -> [Int] -> Bool
(s ⊂ x) b = s == (sum . map (b !!) . takeWhile (< length b)) x
(a → x) b = (b !! a == 0) || all ((== 1) . (b !!)) x
testall :: [[Int] -> Bool] -> Int -> [([Int], [Int])]
testall s n =
[ (b, w)
| b <- tf s
, w <- [wrongness b s]
, length w == n ]
main :: IO ()
main =
let t = testall statements
in do putStrLn "Answer"
mapM_ print $ t 0
putStrLn "Near misses"
mapM_ print $ t 1 |
http://rosettacode.org/wiki/Truth_table | Truth table | A truth table is a display of the inputs to, and the output of a Boolean function organized as a table where each row gives one combination of input values and the corresponding value of the function.
Task
Input a Boolean function from the user as a string then calculate and print a formatted truth table for the given function.
(One can assume that the user input is correct).
Print and show output for Boolean functions of two and three input variables, but any program should not be limited to that many variables in the function.
Either reverse-polish or infix notation expressions are allowed.
Related tasks
Boolean values
Ternary logic
See also
Wolfram MathWorld entry on truth tables.
some "truth table" examples from Google.
| #J | J | truthTable=:3 :0
assert. -. 1 e. 'data expr names table' e.&;: y
names=. ~. (#~ _1 <: nc) ;:expr=. y
data=. #:i.2^#names
(names)=. |:data
(' ',;:inv names,<expr),(1+#@>names,<expr)":data,.".expr
) |
http://rosettacode.org/wiki/Ulam_spiral_(for_primes) | Ulam spiral (for primes) | An Ulam spiral (of primes) is a method of visualizing primes when expressed in a (normally counter-clockwise) outward spiral (usually starting at 1), constructed on a square grid, starting at the "center".
An Ulam spiral is also known as a prime spiral.
The first grid (green) is shown with sequential integers, starting at 1.
In an Ulam spiral of primes, only the primes are shown (usually indicated by some glyph such as a dot or asterisk), and all non-primes as shown as a blank (or some other whitespace).
Of course, the grid and border are not to be displayed (but they are displayed here when using these Wiki HTML tables).
Normally, the spiral starts in the "center", and the 2nd number is to the viewer's right and the number spiral starts from there in a counter-clockwise direction.
There are other geometric shapes that are used as well, including clock-wise spirals.
Also, some spirals (for the 2nd number) is viewed upwards from the 1st number instead of to the right, but that is just a matter of orientation.
Sometimes, the starting number can be specified to show more visual striking patterns (of prime densities).
[A larger than necessary grid (numbers wise) is shown here to illustrate the pattern of numbers on the diagonals (which may be used by the method to orientate the direction of spiral-construction algorithm within the example computer programs)].
Then, in the next phase in the transformation of the Ulam prime spiral, the non-primes are translated to blanks.
In the orange grid below, the primes are left intact, and all non-primes are changed to blanks.
Then, in the final transformation of the Ulam spiral (the yellow grid), translate the primes to a glyph such as a • or some other suitable glyph.
65
64
63
62
61
60
59
58
57
66
37
36
35
34
33
32
31
56
67
38
17
16
15
14
13
30
55
68
39
18
5
4
3
12
29
54
69
40
19
6
1
2
11
28
53
70
41
20
7
8
9
10
27
52
71
42
21
22
23
24
25
26
51
72
43
44
45
46
47
48
49
50
73
74
75
76
77
78
79
80
81
61
59
37
31
67
17
13
5
3
29
19
2
11
53
41
7
71
23
43
47
73
79
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
The Ulam spiral becomes more visually obvious as the grid increases in size.
Task
For any sized N × N grid, construct and show an Ulam spiral (counter-clockwise) of primes starting at some specified initial number (the default would be 1), with some suitably dotty (glyph) representation to indicate primes, and the absence of dots to indicate non-primes.
You should demonstrate the generator by showing at Ulam prime spiral large enough to (almost) fill your terminal screen.
Related tasks
Spiral matrix
Zig-zag matrix
Identity matrix
Sequence of primes by Trial Division
See also
Wikipedia entry: Ulam spiral
MathWorld™ entry: Prime Spiral
| #JavaScript | JavaScript |
<!-- UlamSpiral.html -->
<html>
<head><title>Ulam Spiral</title>
<script src="VOE.js"></script>
<script>
// http://rosettacode.org/wiki/User:AnatolV/Helper_Functions
// Use v.2.0
var pst;
// ***** Additional helper functions
// Pad number from left
function padLeft(n,ns) {
return (" " + n).slice(-ns);
}
// Is number n a prime?
function isPrime(n) {
var n2=Math.sqrt(n);
for(var i=2; i<=n2; i++) {
if(n%i === 0) return false;
}//fend i
return n !== 1;
}
function insm(mat,x,y) {
var xz=mat[0].length, yz=xz;
return(x>=0 && x<xz && y>=0 && y<yz)
}
// *****
function rbCheck() {
if (document.getElementById('rbDef').checked) {pst=0}
if (document.getElementById('rbAst').checked) {pst=1}
if (document.getElementById('rbNum').checked) {pst=2}
}
function rbSet() {
document.getElementById("rbDef").checked = true;
rbCheck();
}
// The Ulam Spiral
function pspUlam() {
var i, j, x, y, xmx, ymx, cnt, dir, M, Mij, sp=" ", sc=3;
// Setting basic vars for canvas and matrix
var cvs = document.getElementById('cvsId');
var ctx = cvs.getContext("2d");
if(pst<0||pst>2) {pst=0}
if(pst==0) {n=100; sc=3} else {n=10; sc=5}
console.log("sc", typeof(sc));
if(n%2==0) {n++};
var n2=n*n, pch, sz=n2.toString().length, pch2=sp.repeat(sz);
var fgc="navy", bgc="white";
// Create matrix, finding number of rows and columns
var M=new Array(n);
for (i=0; i<n; i++) { M[i]=new Array(n);
for (j=0; j<n; j++) {M[i][j]=0} }
var r = M[0].length, c = M.length, k=0, dsz=1;
// Logging init parameters
var ttl="Matrix ("+r+","+c+")";
console.log(" *** Ulam spiral: ",n,"x",n,"p-flag=",pst, "sc", sc);
// Generating and plotting Ulam spiral
x=y=Math.floor(n/2)+1; xmx=ymx=cnt=1; dir="R";
for(var i=1; i<=n2; i++) { //
if(isPrime(i)) // if prime
{ if(!insm(M,x,y)) {break};
if(pst==2) {M[y][x]=i} else {M[y][x]=1};
}
// all numbers
if(dir=="R") {if(xmx>0){x++;xmx--} else {dir="U";ymx=cnt;y--;ymx--} continue};
if(dir=="U") {if(ymx>0){y--;ymx--} else {dir="L";cnt++;xmx=cnt;x--;xmx--} continue};
if(dir=="L") {if(xmx>0){x--;xmx--} else {dir="D";ymx=cnt;y++;ymx--} continue};
if(dir=="D") {if(ymx>0){y++;ymx--} else {dir="R";cnt++;xmx=cnt;x++;xmx--}; continue};
}//fend i
//Plot/Print according to the p-flag(0-real plot,1-"*",2-primes)
if(pst==0) {pmat01(M, fgc, bgc, sc, 0); return};
var logs;
if(pst==1) {for(i=1;i<n;i++) {logs="|";
for(j=1;j<n;j++) { Mij=M[i][j]; if(Mij>0) {pch="*"} else {pch=" "};
logs+=" "+pch;}
logs+="|"; console.log(logs);}//fiend
pmat01(M, fgc, bgc, sc, 0); console.log("sc", sc);
return;
}//ifend
//console.log(" ",pch);} console.log(" ")}; return};
if(pst==2) {for(i=1;i<n;i++) {logs="|";
for(j=1;j<n;j++) {Mij=M[i][j];
if(Mij==0) {pch=pch2}
else {pch=padLeft(Mij,sz)};
logs+=pch; } //" "+
logs+=" |"; console.log(logs);}//fiend
pmat01(M, fgc, bgc, sc, 0); console.log("sc", sc);
return;
}//ifend
}//func end
// ******************************************
</script></head>
<body onload='rbSet();' style="font-family: arial, helvatica, sans-serif;">
<b>Plot/print style:</b>
<input type="radio" onclick="rbCheck();" name="rb" id="rbDef"/><b>Plot</b>
<input type="radio" onclick="rbCheck();" name="rb" id="rbAst"/><b>Print *</b>
<input type="radio" onclick="rbCheck();" name="rb" id="rbNum"/><b>Print numbers</b>
<input type="button" value="Plot it!" onclick="pspUlam();">
<h3>Ulam Spiral</h3>
<canvas id="cvsId" width="300" height="300" style="border: 2px inset;"></canvas>
</body>
</html>
|
http://rosettacode.org/wiki/Truncatable_primes | Truncatable primes | A truncatable prime is a prime number that when you successively remove digits from one end of the prime, you are left with a new prime number.
Examples
The number 997 is called a left-truncatable prime as the numbers 997, 97, and 7 are all prime.
The number 7393 is a right-truncatable prime as the numbers 7393, 739, 73, and 7 formed by removing digits from its right are also prime.
No zeroes are allowed in truncatable primes.
Task
The task is to find the largest left-truncatable and right-truncatable primes less than one million (base 10 is implied).
Related tasks
Find largest left truncatable prime in a given base
Sieve of Eratosthenes
See also
Truncatable Prime from MathWorld.]
| #Common_Lisp | Common Lisp |
(defun start ()
(format t "Largest right-truncatable ~a~%" (max-right-truncatable))
(format t "Largest left-truncatable ~a~%" (max-left-truncatable)))
(defun max-right-truncatable ()
(loop for el in (6-digits-R-truncatables)
maximizing el into max
finally (return max)))
(defun 6-digits-R-truncatables (&optional (lst '(2 3 5 7)) (n 5))
(if (zerop n)
lst
(6-digits-R-truncatables (R-trunc lst) (- n 1))))
(defun R-trunc (lst)
(remove-if (lambda (x) (not (primep x)))
(loop for el in lst
append (mapcar (lambda (x) (+ (* 10 el) x)) '(1 3 7 9)))))
(defun max-left-truncatable ()
(loop for el in (6-digits-L-truncatables)
maximizing el into max
finally (return max)))
(defun 6-digits-L-truncatables (&optional (lst '(3 7)) (n 5))
(if (zerop n)
lst
(6-digits-L-truncatables (L-trunc lst (- 6 n)) (- n 1))))
(defun L-trunc (lst n)
(remove-if (lambda (x) (not (primep x)))
(loop for el in lst
append (mapcar (lambda (x) (+ (* (expt 10 n) x) el)) '(1 2 3 4 5 6 7 8 9)))))
(defun primep (n)
(primep-aux n 2))
(defun primep-aux (n d)
(cond ((> d (sqrt n)) t)
((zerop (rem n d)) nil)
(t (primep-aux n (+ d 1)))))
|
http://rosettacode.org/wiki/Truncate_a_file | Truncate a file | Task
Truncate a file to a specific length. This should be implemented as a routine that takes two parameters: the filename and the required file length (in bytes).
Truncation can be achieved using system or library calls intended for such a task, if such methods exist, or by creating a temporary file of a reduced size and renaming it, after first deleting the original file, if no other method is available. The file may contain non human readable binary data in an unspecified format, so the routine should be "binary safe", leaving the contents of the untruncated part of the file unchanged.
If the specified filename does not exist, or the provided length is not less than the current file length, then the routine should raise an appropriate error condition.
On some systems, the provided file truncation facilities might not change the file or may extend the file, if the specified length is greater than the current length of the file.
This task permits the use of such facilities. However, such behaviour should be noted, or optionally a warning message relating to an non change or increase in file size may be implemented.
| #Raku | Raku | use NativeCall;
sub truncate(Str, int32 --> int32) is native {*}
sub MAIN (Str $file, Int $to) {
given $file.IO {
.e or die "$file doesn't exist";
.w or die "$file isn't writable";
.s >= $to or die "$file is not big enough to truncate";
}
truncate($file, $to) == 0 or die "Truncation was unsuccessful";
} |
http://rosettacode.org/wiki/Truncate_a_file | Truncate a file | Task
Truncate a file to a specific length. This should be implemented as a routine that takes two parameters: the filename and the required file length (in bytes).
Truncation can be achieved using system or library calls intended for such a task, if such methods exist, or by creating a temporary file of a reduced size and renaming it, after first deleting the original file, if no other method is available. The file may contain non human readable binary data in an unspecified format, so the routine should be "binary safe", leaving the contents of the untruncated part of the file unchanged.
If the specified filename does not exist, or the provided length is not less than the current file length, then the routine should raise an appropriate error condition.
On some systems, the provided file truncation facilities might not change the file or may extend the file, if the specified length is greater than the current length of the file.
This task permits the use of such facilities. However, such behaviour should be noted, or optionally a warning message relating to an non change or increase in file size may be implemented.
| #REXX | REXX | /*REXX program truncates a file to a specified (and smaller) number of bytes. */
parse arg siz FID /*obtain required arguments from the CL*/
FID=strip(FID) /*elide FID leading/trailing blanks. */
if siz=='' then call ser "No truncation size was specified (1st argument)."
if FID=='' then call ser "No fileID was specified (2nd argument)."
if \datatype(siz,'W') then call ser "trunc size isn't an integer: " siz
if siz<1 then call ser "trunc size isn't a positive integer: " siz
_=charin(FID,1,siz+1) /*position file and read a wee bit more*/
#=length(_) /*get the length of the part just read.*/
if #==0 then call ser "the specified file doesn't exist: " FID
if #<siz then call ser "the file is smaller than trunc size: " #
call lineout FID /*close the file used, just to be safe.*/
'ERASE' FID /*invoke a command to delete the file */
call lineout FID /*close the file, maybe for REXX's use.*/
call charout FID, left(_,siz), 1 /*write a truncated version of the file*/
call lineout FID /*close the file used, just to be safe.*/
say 'file ' FID " truncated to " siz 'bytes.' /*display some information to terminal.*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
ser: say '***error***' arg(1); exit 13 /*display an error message and exit. */ |
http://rosettacode.org/wiki/Universal_Turing_machine | Universal Turing machine | One of the foundational mathematical constructs behind computer science
is the universal Turing Machine.
(Alan Turing introduced the idea of such a machine in 1936–1937.)
Indeed one way to definitively prove that a language
is turing-complete
is to implement a universal Turing machine in it.
Task
Simulate such a machine capable
of taking the definition of any other Turing machine and executing it.
Of course, you will not have an infinite tape,
but you should emulate this as much as is possible.
The three permissible actions on the tape are "left", "right" and "stay".
To test your universal Turing machine (and prove your programming language
is Turing complete!), you should execute the following two Turing machines
based on the following definitions.
Simple incrementer
States: q0, qf
Initial state: q0
Terminating states: qf
Permissible symbols: B, 1
Blank symbol: B
Rules:
(q0, 1, 1, right, q0)
(q0, B, 1, stay, qf)
The input for this machine should be a tape of 1 1 1
Three-state busy beaver
States: a, b, c, halt
Initial state: a
Terminating states: halt
Permissible symbols: 0, 1
Blank symbol: 0
Rules:
(a, 0, 1, right, b)
(a, 1, 1, left, c)
(b, 0, 1, left, a)
(b, 1, 1, right, b)
(c, 0, 1, left, b)
(c, 1, 1, stay, halt)
The input for this machine should be an empty tape.
Bonus:
5-state, 2-symbol probable Busy Beaver machine from Wikipedia
States: A, B, C, D, E, H
Initial state: A
Terminating states: H
Permissible symbols: 0, 1
Blank symbol: 0
Rules:
(A, 0, 1, right, B)
(A, 1, 1, left, C)
(B, 0, 1, right, C)
(B, 1, 1, right, B)
(C, 0, 1, right, D)
(C, 1, 0, left, E)
(D, 0, 1, left, A)
(D, 1, 1, left, D)
(E, 0, 1, stay, H)
(E, 1, 0, left, A)
The input for this machine should be an empty tape.
This machine runs for more than 47 millions steps.
| #C | C | #include <stdio.h>
#include <stdarg.h>
#include <stdlib.h>
#include <string.h>
enum {
LEFT,
RIGHT,
STAY
};
typedef struct {
int state1;
int symbol1;
int symbol2;
int dir;
int state2;
} transition_t;
typedef struct tape_t tape_t;
struct tape_t {
int symbol;
tape_t *left;
tape_t *right;
};
typedef struct {
int states_len;
char **states;
int final_states_len;
int *final_states;
int symbols_len;
char *symbols;
int blank;
int state;
int tape_len;
tape_t *tape;
int transitions_len;
transition_t ***transitions;
} turing_t;
int state_index (turing_t *t, char *state) {
int i;
for (i = 0; i < t->states_len; i++) {
if (!strcmp(t->states[i], state)) {
return i;
}
}
return 0;
}
int symbol_index (turing_t *t, char symbol) {
int i;
for (i = 0; i < t->symbols_len; i++) {
if (t->symbols[i] == symbol) {
return i;
}
}
return 0;
}
void move (turing_t *t, int dir) {
tape_t *orig = t->tape;
if (dir == RIGHT) {
if (orig && orig->right) {
t->tape = orig->right;
}
else {
t->tape = calloc(1, sizeof (tape_t));
t->tape->symbol = t->blank;
if (orig) {
t->tape->left = orig;
orig->right = t->tape;
}
}
}
else if (dir == LEFT) {
if (orig && orig->left) {
t->tape = orig->left;
}
else {
t->tape = calloc(1, sizeof (tape_t));
t->tape->symbol = t->blank;
if (orig) {
t->tape->right = orig;
orig->left = t->tape;
}
}
}
}
turing_t *create (int states_len, ...) {
va_list args;
va_start(args, states_len);
turing_t *t = malloc(sizeof (turing_t));
t->states_len = states_len;
t->states = malloc(states_len * sizeof (char *));
int i;
for (i = 0; i < states_len; i++) {
t->states[i] = va_arg(args, char *);
}
t->final_states_len = va_arg(args, int);
t->final_states = malloc(t->final_states_len * sizeof (int));
for (i = 0; i < t->final_states_len; i++) {
t->final_states[i] = state_index(t, va_arg(args, char *));
}
t->symbols_len = va_arg(args, int);
t->symbols = malloc(t->symbols_len);
for (i = 0; i < t->symbols_len; i++) {
t->symbols[i] = va_arg(args, int);
}
t->blank = symbol_index(t, va_arg(args, int));
t->state = state_index(t, va_arg(args, char *));
t->tape_len = va_arg(args, int);
t->tape = NULL;
for (i = 0; i < t->tape_len; i++) {
move(t, RIGHT);
t->tape->symbol = symbol_index(t, va_arg(args, int));
}
if (!t->tape_len) {
move(t, RIGHT);
}
while (t->tape->left) {
t->tape = t->tape->left;
}
t->transitions_len = va_arg(args, int);
t->transitions = malloc(t->states_len * sizeof (transition_t **));
for (i = 0; i < t->states_len; i++) {
t->transitions[i] = malloc(t->symbols_len * sizeof (transition_t *));
}
for (i = 0; i < t->transitions_len; i++) {
transition_t *tran = malloc(sizeof (transition_t));
tran->state1 = state_index(t, va_arg(args, char *));
tran->symbol1 = symbol_index(t, va_arg(args, int));
tran->symbol2 = symbol_index(t, va_arg(args, int));
tran->dir = va_arg(args, int);
tran->state2 = state_index(t, va_arg(args, char *));
t->transitions[tran->state1][tran->symbol1] = tran;
}
va_end(args);
return t;
}
void print_state (turing_t *t) {
printf("%-10s ", t->states[t->state]);
tape_t *tape = t->tape;
while (tape->left) {
tape = tape->left;
}
while (tape) {
if (tape == t->tape) {
printf("[%c]", t->symbols[tape->symbol]);
}
else {
printf(" %c ", t->symbols[tape->symbol]);
}
tape = tape->right;
}
printf("\n");
}
void run (turing_t *t) {
int i;
while (1) {
print_state(t);
for (i = 0; i < t->final_states_len; i++) {
if (t->final_states[i] == t->state) {
return;
}
}
transition_t *tran = t->transitions[t->state][t->tape->symbol];
t->tape->symbol = tran->symbol2;
move(t, tran->dir);
t->state = tran->state2;
}
}
int main () {
printf("Simple incrementer\n");
turing_t *t = create(
/* states */ 2, "q0", "qf",
/* final_states */ 1, "qf",
/* symbols */ 2, 'B', '1',
/* blank */ 'B',
/* initial_state */ "q0",
/* initial_tape */ 3, '1', '1', '1',
/* transitions */ 2,
"q0", '1', '1', RIGHT, "q0",
"q0", 'B', '1', STAY, "qf"
);
run(t);
printf("\nThree-state busy beaver\n");
t = create(
/* states */ 4, "a", "b", "c", "halt",
/* final_states */ 1, "halt",
/* symbols */ 2, '0', '1',
/* blank */ '0',
/* initial_state */ "a",
/* initial_tape */ 0,
/* transitions */ 6,
"a", '0', '1', RIGHT, "b",
"a", '1', '1', LEFT, "c",
"b", '0', '1', LEFT, "a",
"b", '1', '1', RIGHT, "b",
"c", '0', '1', LEFT, "b",
"c", '1', '1', STAY, "halt"
);
run(t);
return 0;
printf("\nFive-state two-symbol probable busy beaver\n");
t = create(
/* states */ 6, "A", "B", "C", "D", "E", "H",
/* final_states */ 1, "H",
/* symbols */ 2, '0', '1',
/* blank */ '0',
/* initial_state */ "A",
/* initial_tape */ 0,
/* transitions */ 10,
"A", '0', '1', RIGHT, "B",
"A", '1', '1', LEFT, "C",
"B", '0', '1', RIGHT, "C",
"B", '1', '1', RIGHT, "B",
"C", '0', '1', RIGHT, "D",
"C", '1', '0', LEFT, "E",
"D", '0', '1', LEFT, "A",
"D", '1', '1', LEFT, "D",
"E", '0', '1', STAY, "H",
"E", '1', '0', LEFT, "A"
);
run(t);
}
|
http://rosettacode.org/wiki/Totient_function | Totient function | The totient function is also known as:
Euler's totient function
Euler's phi totient function
phi totient function
Φ function (uppercase Greek phi)
φ function (lowercase Greek phi)
Definitions (as per number theory)
The totient function:
counts the integers up to a given positive integer n that are relatively prime to n
counts the integers k in the range 1 ≤ k ≤ n for which the greatest common divisor gcd(n,k) is equal to 1
counts numbers ≤ n and prime to n
If the totient number (for N) is one less than N, then N is prime.
Task
Create a totient function and:
Find and display (1 per line) for the 1st 25 integers:
the integer (the index)
the totient number for that integer
indicate if that integer is prime
Find and display the count of the primes up to 100
Find and display the count of the primes up to 1,000
Find and display the count of the primes up to 10,000
Find and display the count of the primes up to 100,000 (optional)
Show all output here.
Related task
Perfect totient numbers
Also see
Wikipedia: Euler's totient function.
MathWorld: totient function.
OEIS: Euler totient function phi(n).
| #AArch64_Assembly | AArch64 Assembly |
/* ARM assembly AARCH64 Raspberry PI 3B */
/* program totient.s */
/************************************/
/* Constantes */
/************************************/
.include "../includeConstantesARM64.inc"
.equ MAXI, 25
/*********************************/
/* Initialized data */
/*********************************/
.data
szMessNumber: .asciz " number @ totient @ @ \n"
szCarriageReturn: .asciz "\n"
szMessPrime: .asciz " is prime."
szMessSpace: .asciz " "
szMessCounterPrime: .asciz "Number of primes to @ : @ \n"
szMessOverflow: .asciz "Overflow function isPrime.\n"
/*********************************/
/* UnInitialized data */
/*********************************/
.bss
sZoneConv: .skip 24
/*********************************/
/* code section */
/*********************************/
.text
.global main
main:
mov x4,#1 // start number
1:
mov x0,x4
bl totient // compute totient
mov x5,x0
mov x0,x4
bl isPrime // control if number is prime
mov x6,x0
mov x0,x4 // display result
ldr x1,qAdrsZoneConv
bl conversion10 // call décimal conversion
ldr x0,qAdrszMessNumber
ldr x1,qAdrsZoneConv // insert conversion in message
bl strInsertAtCharInc
mov x7,x0
mov x0,x5
ldr x1,qAdrsZoneConv
bl conversion10 // call décimal conversion
mov x0,x7
ldr x1,qAdrsZoneConv // insert conversion in message
bl strInsertAtCharInc
mov x7,x0
cmp x6,#1
ldr x1,qAdrszMessPrime
ldr x8,qAdrszMessSpace
csel x1,x1,x8,eq
mov x0,x7
bl strInsertAtCharInc
bl affichageMess // display message
add x4,x4,#1 // increment number
cmp x4,#MAXI // maxi ?
ble 1b // and loop
mov x4,#2 // first number
mov x5,#0 // prime counter
ldr x6,iCst1000 // load constantes
ldr x7,iCst10000
ldr x8,iCst100000
2:
mov x0,x4
bl isPrime
cmp x0,#0
beq 3f
add x5,x5,#1
3:
add x4,x4,#1
cmp x4,#100
bne 4f
mov x0,#100
mov x1,x5
bl displayCounter
b 7f
4:
cmp x4,x6 // 1000
bne 5f
mov x0,x6
mov x1,x5
bl displayCounter
b 7f
5:
cmp x4,x7 // 10000
bne 6f
mov x0,x7
mov x1,x5
bl displayCounter
b 7f
6:
cmp x4,x8 // 100000
bne 7f
mov x0,x8
mov x1,x5
bl displayCounter
7:
cmp x4,x8
ble 2b // and loop
100: // standard end of the program
mov x0, #0 // return code
mov x8,EXIT
svc #0 // perform the system call
qAdrszCarriageReturn: .quad szCarriageReturn
qAdrsZoneConv: .quad sZoneConv
qAdrszMessNumber: .quad szMessNumber
qAdrszMessCounterPrime: .quad szMessCounterPrime
qAdrszMessPrime: .quad szMessPrime
qAdrszMessSpace: .quad szMessSpace
iCst1000: .quad 1000
iCst10000: .quad 10000
iCst100000: .quad 100000
/******************************************************************/
/* display counter */
/******************************************************************/
/* x0 contains limit */
/* x1 contains counter */
displayCounter:
stp x1,lr,[sp,-16]! // save registers
stp x2,x3,[sp,-16]! // save registers
mov x2,x1
ldr x1,qAdrsZoneConv
bl conversion10 // call décimal conversion
ldr x0,qAdrszMessCounterPrime
ldr x1,qAdrsZoneConv // insert conversion in message
bl strInsertAtCharInc
mov x3,x0
mov x0,x2
ldr x1,qAdrsZoneConv
bl conversion10 // call décimal conversion
mov x0,x3
ldr x1,qAdrsZoneConv // insert conversion in message
bl strInsertAtCharInc
bl affichageMess
100:
ldp x2,x3,[sp],16 // restaur registers
ldp x1,lr,[sp],16 // restaur registers
ret
/******************************************************************/
/* compute totient of number */
/******************************************************************/
/* x0 contains number */
totient:
stp x1,lr,[sp,-16]! // save registers
stp x2,x3,[sp,-16]! // save registers
stp x4,x5,[sp,-16]! // save registers
mov x4,x0 // totient
mov x5,x0 // save number
mov x1,#0 // for first divisor
1: // begin loop
mul x3,x1,x1 // compute square
cmp x3,x5 // compare number
bgt 4f // end
add x1,x1,#2 // next divisor
udiv x2,x5,x1
msub x3,x1,x2,x5 // compute remainder
cmp x3,#0 // remainder null ?
bne 3f
2: // begin loop 2
udiv x2,x5,x1
msub x3,x1,x2,x5 // compute remainder
cmp x3,#0
csel x5,x2,x5,eq // new value = quotient
beq 2b
udiv x2,x4,x1 // divide totient
sub x4,x4,x2 // compute new totient
3:
cmp x1,#2 // first divisor ?
mov x0,1
csel x1,x0,x1,eq // divisor = 1
b 1b // and loop
4:
cmp x5,#1 // final value > 1
ble 5f
mov x0,x4 // totient
mov x1,x5 // divide by value
udiv x2,x4,x5 // totient divide by value
sub x4,x4,x2 // compute new totient
5:
mov x0,x4
100:
ldp x4,x5,[sp],16 // restaur registers
ldp x2,x3,[sp],16 // restaur registers
ldp x1,lr,[sp],16 // restaur registers
ret
/***************************************************/
/* Verification si un nombre est premier */
/***************************************************/
/* x0 contient le nombre à verifier */
/* x0 retourne 1 si premier 0 sinon */
isPrime:
stp x1,lr,[sp,-16]! // save registres
stp x2,x3,[sp,-16]! // save registres
mov x2,x0
sub x1,x0,#1
cmp x2,0
beq 99f // retourne zéro
cmp x2,2 // pour 1 et 2 retourne 1
ble 2f
mov x0,#2
bl moduloPur64
bcs 100f // erreur overflow
cmp x0,#1
bne 99f // Pas premier
cmp x2,3
beq 2f
mov x0,#3
bl moduloPur64
blt 100f // erreur overflow
cmp x0,#1
bne 99f
cmp x2,5
beq 2f
mov x0,#5
bl moduloPur64
bcs 100f // erreur overflow
cmp x0,#1
bne 99f // Pas premier
cmp x2,7
beq 2f
mov x0,#7
bl moduloPur64
bcs 100f // erreur overflow
cmp x0,#1
bne 99f // Pas premier
cmp x2,11
beq 2f
mov x0,#11
bl moduloPur64
bcs 100f // erreur overflow
cmp x0,#1
bne 99f // Pas premier
cmp x2,13
beq 2f
mov x0,#13
bl moduloPur64
bcs 100f // erreur overflow
cmp x0,#1
bne 99f // Pas premier
cmp x2,17
beq 2f
mov x0,#17
bl moduloPur64
bcs 100f // erreur overflow
cmp x0,#1
bne 99f // Pas premier
2:
cmn x0,0 // carry à zero pas d'erreur
mov x0,1 // premier
b 100f
99:
cmn x0,0 // carry à zero pas d'erreur
mov x0,#0 // Pas premier
100:
ldp x2,x3,[sp],16 // restaur des 2 registres
ldp x1,lr,[sp],16 // restaur des 2 registres
ret // retour adresse lr x30
/**************************************************************/
/********************************************************/
/* Calcul modulo de b puissance e modulo m */
/* Exemple 4 puissance 13 modulo 497 = 445 */
/********************************************************/
/* x0 nombre */
/* x1 exposant */
/* x2 modulo */
moduloPur64:
stp x1,lr,[sp,-16]! // save registres
stp x3,x4,[sp,-16]! // save registres
stp x5,x6,[sp,-16]! // save registres
stp x7,x8,[sp,-16]! // save registres
stp x9,x10,[sp,-16]! // save registres
cbz x0,100f
cbz x1,100f
mov x8,x0
mov x7,x1
mov x6,1 // resultat
udiv x4,x8,x2
msub x9,x4,x2,x8 // contient le reste
1:
tst x7,1
beq 2f
mul x4,x9,x6
umulh x5,x9,x6
//cbnz x5,99f
mov x6,x4
mov x0,x6
mov x1,x5
bl divisionReg128U
cbnz x1,99f // overflow
mov x6,x3
2:
mul x8,x9,x9
umulh x5,x9,x9
mov x0,x8
mov x1,x5
bl divisionReg128U
cbnz x1,99f // overflow
mov x9,x3
lsr x7,x7,1
cbnz x7,1b
mov x0,x6 // result
cmn x0,0 // carry à zero pas d'erreur
b 100f
99:
ldr x0,qAdrszMessOverflow
bl affichageMess
cmp x0,0 // carry à un car erreur
mov x0,-1 // code erreur
100:
ldp x9,x10,[sp],16 // restaur des 2 registres
ldp x7,x8,[sp],16 // restaur des 2 registres
ldp x5,x6,[sp],16 // restaur des 2 registres
ldp x3,x4,[sp],16 // restaur des 2 registres
ldp x1,lr,[sp],16 // restaur des 2 registres
ret // retour adresse lr x30
qAdrszMessOverflow: .quad szMessOverflow
/***************************************************/
/* division d un nombre de 128 bits par un nombre de 64 bits */
/***************************************************/
/* x0 contient partie basse dividende */
/* x1 contient partie haute dividente */
/* x2 contient le diviseur */
/* x0 retourne partie basse quotient */
/* x1 retourne partie haute quotient */
/* x3 retourne le reste */
divisionReg128U:
stp x6,lr,[sp,-16]! // save registres
stp x4,x5,[sp,-16]! // save registres
mov x5,#0 // raz du reste R
mov x3,#128 // compteur de boucle
mov x4,#0 // dernier bit
1:
lsl x5,x5,#1 // on decale le reste de 1
tst x1,1<<63 // test du bit le plus à gauche
lsl x1,x1,#1 // on decale la partie haute du quotient de 1
beq 2f
orr x5,x5,#1 // et on le pousse dans le reste R
2:
tst x0,1<<63
lsl x0,x0,#1 // puis on decale la partie basse
beq 3f
orr x1,x1,#1 // et on pousse le bit de gauche dans la partie haute
3:
orr x0,x0,x4 // position du dernier bit du quotient
mov x4,#0 // raz du bit
cmp x5,x2
blt 4f
sub x5,x5,x2 // on enleve le diviseur du reste
mov x4,#1 // dernier bit à 1
4:
// et boucle
subs x3,x3,#1
bgt 1b
lsl x1,x1,#1 // on decale le quotient de 1
tst x0,1<<63
lsl x0,x0,#1 // puis on decale la partie basse
beq 5f
orr x1,x1,#1
5:
orr x0,x0,x4 // position du dernier bit du quotient
mov x3,x5
100:
ldp x4,x5,[sp],16 // restaur des 2 registres
ldp x6,lr,[sp],16 // restaur des 2 registres
ret // retour adresse lr x30
/***************************************************/
/* ROUTINES INCLUDE */
/***************************************************/
.include "../includeARM64.inc"
|
http://rosettacode.org/wiki/Topswops | Topswops | Topswops is a card game created by John Conway in the 1970's.
Assume you have a particular permutation of a set of n cards numbered 1..n on both of their faces, for example the arrangement of four cards given by [2, 4, 1, 3] where the leftmost card is on top.
A round is composed of reversing the first m cards where m is the value of the topmost card.
Rounds are repeated until the topmost card is the number 1 and the number of swaps is recorded.
For our example the swaps produce:
[2, 4, 1, 3] # Initial shuffle
[4, 2, 1, 3]
[3, 1, 2, 4]
[2, 1, 3, 4]
[1, 2, 3, 4]
For a total of four swaps from the initial ordering to produce the terminating case where 1 is on top.
For a particular number n of cards, topswops(n) is the maximum swaps needed for any starting permutation of the n cards.
Task
The task is to generate and show here a table of n vs topswops(n) for n in the range 1..10 inclusive.
Note
Topswops is also known as Fannkuch from the German word Pfannkuchen meaning pancake.
Related tasks
Number reversal game
Sorting algorithms/Pancake sort
| #360_Assembly | 360 Assembly | * Topswops optimized 12/07/2016
TOPSWOPS CSECT
USING TOPSWOPS,R13 base register
B 72(R15) skip savearea
DC 17F'0' savearea
STM R14,R12,12(R13) prolog
ST R13,4(R15) " <-
ST R15,8(R13) " ->
LR R13,R15 " addressability
MVC N,=F'1' n=1
LOOPN L R4,N n; do n=1 to 10 ===-------------==*
C R4,=F'10' " *
BH ELOOPN . *
MVC P(40),PINIT p=pinit
MVC COUNTM,=F'0' countm=0
REPEAT MVC CARDS(40),P cards=p -------------------------+
SR R11,R11 count=0 |
WHILE CLC CARDS,=F'1' do while cards(1)^=1 ---------+
BE EWHILE . |
MVC M,CARDS m=cards(1)
L R2,M m
SRA R2,1 m/2
ST R2,MD2 md2=m/2
L R3,M @card(mm)=m
SLA R3,2 *4
LA R3,CARDS-4(R3) @card(mm)
LA R2,CARDS @card(i)=0
LA R6,1 i=1
LOOPI C R6,MD2 do i=1 to m/2 -------------+
BH ELOOPI . |
L R0,0(R2) swap r0=cards(i)
MVC 0(4,R2),0(R3) swap cards(i)=cards(mm)
ST R0,0(R3) swap cards(mm)=r0
AH R2,=H'4' @card(i)=@card(i)+4
SH R3,=H'4' @card(mm)=@card(mm)-4
LA R6,1(R6) i=i+1 |
B LOOPI ----------------------------+
ELOOPI LA R11,1(R11) count=count+1 |
B WHILE -------------------------------+
EWHILE C R11,COUNTM if count>countm
BNH NOTGT then
ST R11,COUNTM countm=count
NOTGT BAL R14,NEXTPERM call nextperm
LTR R0,R0 until nextperm=0 |
BNZ REPEAT ---------------------------------+
L R1,N n
XDECO R1,XDEC edit n
MVC PG(2),XDEC+10 output n
MVI PG+2,C':' output ':'
L R1,COUNTM countm
XDECO R1,XDEC edit countm
MVC PG+3(4),XDEC+8 output countm
XPRNT PG,L'PG print buffer
L R1,N n *
LA R1,1(R1) +1 *
ST R1,N n=n+1 *
B LOOPN ===------------------------------==*
ELOOPN L R13,4(0,R13) epilog
LM R14,R12,12(R13) " restore
XR R15,R15 " rc=0
BR R14 exit
PINIT DC F'1',F'2',F'3',F'4',F'5',F'6',F'7',F'8',F'9',F'10'
CARDS DS 10F cards
P DS 10F p
COUNTM DS F countm
M DS F m
N DS F n
MD2 DS F m/2
PG DC CL20' ' buffer
XDEC DS CL12 temp
*------- ---- nextperm ----------{-----------------------------------
NEXTPERM L R9,N nn=n
SR R8,R8 jj=0
LR R7,R9 nn
BCTR R7,0 j=nn-1
LTR R7,R7 if j=0
BZ ELOOPJ1 then skip do loop
LOOPJ1 LR R1,R7 do j=nn-1 to 1 by -1; j ----+
SLA R1,2 . |
L R2,P-4(R1) p(j)
C R2,P(R1) if p(j)<p(j+1)
BNL PJGEPJP then
LR R8,R7 jj=j
B ELOOPJ1 leave j |
PJGEPJP BCT R7,LOOPJ1 j=j-1 ---------------------+
ELOOPJ1 LA R7,1(R8) j=jj+1
LOOPJ2 CR R7,R9 do j=jj+1 while j<nn ------+
BNL ELOOPJ2 . |
LR R2,R7 j
SLA R2,2 .
LR R3,R9 nn
SLA R3,2 .
L R0,P-4(R2) swap p(j),p(nn)
L R1,P-4(R3) "
ST R0,P-4(R3) "
ST R1,P-4(R2) "
BCTR R9,0 nn=nn-1
LA R7,1(R7) j=j+1 |
B LOOPJ2 ----------------------------+
ELOOPJ2 LTR R8,R8 if jj=0
BNZ JJNE0 then
LA R0,0 return(0)
BR R14 "
JJNE0 LA R7,1(R8) j=jj+1
LR R2,R7 j
SLA R2,2 r@p(j)
LR R3,R8 jj
SLA R3,2 r@p(jj)
LOOPJ3 L R0,P-4(R2) p(j) ----------------------+
C R0,P-4(R3) do j=jj+1 while p(j)<p(jj) |
BNL ELOOPJ3
LA R2,4(R2) r@p(j)=r@p(j)+4
LA R7,1(R7) j=j+1 |
B LOOPJ3 ----------------------------+
ELOOPJ3 L R1,P-4(R3) swap p(j),p(jj)
ST R0,P-4(R3) "
ST R1,P-4(R2) "
LA R0,1 return(1)
BR R14 ---------------}-----------------------------------
YREGS
END TOPSWOPS |
http://rosettacode.org/wiki/Topswops | Topswops | Topswops is a card game created by John Conway in the 1970's.
Assume you have a particular permutation of a set of n cards numbered 1..n on both of their faces, for example the arrangement of four cards given by [2, 4, 1, 3] where the leftmost card is on top.
A round is composed of reversing the first m cards where m is the value of the topmost card.
Rounds are repeated until the topmost card is the number 1 and the number of swaps is recorded.
For our example the swaps produce:
[2, 4, 1, 3] # Initial shuffle
[4, 2, 1, 3]
[3, 1, 2, 4]
[2, 1, 3, 4]
[1, 2, 3, 4]
For a total of four swaps from the initial ordering to produce the terminating case where 1 is on top.
For a particular number n of cards, topswops(n) is the maximum swaps needed for any starting permutation of the n cards.
Task
The task is to generate and show here a table of n vs topswops(n) for n in the range 1..10 inclusive.
Note
Topswops is also known as Fannkuch from the German word Pfannkuchen meaning pancake.
Related tasks
Number reversal game
Sorting algorithms/Pancake sort
| #Ada | Ada | with Ada.Integer_Text_IO, Generic_Perm;
procedure Topswaps is
function Topswaps(Size: Positive) return Natural is
package Perms is new Generic_Perm(Size);
P: Perms.Permutation;
Done: Boolean;
Max: Natural;
function Swapper_Calls(P: Perms.Permutation) return Natural is
Q: Perms.Permutation := P;
I: Perms.Element := P(1);
begin
if I = 1 then
return 0;
else
for Idx in 1 .. I loop
Q(Idx) := P(I-Idx+1);
end loop;
return 1 + Swapper_Calls(Q);
end if;
end Swapper_Calls;
begin
Perms.Set_To_First(P, Done);
Max:= Swapper_Calls(P);
while not Done loop
Perms.Go_To_Next(P, Done);
Max := natural'Max(Max, Swapper_Calls(P));
end loop;
return Max;
end Topswaps;
begin
for I in 1 .. 10 loop
Ada.Integer_Text_IO.Put(Item => Topswaps(I), Width => 3);
end loop;
end Topswaps; |
http://rosettacode.org/wiki/Trigonometric_functions | Trigonometric functions | Task
If your language has a library or built-in functions for trigonometry, show examples of:
sine
cosine
tangent
inverses (of the above)
using the same angle in radians and degrees.
For the non-inverse functions, each radian/degree pair should use arguments that evaluate to the same angle (that is, it's not necessary to use the same angle for all three regular functions as long as the two sine calls use the same angle).
For the inverse functions, use the same number and convert its answer to radians and degrees.
If your language does not have trigonometric functions available or only has some available, write functions to calculate the functions based on any known approximation or identity.
| #ALGOL_W | ALGOL W | begin
% Algol W only supplies sin, cos and arctan as standard. We can define %
% arcsin, arccos and tan functions using these. The standard functions %
% use radians so we also provide versions that use degrees %
% convert degrees to radians %
real procedure toRadians( real value x ) ; pi * ( x / 180 );
% convert radians to degrees %
real procedure toDegrees( real value x ) ; 180 * ( x / pi );
% tan of an angle in radians %
real procedure tan( real value x ) ; sin( x ) / cos( x );
% arcsin in radians %
real procedure arcsin( real value x ) ; arctan( x / sqrt( 1 - ( x * x ) ) );
% arccos in radians %
real procedure arccos( real value x ) ; arctan( sqrt( 1 - ( x * x ) ) / x );
% sin of an angle in degrees %
real procedure sinD( real value x ) ; sin( toRadians( x ) );
% cos of an angle in degrees %
real procedure cosD( real value x ) ; cos( toRadians( x ) );
% tan of an angle in degrees %
real procedure tanD( real value x ) ; tan( toRadians( x ) );
% arctan in degrees %
real procedure arctanD( real value x ) ; toDegrees( arctan( x ) );
% arcsin in degrees %
real procedure arcsinD( real value x ) ; toDegrees( arcsin( x ) );
% arccos in degrees %
real procedure arccosD( real value x ) ; toDegrees( arccos( x ) );
% test the procedures %
begin
real piOver4, piOver3, oneOverRoot2, root3Over2;
piOver3 := pi / 3; piOver4 := pi / 4;
oneOverRoot2 := 1.0 / sqrt( 2 ); root3Over2 := sqrt( 3 ) / 2;
r_w := 12; r_d := 5; r_format := "A"; s_w := 0; % set output format %
write( "PI/4: ", piOver4, " 1/root(2): ", oneOverRoot2 );
write();
write( "sin 45 degrees: ", sinD( 45 ), " sin pi/4 radians: ", sin( piOver4 ) );
write( "cos 45 degrees: ", cosD( 45 ), " cos pi/4 radians: ", cos( piOver4 ) );
write( "tan 45 degrees: ", tanD( 45 ), " tan pi/4 radians: ", tan( piOver4 ) );
write();
write( "arcsin( sin( pi/4 radians ) ): ", arcsin( sin( piOver4 ) ) );
write( "arccos( cos( pi/4 radians ) ): ", arccos( cos( piOver4 ) ) );
write( "arctan( tan( pi/4 radians ) ): ", arctan( tan( piOver4 ) ) );
write();
write( "PI/3: ", piOver4, " root(3)/2: ", root3Over2 );
write();
write( "sin 60 degrees: ", sinD( 60 ), " sin pi/3 radians: ", sin( piOver3 ) );
write( "cos 60 degrees: ", cosD( 60 ), " cos pi/3 radians: ", cos( piOver3 ) );
write( "tan 60 degrees: ", tanD( 60 ), " tan pi/3 radians: ", tan( piOver3 ) );
write();
write( "arcsin( sin( 60 degrees ) ): ", arcsinD( sinD( 60 ) ) );
write( "arccos( cos( 60 degrees ) ): ", arccosD( cosD( 60 ) ) );
write( "arctan( tan( 60 degrees ) ): ", arctanD( tanD( 60 ) ) );
end
end. |
http://rosettacode.org/wiki/Trabb_Pardo%E2%80%93Knuth_algorithm | Trabb Pardo–Knuth algorithm | The TPK algorithm is an early example of a programming chrestomathy.
It was used in Donald Knuth and Luis Trabb Pardo's Stanford tech report The Early Development of Programming Languages.
The report traces the early history of work in developing computer languages in the 1940s and 1950s, giving several translations of the algorithm.
From the wikipedia entry:
ask for 11 numbers to be read into a sequence S
reverse sequence S
for each item in sequence S
result := call a function to do an operation
if result overflows
alert user
else
print result
The task is to implement the algorithm:
Use the function:
f
(
x
)
=
|
x
|
0.5
+
5
x
3
{\displaystyle f(x)=|x|^{0.5}+5x^{3}}
The overflow condition is an answer of greater than 400.
The 'user alert' should not stop processing of other items of the sequence.
Print a prompt before accepting eleven, textual, numeric inputs.
You may optionally print the item as well as its associated result, but the results must be in reverse order of input.
The sequence S may be 'implied' and so not shown explicitly.
Print and show the program in action from a typical run here. (If the output is graphical rather than text then either add a screendump or describe textually what is displayed).
| #ASIC | ASIC |
REM Trabb Pardo-Knuth algorithm
REM Used "magic numbers" because of strict specification of the algorithm.
DIM S@(10)
PRINT "Enter 11 numbers."
FOR I = 0 TO 10
I1= I + 1
PRINT I1;
PRINT " => ";
INPUT TMP@
S@(I) = TMP@
NEXT I
PRINT
REM Reverse
HALFIMAX = 10 / 2
FOR I = 0 TO HALFIMAX
IREV = 10 - I
TMP@ = S@(I)
S@(I) = S@(IREV)
S@(IREV) = TMP@
NEXT I
REM Results
REM Leading spaces in printed numbers are removed
CLS
FOR I = 0 TO 10
PRINT "f(";
STMP$ = STR$(S@(I))
STMP$ = LTRIM$(STMP$)
PRINT STMP$;
PRINT ") = ";
N@ = S@(I)
GOSUB CALCF:
R@ = F@
IF R@ > 400 THEN
PRINT "overflow"
ELSE
STMP$ = STR$(R@)
STMP$ = LTRIM$(STMP$)
PRINT STMP$
ENDIF
NEXT I
END
CALCF:
REM Calculates f(N@)
REM Result in F@
X@ = ABS(N@)
GOSUB CALCSQRT:
F@ = 5.0 * N@
F@ = F@ * N@
F@ = F@ * N@
F@ = F@ + SQRT@
RETURN
CALCSQRT:
REM Calculates approximate (+- 0.00001) square root of X@ for X@ >= 0 (bisection method)
REM Result in SQRT@
A@ = 0.0
IF X@ >= 1.0 THEN
B@ = X@
ELSE
B@ = 1.0
ENDIF
L@ = B@ - A@
L@ = ABS(L@)
WHILE L@ > 0.00001
MIDDLE@ = A@ + B@
MIDDLE@ = MIDDLE@ / 2
MIDDLETO2@ = MIDDLE@ * MIDDLE@
IF MIDDLETO2@ < X@ THEN
A@ = MIDDLE@
ELSE
B@ = MIDDLE@
ENDIF
L@ = B@ - A@
L@ = ABS(L@)
WEND
SQRT@ = MIDDLE@
RETURN
|
http://rosettacode.org/wiki/Tree_from_nesting_levels | Tree from nesting levels | Given a flat list of integers greater than zero, representing object nesting
levels, e.g. [1, 2, 4],
generate a tree formed from nested lists of those nesting level integers where:
Every int appears, in order, at its depth of nesting.
If the next level int is greater than the previous then it appears in a sub-list of the list containing the previous item
The generated tree data structure should ideally be in a languages nested list format that can
be used for further calculations rather than something just calculated for printing.
An input of [1, 2, 4] should produce the equivalent of: [1, [2, [[4]]]]
where 1 is at depth1, 2 is two deep and 4 is nested 4 deep.
[1, 2, 4, 2, 2, 1] should produce [1, [2, [[4]], 2, 2], 1].
All the nesting integers are in the same order but at the correct nesting
levels.
Similarly [3, 1, 3, 1] should generate [[[3]], 1, [[3]], 1]
Task
Generate and show here the results for the following inputs:
[]
[1, 2, 4]
[3, 1, 3, 1]
[1, 2, 3, 1]
[3, 2, 1, 3]
[3, 3, 3, 1, 1, 3, 3, 3]
| #Wren | Wren | import "/seq" for Stack
import "/fmt" for Fmt
var toTree = Fn.new { |list|
var nested = []
var s = Stack.new()
s.push(nested)
for (n in list) {
while (n != s.count) {
if (n > s.count) {
var inner = []
s.peek().add(inner)
s.push(inner)
} else {
s.pop()
}
}
s.peek().add(n)
}
return nested
}
var tests = [
[],
[1, 2, 4],
[3, 1, 3, 1],
[1, 2, 3, 1],
[3, 2, 1, 3],
[3, 3, 3, 1, 1, 3, 3, 3]
]
for (test in tests) {
var nest = toTree.call(test)
Fmt.print("$24n => $n", test, nest)
} |
http://rosettacode.org/wiki/Twelve_statements | Twelve statements | This puzzle is borrowed from math-frolic.blogspot.
Given the following twelve statements, which of them are true?
1. This is a numbered list of twelve statements.
2. Exactly 3 of the last 6 statements are true.
3. Exactly 2 of the even-numbered statements are true.
4. If statement 5 is true, then statements 6 and 7 are both true.
5. The 3 preceding statements are all false.
6. Exactly 4 of the odd-numbered statements are true.
7. Either statement 2 or 3 is true, but not both.
8. If statement 7 is true, then 5 and 6 are both true.
9. Exactly 3 of the first 6 statements are true.
10. The next two statements are both true.
11. Exactly 1 of statements 7, 8 and 9 are true.
12. Exactly 4 of the preceding statements are true.
Task
When you get tired of trying to figure it out in your head,
write a program to solve it, and print the correct answer or answers.
Extra credit
Print out a table of near misses, that is, solutions that are contradicted by only a single statement.
| #J | J | apply
128!:2
NB. example
'*:' apply 1 2 3
1 4 9 |
http://rosettacode.org/wiki/Twelve_statements | Twelve statements | This puzzle is borrowed from math-frolic.blogspot.
Given the following twelve statements, which of them are true?
1. This is a numbered list of twelve statements.
2. Exactly 3 of the last 6 statements are true.
3. Exactly 2 of the even-numbered statements are true.
4. If statement 5 is true, then statements 6 and 7 are both true.
5. The 3 preceding statements are all false.
6. Exactly 4 of the odd-numbered statements are true.
7. Either statement 2 or 3 is true, but not both.
8. If statement 7 is true, then 5 and 6 are both true.
9. Exactly 3 of the first 6 statements are true.
10. The next two statements are both true.
11. Exactly 1 of statements 7, 8 and 9 are true.
12. Exactly 4 of the preceding statements are true.
Task
When you get tired of trying to figure it out in your head,
write a program to solve it, and print the correct answer or answers.
Extra credit
Print out a table of near misses, that is, solutions that are contradicted by only a single statement.
| #Java | Java |
public class LogicPuzzle
{
boolean S[] = new boolean[13];
int Count = 0;
public boolean check2 ()
{
int count = 0;
for (int k = 7; k <= 12; k++)
if (S[k]) count++;
return S[2] == (count == 3);
}
public boolean check3 ()
{
int count = 0;
for (int k = 2; k <= 12; k += 2)
if (S[k]) count++;
return S[3] == (count == 2);
}
public boolean check4 ()
{
return S[4] == ( !S[5] || S[6] && S[7]);
}
public boolean check5 ()
{
return S[5] == ( !S[2] && !S[3] && !S[4]);
}
public boolean check6 ()
{
int count = 0;
for (int k = 1; k <= 11; k += 2)
if (S[k]) count++;
return S[6] == (count == 4);
}
public boolean check7 ()
{
return S[7] == ((S[2] || S[3]) && !(S[2] && S[3]));
}
public boolean check8 ()
{
return S[8] == ( !S[7] || S[5] && S[6]);
}
public boolean check9 ()
{
int count = 0;
for (int k = 1; k <= 6; k++)
if (S[k]) count++;
return S[9] == (count == 3);
}
public boolean check10 ()
{
return S[10] == (S[11] && S[12]);
}
public boolean check11 ()
{
int count = 0;
for (int k = 7; k <= 9; k++)
if (S[k]) count++;
return S[11] == (count == 1);
}
public boolean check12 ()
{
int count = 0;
for (int k = 1; k <= 11; k++)
if (S[k]) count++;
return S[12] == (count == 4);
}
public void check ()
{
if (check2() && check3() && check4() && check5() && check6()
&& check7() && check8() && check9() && check10() && check11()
&& check12())
{
for (int k = 1; k <= 12; k++)
if (S[k]) System.out.print(k + " ");
System.out.println();
Count++;
}
}
public void recurseAll (int k)
{
if (k == 13)
check();
else
{
S[k] = false;
recurseAll(k + 1);
S[k] = true;
recurseAll(k + 1);
}
}
public static void main (String args[])
{
LogicPuzzle P = new LogicPuzzle();
P.S[1] = true;
P.recurseAll(2);
System.out.println();
System.out.println(P.Count + " Solutions found.");
}
}
|
http://rosettacode.org/wiki/Truth_table | Truth table | A truth table is a display of the inputs to, and the output of a Boolean function organized as a table where each row gives one combination of input values and the corresponding value of the function.
Task
Input a Boolean function from the user as a string then calculate and print a formatted truth table for the given function.
(One can assume that the user input is correct).
Print and show output for Boolean functions of two and three input variables, but any program should not be limited to that many variables in the function.
Either reverse-polish or infix notation expressions are allowed.
Related tasks
Boolean values
Ternary logic
See also
Wolfram MathWorld entry on truth tables.
some "truth table" examples from Google.
| #Java | Java | import java.util.ArrayList;
import java.util.HashMap;
import java.util.Iterator;
import java.util.LinkedHashSet;
import java.util.List;
import java.util.Map;
import java.util.Set;
import java.util.Stack;
public class TruthTable {
public static void main( final String... args ) {
System.out.println( new TruthTable( args ) );
}
private interface Operator {
boolean evaluate( Stack<Boolean> s );
}
/**
* Supported operators and what they do. For more ops, add entries here.
*/
private static final Map<String,Operator> operators = new HashMap<String,Operator>() {{
// Can't use && or || because shortcut evaluation may mean the stack is not popped enough
put( "&", stack -> Boolean.logicalAnd( stack.pop(), stack.pop() ) );
put( "|", stack -> Boolean.logicalOr( stack.pop(), stack.pop() ) );
put( "!", stack -> ! stack.pop() );
put( "^", stack -> ! stack.pop().equals ( stack.pop() ) );
}};
private final List<String> variables;
private final String[] symbols;
/**
* Constructs a truth table for the symbols in an expression.
*/
public TruthTable( final String... symbols ) {
final Set<String> variables = new LinkedHashSet<>();
for ( final String symbol : symbols ) {
if ( ! operators.containsKey( symbol ) ) {
variables.add( symbol );
}
}
this.variables = new ArrayList<>( variables );
this.symbols = symbols;
}
@Override
public String toString () {
final StringBuilder result = new StringBuilder();
for ( final String variable : variables ) {
result.append( variable ).append( ' ' );
}
result.append( ' ' );
for ( final String symbol : symbols ) {
result.append( symbol ).append ( ' ' );
}
result.append( '\n' );
for ( final List<Boolean> values : enumerate( variables.size () ) ) {
final Iterator<String> i = variables.iterator();
for ( final Boolean value : values ) {
result.append(
String.format(
"%-" + i.next().length() + "c ",
value ? 'T' : 'F'
)
);
}
result.append( ' ' )
.append( evaluate( values ) ? 'T' : 'F' )
.append( '\n' );
}
return result.toString ();
}
/**
* Recursively generates T/F values
*/
private static List<List<Boolean>> enumerate( final int size ) {
if ( 1 == size )
return new ArrayList<List<Boolean>>() {{
add( new ArrayList<Boolean>() {{ add(false); }} );
add( new ArrayList<Boolean>() {{ add(true); }} );
}};
return new ArrayList<List<Boolean>>() {{
for ( final List<Boolean> head : enumerate( size - 1 ) ) {
add( new ArrayList<Boolean>( head ) {{ add(false); }} );
add( new ArrayList<Boolean>( head ) {{ add(true); }} );
}
}};
}
/**
* Evaluates the expression for a set of values.
*/
private boolean evaluate( final List<Boolean> enumeration ) {
final Iterator<Boolean> i = enumeration.iterator();
final Map<String,Boolean> values = new HashMap<>();
final Stack<Boolean> stack = new Stack<>();
variables.forEach ( v -> values.put( v, i.next() ) );
for ( final String symbol : symbols ) {
final Operator op = operators.get ( symbol );
// Reverse Polish notation makes this bit easy
stack.push(
null == op
? values.get ( symbol )
: op.evaluate ( stack )
);
}
return stack.pop();
}
} |
http://rosettacode.org/wiki/Ulam_spiral_(for_primes) | Ulam spiral (for primes) | An Ulam spiral (of primes) is a method of visualizing primes when expressed in a (normally counter-clockwise) outward spiral (usually starting at 1), constructed on a square grid, starting at the "center".
An Ulam spiral is also known as a prime spiral.
The first grid (green) is shown with sequential integers, starting at 1.
In an Ulam spiral of primes, only the primes are shown (usually indicated by some glyph such as a dot or asterisk), and all non-primes as shown as a blank (or some other whitespace).
Of course, the grid and border are not to be displayed (but they are displayed here when using these Wiki HTML tables).
Normally, the spiral starts in the "center", and the 2nd number is to the viewer's right and the number spiral starts from there in a counter-clockwise direction.
There are other geometric shapes that are used as well, including clock-wise spirals.
Also, some spirals (for the 2nd number) is viewed upwards from the 1st number instead of to the right, but that is just a matter of orientation.
Sometimes, the starting number can be specified to show more visual striking patterns (of prime densities).
[A larger than necessary grid (numbers wise) is shown here to illustrate the pattern of numbers on the diagonals (which may be used by the method to orientate the direction of spiral-construction algorithm within the example computer programs)].
Then, in the next phase in the transformation of the Ulam prime spiral, the non-primes are translated to blanks.
In the orange grid below, the primes are left intact, and all non-primes are changed to blanks.
Then, in the final transformation of the Ulam spiral (the yellow grid), translate the primes to a glyph such as a • or some other suitable glyph.
65
64
63
62
61
60
59
58
57
66
37
36
35
34
33
32
31
56
67
38
17
16
15
14
13
30
55
68
39
18
5
4
3
12
29
54
69
40
19
6
1
2
11
28
53
70
41
20
7
8
9
10
27
52
71
42
21
22
23
24
25
26
51
72
43
44
45
46
47
48
49
50
73
74
75
76
77
78
79
80
81
61
59
37
31
67
17
13
5
3
29
19
2
11
53
41
7
71
23
43
47
73
79
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
The Ulam spiral becomes more visually obvious as the grid increases in size.
Task
For any sized N × N grid, construct and show an Ulam spiral (counter-clockwise) of primes starting at some specified initial number (the default would be 1), with some suitably dotty (glyph) representation to indicate primes, and the absence of dots to indicate non-primes.
You should demonstrate the generator by showing at Ulam prime spiral large enough to (almost) fill your terminal screen.
Related tasks
Spiral matrix
Zig-zag matrix
Identity matrix
Sequence of primes by Trial Division
See also
Wikipedia entry: Ulam spiral
MathWorld™ entry: Prime Spiral
| #jq | jq | def array($init): [range(0; .) | $init];
# Test if input is a one-character string holding a digit
def isDigit:
type=="string" and length==1 and explode[0] as $c | (48 <= $c and $c <= 57);
def lpad($len): tostring | ($len - length) as $l | (" " * $l)[:$l] + .;
def generate($n; $i; $c):
if $n <= 1 then "'n' must be more than 1." | error
else { s: ($n|array(array(""))),
dir: "right",
y: (($n/2)|floor) }
| .x = if ($n % 2 == 0) then .y - 1 else .y end # shift left for even n
| reduce range($i; $n * $n + $i) as $j (.;
.s[.y][.x] = (
if $j | is_prime
then (if $c|isDigit then $j|lpad(4) else " \($c) " end)
else " ---"
end)
| if .dir == "right"
then if (.x <= $n - 1 and .s[.y - 1][.x] == "" and $j > i) then .dir = "up" else . end
elif .dir == "up"
then if (.s[.y][.x - 1] == "") then .dir = "left" else . end
elif .dir == "left"
then if (.x == 0 or .s[.y + 1][.x] == "") then .dir = "down" else . end
elif .dir == "down"
then if (.s[.y][.x + 1] == "") then .dir = "right" else . end
else .
end
| if .dir == "right" then .x += 1
elif .dir == "up" then .y += -1
elif .dir == "left" then .x += -1
elif .dir == "down" then .y += 1
else .
end )
| .s[] | join(" ")
end ;
# with digits
generate(9; 1; "0"), "",
# with *
generate(9; 1; "*")
|
http://rosettacode.org/wiki/Truncatable_primes | Truncatable primes | A truncatable prime is a prime number that when you successively remove digits from one end of the prime, you are left with a new prime number.
Examples
The number 997 is called a left-truncatable prime as the numbers 997, 97, and 7 are all prime.
The number 7393 is a right-truncatable prime as the numbers 7393, 739, 73, and 7 formed by removing digits from its right are also prime.
No zeroes are allowed in truncatable primes.
Task
The task is to find the largest left-truncatable and right-truncatable primes less than one million (base 10 is implied).
Related tasks
Find largest left truncatable prime in a given base
Sieve of Eratosthenes
See also
Truncatable Prime from MathWorld.]
| #D | D | import std.stdio, std.math, std.string, std.conv, std.algorithm,
std.range;
bool isPrime(in int n) pure nothrow {
if (n <= 1)
return false;
foreach (immutable i; 2 .. cast(int)sqrt(real(n)) + 1)
if (!(n % i))
return false;
return true;
}
bool isTruncatablePrime(bool left)(in int n) pure {
immutable s = n.text;
if (s.canFind('0'))
return false;
foreach (immutable i; 0 .. s.length)
static if (left) {
if (!s[i .. $].to!int.isPrime)
return false;
} else {
if (!s[0 .. i + 1].to!int.isPrime)
return false;
}
return true;
}
void main() {
enum n = 1_000_000;
writeln("Largest left-truncatable prime in 2 .. ", n, ": ",
iota(n, 1, -1).filter!(isTruncatablePrime!true).front);
writeln("Largest right-truncatable prime in 2 .. ", n, ": ",
iota(n, 1, -1).filter!(isTruncatablePrime!false).front);
} |
http://rosettacode.org/wiki/Truncate_a_file | Truncate a file | Task
Truncate a file to a specific length. This should be implemented as a routine that takes two parameters: the filename and the required file length (in bytes).
Truncation can be achieved using system or library calls intended for such a task, if such methods exist, or by creating a temporary file of a reduced size and renaming it, after first deleting the original file, if no other method is available. The file may contain non human readable binary data in an unspecified format, so the routine should be "binary safe", leaving the contents of the untruncated part of the file unchanged.
If the specified filename does not exist, or the provided length is not less than the current file length, then the routine should raise an appropriate error condition.
On some systems, the provided file truncation facilities might not change the file or may extend the file, if the specified length is greater than the current length of the file.
This task permits the use of such facilities. However, such behaviour should be noted, or optionally a warning message relating to an non change or increase in file size may be implemented.
| #Ring | Ring |
file = "C:\Ring\ReadMe.txt"
fp = read(file)
fpstr = left(fp, 100)
see fpstr + nl
write(file, fpstr)
|
http://rosettacode.org/wiki/Truncate_a_file | Truncate a file | Task
Truncate a file to a specific length. This should be implemented as a routine that takes two parameters: the filename and the required file length (in bytes).
Truncation can be achieved using system or library calls intended for such a task, if such methods exist, or by creating a temporary file of a reduced size and renaming it, after first deleting the original file, if no other method is available. The file may contain non human readable binary data in an unspecified format, so the routine should be "binary safe", leaving the contents of the untruncated part of the file unchanged.
If the specified filename does not exist, or the provided length is not less than the current file length, then the routine should raise an appropriate error condition.
On some systems, the provided file truncation facilities might not change the file or may extend the file, if the specified length is greater than the current length of the file.
This task permits the use of such facilities. However, such behaviour should be noted, or optionally a warning message relating to an non change or increase in file size may be implemented.
| #Ruby | Ruby | # Open a file for writing, and truncate it to 1234 bytes.
File.open("file", "ab") do |f|
f.truncate(1234)
f << "Killroy was here" # write to file
end # file is closed now.
# Just truncate a file to 567 bytes.
File.truncate("file", 567) |
http://rosettacode.org/wiki/Truncate_a_file | Truncate a file | Task
Truncate a file to a specific length. This should be implemented as a routine that takes two parameters: the filename and the required file length (in bytes).
Truncation can be achieved using system or library calls intended for such a task, if such methods exist, or by creating a temporary file of a reduced size and renaming it, after first deleting the original file, if no other method is available. The file may contain non human readable binary data in an unspecified format, so the routine should be "binary safe", leaving the contents of the untruncated part of the file unchanged.
If the specified filename does not exist, or the provided length is not less than the current file length, then the routine should raise an appropriate error condition.
On some systems, the provided file truncation facilities might not change the file or may extend the file, if the specified length is greater than the current length of the file.
This task permits the use of such facilities. However, such behaviour should be noted, or optionally a warning message relating to an non change or increase in file size may be implemented.
| #Rust | Rust | use std::path::Path;
use std::fs;
fn truncate_file<P: AsRef<Path>>(filename: P, filesize: usize) -> Result<(), Error> {
use Error::*;
let file = fs::read(&filename).or(Err(NotFound))?;
if filesize > file.len() {
return Err(FilesizeTooSmall)
}
fs::write(&filename, &file[..filesize]).or(Err(UnableToWrite))?;
Ok(())
}
#[derive(Debug)]
enum Error {
/// File not found
NotFound,
/// Truncated size would be larger than the current size
FilesizeTooSmall,
/// Likely due to having read but not write permissions
UnableToWrite,
} |
http://rosettacode.org/wiki/Universal_Turing_machine | Universal Turing machine | One of the foundational mathematical constructs behind computer science
is the universal Turing Machine.
(Alan Turing introduced the idea of such a machine in 1936–1937.)
Indeed one way to definitively prove that a language
is turing-complete
is to implement a universal Turing machine in it.
Task
Simulate such a machine capable
of taking the definition of any other Turing machine and executing it.
Of course, you will not have an infinite tape,
but you should emulate this as much as is possible.
The three permissible actions on the tape are "left", "right" and "stay".
To test your universal Turing machine (and prove your programming language
is Turing complete!), you should execute the following two Turing machines
based on the following definitions.
Simple incrementer
States: q0, qf
Initial state: q0
Terminating states: qf
Permissible symbols: B, 1
Blank symbol: B
Rules:
(q0, 1, 1, right, q0)
(q0, B, 1, stay, qf)
The input for this machine should be a tape of 1 1 1
Three-state busy beaver
States: a, b, c, halt
Initial state: a
Terminating states: halt
Permissible symbols: 0, 1
Blank symbol: 0
Rules:
(a, 0, 1, right, b)
(a, 1, 1, left, c)
(b, 0, 1, left, a)
(b, 1, 1, right, b)
(c, 0, 1, left, b)
(c, 1, 1, stay, halt)
The input for this machine should be an empty tape.
Bonus:
5-state, 2-symbol probable Busy Beaver machine from Wikipedia
States: A, B, C, D, E, H
Initial state: A
Terminating states: H
Permissible symbols: 0, 1
Blank symbol: 0
Rules:
(A, 0, 1, right, B)
(A, 1, 1, left, C)
(B, 0, 1, right, C)
(B, 1, 1, right, B)
(C, 0, 1, right, D)
(C, 1, 0, left, E)
(D, 0, 1, left, A)
(D, 1, 1, left, D)
(E, 0, 1, stay, H)
(E, 1, 0, left, A)
The input for this machine should be an empty tape.
This machine runs for more than 47 millions steps.
| #C.23 | C# | using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Linq;
using System.Text;
using System.Threading;
using System.Threading.Tasks;
public class TuringMachine
{
public static async Task Main() {
var fiveStateBusyBeaver = new TuringMachine("A", '0', "H").WithTransitions(
("A", '0', '1', Right, "B"),
("A", '1', '1', Left, "C"),
("B", '0', '1', Right, "C"),
("B", '1', '1', Right, "B"),
("C", '0', '1', Right, "D"),
("C", '1', '0', Left, "E"),
("D", '0', '1', Left, "A"),
("D", '1', '1', Left, "D"),
("E", '0', '1', Stay, "H"),
("E", '1', '0', Left, "A")
);
var busyBeaverTask = fiveStateBusyBeaver.TimeAsync();
var incrementer = new TuringMachine("q0", 'B', "qf").WithTransitions(
("q0", '1', '1', Right, "q0"),
("q0", 'B', '1', Stay, "qf")
)
.WithInput("111");
foreach (var _ in incrementer.Run()) PrintLine(incrementer);
PrintResults(incrementer);
var threeStateBusyBeaver = new TuringMachine("a", '0', "halt").WithTransitions(
("a", '0', '1', Right, "b"),
("a", '1', '1', Left, "c"),
("b", '0', '1', Left, "a"),
("b", '1', '1', Right, "b"),
("c", '0', '1', Left, "b"),
("c", '1', '1', Stay, "halt")
);
foreach (var _ in threeStateBusyBeaver.Run()) PrintLine(threeStateBusyBeaver);
PrintResults(threeStateBusyBeaver);
var sorter = new TuringMachine("A", '*', "X").WithTransitions(
("A", 'a', 'a', Right, "A"),
("A", 'b', 'B', Right, "B"),
("A", '*', '*', Left, "E"),
("B", 'a', 'a', Right, "B"),
("B", 'b', 'b', Right, "B"),
("B", '*', '*', Left, "C"),
("C", 'a', 'b', Left, "D"),
("C", 'b', 'b', Left, "C"),
("C", 'B', 'b', Left, "E"),
("D", 'a', 'a', Left, "D"),
("D", 'b', 'b', Left, "D"),
("D", 'B', 'a', Right, "A"),
("E", 'a', 'a', Left, "E"),
("E", '*', '*', Right, "X")
)
.WithInput("babbababaa");
sorter.Run().Last();
Console.WriteLine("Sorted: " + sorter.TapeString);
PrintResults(sorter);
sorter.Reset().WithInput("bbbababaaabba");
sorter.Run().Last();
Console.WriteLine("Sorted: " + sorter.TapeString);
PrintResults(sorter);
Console.WriteLine(await busyBeaverTask);
PrintResults(fiveStateBusyBeaver);
void PrintLine(TuringMachine tm) => Console.WriteLine(tm.TapeString + "\tState " + tm.State);
void PrintResults(TuringMachine tm) {
Console.WriteLine($"End state: {tm.State} = {(tm.Success ? "Success" : "Failure")}");
Console.WriteLine(tm.Steps + " steps");
Console.WriteLine("tape length: " + tm.TapeLength);
Console.WriteLine();
}
}
public const int Left = -1, Stay = 0, Right = 1;
private readonly Tape tape;
private readonly string initialState;
private readonly HashSet<string> terminatingStates;
private Dictionary<(string state, char read), (char write, int move, string toState)> transitions;
public TuringMachine(string initialState, char blankSymbol, params string[] terminatingStates) {
State = this.initialState = initialState;
tape = new Tape(blankSymbol);
this.terminatingStates = terminatingStates.ToHashSet();
}
public TuringMachine WithTransitions(
params (string state, char read, char write, int move, string toState)[] transitions)
{
this.transitions = transitions.ToDictionary(k => (k.state, k.read), k => (k.write, k.move, k.toState));
return this;
}
public TuringMachine Reset() {
State = initialState;
Steps = 0;
tape.Reset();
return this;
}
public TuringMachine WithInput(string input) {
tape.Input(input);
return this;
}
public int Steps { get; private set; }
public string State { get; private set; }
public bool Success => terminatingStates.Contains(State);
public int TapeLength => tape.Length;
public string TapeString => tape.ToString();
public IEnumerable<string> Run() {
yield return State;
while (Step()) yield return State;
}
public async Task<TimeSpan> TimeAsync(CancellationToken cancel = default) {
var chrono = Stopwatch.StartNew();
await RunAsync(cancel);
chrono.Stop();
return chrono.Elapsed;
}
public Task RunAsync(CancellationToken cancel = default)
=> Task.Run(() => {
while (Step()) cancel.ThrowIfCancellationRequested();
});
private bool Step() {
if (!transitions.TryGetValue((State, tape.Current), out var action)) return false;
tape.Current = action.write;
tape.Move(action.move);
State = action.toState;
Steps++;
return true;
}
private class Tape
{
private List<char> forwardTape = new List<char>(), backwardTape = new List<char>();
private int head = 0;
private char blank;
public Tape(char blankSymbol) => forwardTape.Add(blank = blankSymbol);
public void Reset() {
backwardTape.Clear();
forwardTape.Clear();
head = 0;
forwardTape.Add(blank);
}
public void Input(string input) {
Reset();
forwardTape.Clear();
forwardTape.AddRange(input);
}
public void Move(int direction) {
head += direction;
if (head >= 0 && forwardTape.Count <= head) forwardTape.Add(blank);
if (head < 0 && backwardTape.Count <= ~head) backwardTape.Add(blank);
}
public char Current {
get => head < 0 ? backwardTape[~head] : forwardTape[head];
set {
if (head < 0) backwardTape[~head] = value;
else forwardTape[head] = value;
}
}
public int Length => backwardTape.Count + forwardTape.Count;
public override string ToString() {
int h = (head < 0 ? ~head : backwardTape.Count + head) * 2 + 1;
var builder = new StringBuilder(" ", Length * 2 + 1);
if (backwardTape.Count > 0) {
builder.Append(string.Join(" ", backwardTape)).Append(" ");
if (head < 0) (builder[h + 1], builder[h - 1]) = ('(', ')');
for (int l = 0, r = builder.Length - 1; l < r; l++, r--) (builder[l], builder[r]) = (builder[r], builder[l]);
}
builder.Append(string.Join(" ", forwardTape)).Append(" ");
if (head >= 0) (builder[h - 1], builder[h + 1]) = ('(', ')');
return builder.ToString();
}
}
} |
http://rosettacode.org/wiki/Totient_function | Totient function | The totient function is also known as:
Euler's totient function
Euler's phi totient function
phi totient function
Φ function (uppercase Greek phi)
φ function (lowercase Greek phi)
Definitions (as per number theory)
The totient function:
counts the integers up to a given positive integer n that are relatively prime to n
counts the integers k in the range 1 ≤ k ≤ n for which the greatest common divisor gcd(n,k) is equal to 1
counts numbers ≤ n and prime to n
If the totient number (for N) is one less than N, then N is prime.
Task
Create a totient function and:
Find and display (1 per line) for the 1st 25 integers:
the integer (the index)
the totient number for that integer
indicate if that integer is prime
Find and display the count of the primes up to 100
Find and display the count of the primes up to 1,000
Find and display the count of the primes up to 10,000
Find and display the count of the primes up to 100,000 (optional)
Show all output here.
Related task
Perfect totient numbers
Also see
Wikipedia: Euler's totient function.
MathWorld: totient function.
OEIS: Euler totient function phi(n).
| #Ada | Ada | with Ada.Text_IO;
with Ada.Integer_Text_IO;
procedure Totient is
function Totient (N : in Integer) return Integer is
Tot : Integer := N;
I : Integer;
N2 : Integer := N;
begin
I := 2;
while I * I <= N2 loop
if N2 mod I = 0 then
while N2 mod I = 0 loop
N2 := N2 / I;
end loop;
Tot := Tot - Tot / I;
end if;
if I = 2 then
I := 1;
end if;
I := I + 2;
end loop;
if N2 > 1 then
Tot := Tot - Tot / N2;
end if;
return Tot;
end Totient;
Count : Integer := 0;
Tot : Integer;
Placeholder : String := " n Phi Is_Prime";
Image_N : String renames Placeholder ( 1 .. 3);
Image_Phi : String renames Placeholder ( 6 .. 8);
Image_Prime : String renames Placeholder (11 .. 17);
use Ada.Text_IO;
use Ada.Integer_Text_IO;
begin
Put_Line (Placeholder);
for N in 1 .. 25 loop
Tot := Totient (N);
if N - 1 = Tot then
Count := Count + 1;
end if;
Put (Image_N, N);
Put (Image_Phi, Tot);
Image_Prime := (if N - 1 = Tot then " True" else " False");
Put_Line (Placeholder);
end loop;
New_Line;
Put_Line ("Number of primes up to " & Integer'(25)'Image &" =" & Count'Image);
for N in 26 .. 100_000 loop
Tot := Totient (N);
if Tot = N - 1 then
Count := Count + 1;
end if;
if N = 100 or N = 1_000 or N mod 10_000 = 0 then
Put_Line ("Number of primes up to " & N'Image & " =" & Count'Image);
end if;
end loop;
end Totient; |
http://rosettacode.org/wiki/Topswops | Topswops | Topswops is a card game created by John Conway in the 1970's.
Assume you have a particular permutation of a set of n cards numbered 1..n on both of their faces, for example the arrangement of four cards given by [2, 4, 1, 3] where the leftmost card is on top.
A round is composed of reversing the first m cards where m is the value of the topmost card.
Rounds are repeated until the topmost card is the number 1 and the number of swaps is recorded.
For our example the swaps produce:
[2, 4, 1, 3] # Initial shuffle
[4, 2, 1, 3]
[3, 1, 2, 4]
[2, 1, 3, 4]
[1, 2, 3, 4]
For a total of four swaps from the initial ordering to produce the terminating case where 1 is on top.
For a particular number n of cards, topswops(n) is the maximum swaps needed for any starting permutation of the n cards.
Task
The task is to generate and show here a table of n vs topswops(n) for n in the range 1..10 inclusive.
Note
Topswops is also known as Fannkuch from the German word Pfannkuchen meaning pancake.
Related tasks
Number reversal game
Sorting algorithms/Pancake sort
| #AutoHotkey | AutoHotkey | Topswops(Obj, n){
R := []
for i, val in obj{
if (i <=n)
res := val (A_Index=1?"":",") res
else
res .= "," val
}
Loop, Parse, res, `,
R[A_Index]:= A_LoopField
return R
} |
http://rosettacode.org/wiki/Topswops | Topswops | Topswops is a card game created by John Conway in the 1970's.
Assume you have a particular permutation of a set of n cards numbered 1..n on both of their faces, for example the arrangement of four cards given by [2, 4, 1, 3] where the leftmost card is on top.
A round is composed of reversing the first m cards where m is the value of the topmost card.
Rounds are repeated until the topmost card is the number 1 and the number of swaps is recorded.
For our example the swaps produce:
[2, 4, 1, 3] # Initial shuffle
[4, 2, 1, 3]
[3, 1, 2, 4]
[2, 1, 3, 4]
[1, 2, 3, 4]
For a total of four swaps from the initial ordering to produce the terminating case where 1 is on top.
For a particular number n of cards, topswops(n) is the maximum swaps needed for any starting permutation of the n cards.
Task
The task is to generate and show here a table of n vs topswops(n) for n in the range 1..10 inclusive.
Note
Topswops is also known as Fannkuch from the German word Pfannkuchen meaning pancake.
Related tasks
Number reversal game
Sorting algorithms/Pancake sort
| #C | C | #include <stdio.h>
#include <string.h>
typedef struct { char v[16]; } deck;
typedef unsigned int uint;
uint n, d, best[16];
void tryswaps(deck *a, uint f, uint s) {
# define A a->v
# define B b.v
if (d > best[n]) best[n] = d;
while (1) {
if ((A[s] == s || (A[s] == -1 && !(f & 1U << s)))
&& (d + best[s] >= best[n] || A[s] == -1))
break;
if (d + best[s] <= best[n]) return;
if (!--s) return;
}
d++;
deck b = *a;
for (uint i = 1, k = 2; i <= s; k <<= 1, i++) {
if (A[i] != i && (A[i] != -1 || (f & k)))
continue;
for (uint j = B[0] = i; j--;) B[i - j] = A[j];
tryswaps(&b, f | k, s);
}
d--;
}
int main(void) {
deck x;
memset(&x, -1, sizeof(x));
x.v[0] = 0;
for (n = 1; n < 13; n++) {
tryswaps(&x, 1, n - 1);
printf("%2d: %d\n", n, best[n]);
}
return 0;
} |
http://rosettacode.org/wiki/Trigonometric_functions | Trigonometric functions | Task
If your language has a library or built-in functions for trigonometry, show examples of:
sine
cosine
tangent
inverses (of the above)
using the same angle in radians and degrees.
For the non-inverse functions, each radian/degree pair should use arguments that evaluate to the same angle (that is, it's not necessary to use the same angle for all three regular functions as long as the two sine calls use the same angle).
For the inverse functions, use the same number and convert its answer to radians and degrees.
If your language does not have trigonometric functions available or only has some available, write functions to calculate the functions based on any known approximation or identity.
| #Arturo | Arturo | pi: 4*atan 1.0
radians: pi/4
degrees: 45.0
print "sine"
print [sin radians, sin degrees*pi/180]
print "cosine"
print [cos radians, cos degrees*pi/180]
print "tangent"
print [tan radians, tan degrees*pi/180]
print "arcsine"
print [asin sin radians, (asin sin radians)*180/pi]
print "arccosine"
print [acos cos radians, (acos cos radians)*180/pi]
print "arctangent"
print [atan tan radians, (atan tan radians)*180/pi] |
http://rosettacode.org/wiki/Trabb_Pardo%E2%80%93Knuth_algorithm | Trabb Pardo–Knuth algorithm | The TPK algorithm is an early example of a programming chrestomathy.
It was used in Donald Knuth and Luis Trabb Pardo's Stanford tech report The Early Development of Programming Languages.
The report traces the early history of work in developing computer languages in the 1940s and 1950s, giving several translations of the algorithm.
From the wikipedia entry:
ask for 11 numbers to be read into a sequence S
reverse sequence S
for each item in sequence S
result := call a function to do an operation
if result overflows
alert user
else
print result
The task is to implement the algorithm:
Use the function:
f
(
x
)
=
|
x
|
0.5
+
5
x
3
{\displaystyle f(x)=|x|^{0.5}+5x^{3}}
The overflow condition is an answer of greater than 400.
The 'user alert' should not stop processing of other items of the sequence.
Print a prompt before accepting eleven, textual, numeric inputs.
You may optionally print the item as well as its associated result, but the results must be in reverse order of input.
The sequence S may be 'implied' and so not shown explicitly.
Print and show the program in action from a typical run here. (If the output is graphical rather than text then either add a screendump or describe textually what is displayed).
| #AutoIt | AutoIt | ; Trabb Pardo–Knuth algorithm
; by James1337 (autoit.de)
; AutoIt Version: 3.3.8.1
Local $S, $i, $y
Do
$S = InputBox("Trabb Pardo–Knuth algorithm", "Please enter 11 numbers:", "1 2 3 4 5 6 7 8 9 10 11")
If @error Then Exit
$S = StringSplit($S, " ")
Until ($S[0] = 11)
For $i = 11 To 1 Step -1
$y = f($S[$i])
If ($y > 400) Then
ConsoleWrite("f(" & $S[$i] & ") = Overflow!" & @CRLF)
Else
ConsoleWrite("f(" & $S[$i] & ") = " & $y & @CRLF)
EndIf
Next
Func f($x)
Return Sqrt(Abs($x)) + 5*$x^3
EndFunc |
http://rosettacode.org/wiki/Twelve_statements | Twelve statements | This puzzle is borrowed from math-frolic.blogspot.
Given the following twelve statements, which of them are true?
1. This is a numbered list of twelve statements.
2. Exactly 3 of the last 6 statements are true.
3. Exactly 2 of the even-numbered statements are true.
4. If statement 5 is true, then statements 6 and 7 are both true.
5. The 3 preceding statements are all false.
6. Exactly 4 of the odd-numbered statements are true.
7. Either statement 2 or 3 is true, but not both.
8. If statement 7 is true, then 5 and 6 are both true.
9. Exactly 3 of the first 6 statements are true.
10. The next two statements are both true.
11. Exactly 1 of statements 7, 8 and 9 are true.
12. Exactly 4 of the preceding statements are true.
Task
When you get tired of trying to figure it out in your head,
write a program to solve it, and print the correct answer or answers.
Extra credit
Print out a table of near misses, that is, solutions that are contradicted by only a single statement.
| #jq | jq | def indexed(filter):
. as $in
| reduce range(0;length) as $i ([]; if ($i | filter) then . + [$in[$i]] else . end);
def count(value): map(select(. == value)) | length;
# The truth or falsity of the 12 statements can be captured in an array of size 12:
def generate(k):
if k == 1 then [true], [false]
else generate(1) + generate(k-1)
end;
# Input: a boolean array
def evaluate:
[ (length == 12), #1
((.[6:] | count(true)) == 3), #2
((indexed(. % 2 == 1) | count(true)) == 2), #3
(if .[4] then .[5] and .[6] else true end), #4
((.[1:4] | count(false)) == 3), #5
((indexed(. % 2 == 0) | count(true)) == 4), #6
(([.[1], .[2]] | count(true)) == 1), #7
(if .[6] then .[4] and .[5] else true end), #8
((.[0:6] | count(true)) == 3), #9
(.[10] and .[11]), #10
((.[6:9] | count(true)) == 1), #11
((.[0:11] | count(true)) == 4) #12
];
# The following query generates the solution to the problem:
# generate(12) | . as $vector | if evaluate == $vector then $vector else empty end
# Running "task" as defined next would generate
# both the general solution as well as the off-by-one solutions:
def task:
# count agreements
def agreed(x;y): reduce range(0;x|length) as $i (0; if x[$i] == y[$i] then .+1 else . end);
reduce generate(12) as $vector
([]; ($vector | evaluate) as $e
| agreed($vector; $e) as $agreed
| if $agreed == 12 then [[12,$vector]] + .
elif $agreed == 11 then . + [[11, $vector]]
else .
end);
# Since the solutions have been given elsewhere, we simply count the
# number of exact and off-by-one solutions:
task | length |
http://rosettacode.org/wiki/Twelve_statements | Twelve statements | This puzzle is borrowed from math-frolic.blogspot.
Given the following twelve statements, which of them are true?
1. This is a numbered list of twelve statements.
2. Exactly 3 of the last 6 statements are true.
3. Exactly 2 of the even-numbered statements are true.
4. If statement 5 is true, then statements 6 and 7 are both true.
5. The 3 preceding statements are all false.
6. Exactly 4 of the odd-numbered statements are true.
7. Either statement 2 or 3 is true, but not both.
8. If statement 7 is true, then 5 and 6 are both true.
9. Exactly 3 of the first 6 statements are true.
10. The next two statements are both true.
11. Exactly 1 of statements 7, 8 and 9 are true.
12. Exactly 4 of the preceding statements are true.
Task
When you get tired of trying to figure it out in your head,
write a program to solve it, and print the correct answer or answers.
Extra credit
Print out a table of near misses, that is, solutions that are contradicted by only a single statement.
| #Julia | Julia | using Printf
function showflaggedbits{T<:BitArray{1}}(a::T, f::T)
tf = map(x->x ? "T" : "F", a)
flg = map(x->x ? "*" : " ", f)
join(tf .* flg, " ")
end
const props = [s -> length(s) == 12,
s -> sum(s[7:12]) == 3,
s -> sum(s[2:2:end]) == 2,
s -> !s[5] || (s[6] & s[7]),
s -> !any(s[2:4]),
s -> sum(s[1:2:end]) == 4,
s -> s[2] $ s[3],
s -> !s[7] || (s[5] & s[6]),
s -> sum(s[1:6]) == 3,
s -> s[11] & s[12],
s -> sum(s[7:9]) == 1,
s -> sum(s[1:end-1]) == 4]
const NDIG = length(props)
NDIG < WORD_SIZE || println("WARNING, too many propositions!")
mhist = zeros(Int, NDIG+1)
println("Checking the ", NDIG, " statements against all possibilities.\n")
print(" "^15)
for i in 1:NDIG
print(@sprintf "%3d" i)
end
println()
for i in 0:(2^NDIG-1)
s = bitpack(digits(i, 2, NDIG))
t = bitpack([p(s) for p in props])
misses = s$t
mcnt = sum(misses)
mhist[NDIG-mcnt+1] += 1
mcnt < 2 || mcnt == NDIG || continue
if mcnt == 0
print(" Exact Match: ")
elseif mcnt == NDIG
print(" Total Miss: ")
else
print(" Near Miss: ")
end
println(showflaggedbits(t, misses))
end
println()
println("Distribution of matches")
println(" Matches Cases")
for i in (NDIG+1):-1:1
println(@sprintf " %2d => %4d" i-1 mhist[i])
end
|
http://rosettacode.org/wiki/Truth_table | Truth table | A truth table is a display of the inputs to, and the output of a Boolean function organized as a table where each row gives one combination of input values and the corresponding value of the function.
Task
Input a Boolean function from the user as a string then calculate and print a formatted truth table for the given function.
(One can assume that the user input is correct).
Print and show output for Boolean functions of two and three input variables, but any program should not be limited to that many variables in the function.
Either reverse-polish or infix notation expressions are allowed.
Related tasks
Boolean values
Ternary logic
See also
Wolfram MathWorld entry on truth tables.
some "truth table" examples from Google.
| #JavaScript | JavaScript | <!DOCTYPE html><html><head><title>Truth table</title><script>
var elem,expr,vars;
function isboolop(chr){return "&|!^".indexOf(chr)!=-1;}
function varsindexof(chr){
var i;
for(i=0;i<vars.length;i++){if(vars[i][0]==chr)return i;}
return -1;
}
function printtruthtable(){
var i,str;
elem=document.createElement("pre");
expr=prompt("Boolean expression:\nAccepts single-character variables (except for \"T\" and \"F\", which specify explicit true or false values), postfix, with \"&|!^\" for and, or, not, xor, respectively; optionally seperated by whitespace.").replace(/\s/g,"");
vars=[];
for(i=0;i<expr.length;i++)if(!isboolop(expr[i])&&expr[i]!="T"&&expr[i]!="F"&&varsindexof(expr[i])==-1)vars.push([expr[i],-1]);
if(vars.length==0)return;
str="";
for(i=0;i<vars.length;i++)str+=vars[i][0]+" ";
elem.innerHTML="<b>"+str+expr+"</b>\n";
vars[0][1]=false;
truthpartfor(1);
vars[0][1]=true;
truthpartfor(1);
vars[0][1]=-1;
document.body.appendChild(elem);
}
function truthpartfor(index){
if(index==vars.length){
var str,i;
str="";
for(i=0;i<index;i++)str+=(vars[i][1]?"<b>T</b>":"F")+" ";
elem.innerHTML+=str+(parsebool()?"<b>T</b>":"F")+"\n";
return;
}
vars[index][1]=false;
truthpartfor(index+1);
vars[index][1]=true;
truthpartfor(index+1);
vars[index][1]=-1;
}
function parsebool(){
var stack,i,idx;
console.log(vars);
stack=[];
for(i=0;i<expr.length;i++){
if(expr[i]=="T")stack.push(true);
else if(expr[i]=="F")stack.push(false);
else if((idx=varsindexof(expr[i]))!=-1)stack.push(vars[idx][1]);
else if(isboolop(expr[i])){
switch(expr[i]){
case "&":stack.push(stack.pop()&stack.pop());break;
case "|":stack.push(stack.pop()|stack.pop());break;
case "!":stack.push(!stack.pop());break;
case "^":stack.push(stack.pop()^stack.pop());break;
}
} else alert("Non-conformant character "+expr[i]+" in expression. Should not be possible.");
console.log(stack);
}
return stack[0];
}
</script></head><body onload="printtruthtable()"></body></html> |
http://rosettacode.org/wiki/Ulam_spiral_(for_primes) | Ulam spiral (for primes) | An Ulam spiral (of primes) is a method of visualizing primes when expressed in a (normally counter-clockwise) outward spiral (usually starting at 1), constructed on a square grid, starting at the "center".
An Ulam spiral is also known as a prime spiral.
The first grid (green) is shown with sequential integers, starting at 1.
In an Ulam spiral of primes, only the primes are shown (usually indicated by some glyph such as a dot or asterisk), and all non-primes as shown as a blank (or some other whitespace).
Of course, the grid and border are not to be displayed (but they are displayed here when using these Wiki HTML tables).
Normally, the spiral starts in the "center", and the 2nd number is to the viewer's right and the number spiral starts from there in a counter-clockwise direction.
There are other geometric shapes that are used as well, including clock-wise spirals.
Also, some spirals (for the 2nd number) is viewed upwards from the 1st number instead of to the right, but that is just a matter of orientation.
Sometimes, the starting number can be specified to show more visual striking patterns (of prime densities).
[A larger than necessary grid (numbers wise) is shown here to illustrate the pattern of numbers on the diagonals (which may be used by the method to orientate the direction of spiral-construction algorithm within the example computer programs)].
Then, in the next phase in the transformation of the Ulam prime spiral, the non-primes are translated to blanks.
In the orange grid below, the primes are left intact, and all non-primes are changed to blanks.
Then, in the final transformation of the Ulam spiral (the yellow grid), translate the primes to a glyph such as a • or some other suitable glyph.
65
64
63
62
61
60
59
58
57
66
37
36
35
34
33
32
31
56
67
38
17
16
15
14
13
30
55
68
39
18
5
4
3
12
29
54
69
40
19
6
1
2
11
28
53
70
41
20
7
8
9
10
27
52
71
42
21
22
23
24
25
26
51
72
43
44
45
46
47
48
49
50
73
74
75
76
77
78
79
80
81
61
59
37
31
67
17
13
5
3
29
19
2
11
53
41
7
71
23
43
47
73
79
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
The Ulam spiral becomes more visually obvious as the grid increases in size.
Task
For any sized N × N grid, construct and show an Ulam spiral (counter-clockwise) of primes starting at some specified initial number (the default would be 1), with some suitably dotty (glyph) representation to indicate primes, and the absence of dots to indicate non-primes.
You should demonstrate the generator by showing at Ulam prime spiral large enough to (almost) fill your terminal screen.
Related tasks
Spiral matrix
Zig-zag matrix
Identity matrix
Sequence of primes by Trial Division
See also
Wikipedia entry: Ulam spiral
MathWorld™ entry: Prime Spiral
| #Julia | Julia | using Primes
function ulamspiral(ord::Int)
# Possible directions
dirs = [[0, 1], [-1, 0], [0, -1], [1, 0]]
# fdir = ["→", "↑", "←", "↓"] # for debug pourpose
cur = maxsteps = 1 # starting direction & starting max steps
steps = n = 0 # starting steps & starting number in cell
pos = [ord ÷ 2 + 1, isodd(ord) ? ord ÷ 2 + 1 : ord ÷ 2] # starting position
M = Matrix{Bool}(ord, ord) # result matrix
while n < ord ^ 2 # main loop (stop when the matrix is filled)
n += 1
M[pos[1], pos[2]] = isprime(n)
steps += 1
# Debug print
# @printf("M[%i, %i] = %5s (%2i), step %i/%i, nxt %s\n", pos[1], pos[2], isprime(n), n, steps, maxsteps, fdir[cur])
pos .+= dirs[cur] # increment position
if steps == maxsteps # if reached max number of steps in that direction...
steps = 0 # ...reset steps
if iseven(cur) maxsteps += 1 end # if the current direction is even increase the number of steps
cur += 1 # change direction
if cur > 4 cur -= 4 end # correct overflow
end
end
return M
end
mprint(m::Matrix) = for i in 1:size(m, 1) println(join(el ? " ∙ " : " " for el in m[i, :]), '\n') end
M = ulamspiral(9)
mprint(M) |
http://rosettacode.org/wiki/Truncatable_primes | Truncatable primes | A truncatable prime is a prime number that when you successively remove digits from one end of the prime, you are left with a new prime number.
Examples
The number 997 is called a left-truncatable prime as the numbers 997, 97, and 7 are all prime.
The number 7393 is a right-truncatable prime as the numbers 7393, 739, 73, and 7 formed by removing digits from its right are also prime.
No zeroes are allowed in truncatable primes.
Task
The task is to find the largest left-truncatable and right-truncatable primes less than one million (base 10 is implied).
Related tasks
Find largest left truncatable prime in a given base
Sieve of Eratosthenes
See also
Truncatable Prime from MathWorld.]
| #EchoLisp | EchoLisp |
;; does p include a 0 in its decimal representation ?
(define (nozero? n) (= -1 (string-index (number->string n) "0")))
;; right truncate : p and successive quotients by 10 (integer division) must be primes
(define (right-trunc p) (unless (zero? p)
(and (prime? p) (right-trunc (quotient p 10)))))
(remember 'right-trunc)
;; left truncate : p and successive modulo by 10, 100, .. must be prime
(define (left-trunc p (mod 1000000))
(unless (< mod 1)
(and (prime? p) (nozero? p) (left-trunc (modulo p mod) (/ mod 10)))))
;; start from 999999. stop on first found
(define (fact-trunc trunc)
(for ((p (in-range 999999 100000 -1))) #:break (when (trunc p) (writeln p) #t)))
|
http://rosettacode.org/wiki/Truncate_a_file | Truncate a file | Task
Truncate a file to a specific length. This should be implemented as a routine that takes two parameters: the filename and the required file length (in bytes).
Truncation can be achieved using system or library calls intended for such a task, if such methods exist, or by creating a temporary file of a reduced size and renaming it, after first deleting the original file, if no other method is available. The file may contain non human readable binary data in an unspecified format, so the routine should be "binary safe", leaving the contents of the untruncated part of the file unchanged.
If the specified filename does not exist, or the provided length is not less than the current file length, then the routine should raise an appropriate error condition.
On some systems, the provided file truncation facilities might not change the file or may extend the file, if the specified length is greater than the current length of the file.
This task permits the use of such facilities. However, such behaviour should be noted, or optionally a warning message relating to an non change or increase in file size may be implemented.
| #Scala | Scala | import java.io.FileOutputStream
object TruncFile extends App {
if (args.length < 2) println("Usage: java TruncFile fileName newSize")
else { //turn on "append" so it doesn't clear the file
val outChan = new FileOutputStream(args(0), true).getChannel()
val newSize = args(1).toLong
outChan.truncate(newSize)
}
} |
http://rosettacode.org/wiki/Truncate_a_file | Truncate a file | Task
Truncate a file to a specific length. This should be implemented as a routine that takes two parameters: the filename and the required file length (in bytes).
Truncation can be achieved using system or library calls intended for such a task, if such methods exist, or by creating a temporary file of a reduced size and renaming it, after first deleting the original file, if no other method is available. The file may contain non human readable binary data in an unspecified format, so the routine should be "binary safe", leaving the contents of the untruncated part of the file unchanged.
If the specified filename does not exist, or the provided length is not less than the current file length, then the routine should raise an appropriate error condition.
On some systems, the provided file truncation facilities might not change the file or may extend the file, if the specified length is greater than the current length of the file.
This task permits the use of such facilities. However, such behaviour should be noted, or optionally a warning message relating to an non change or increase in file size may be implemented.
| #Sidef | Sidef | func truncate(filename, len) {
var file = File(filename);
len > file.size ->
&& die "The provided length is greater than the length of the file";
file.truncate(len);
}
# truncate "file.ext" to 1234 bytes
truncate("file.ext", 1234); |
http://rosettacode.org/wiki/Universal_Turing_machine | Universal Turing machine | One of the foundational mathematical constructs behind computer science
is the universal Turing Machine.
(Alan Turing introduced the idea of such a machine in 1936–1937.)
Indeed one way to definitively prove that a language
is turing-complete
is to implement a universal Turing machine in it.
Task
Simulate such a machine capable
of taking the definition of any other Turing machine and executing it.
Of course, you will not have an infinite tape,
but you should emulate this as much as is possible.
The three permissible actions on the tape are "left", "right" and "stay".
To test your universal Turing machine (and prove your programming language
is Turing complete!), you should execute the following two Turing machines
based on the following definitions.
Simple incrementer
States: q0, qf
Initial state: q0
Terminating states: qf
Permissible symbols: B, 1
Blank symbol: B
Rules:
(q0, 1, 1, right, q0)
(q0, B, 1, stay, qf)
The input for this machine should be a tape of 1 1 1
Three-state busy beaver
States: a, b, c, halt
Initial state: a
Terminating states: halt
Permissible symbols: 0, 1
Blank symbol: 0
Rules:
(a, 0, 1, right, b)
(a, 1, 1, left, c)
(b, 0, 1, left, a)
(b, 1, 1, right, b)
(c, 0, 1, left, b)
(c, 1, 1, stay, halt)
The input for this machine should be an empty tape.
Bonus:
5-state, 2-symbol probable Busy Beaver machine from Wikipedia
States: A, B, C, D, E, H
Initial state: A
Terminating states: H
Permissible symbols: 0, 1
Blank symbol: 0
Rules:
(A, 0, 1, right, B)
(A, 1, 1, left, C)
(B, 0, 1, right, C)
(B, 1, 1, right, B)
(C, 0, 1, right, D)
(C, 1, 0, left, E)
(D, 0, 1, left, A)
(D, 1, 1, left, D)
(E, 0, 1, stay, H)
(E, 1, 0, left, A)
The input for this machine should be an empty tape.
This machine runs for more than 47 millions steps.
| #C.2B.2B | C++ |
#include <vector>
#include <string>
#include <iostream>
#include <algorithm>
#include <fstream>
#include <iomanip>
//--------------------------------------------------------------------------------------------------
typedef unsigned int uint;
using namespace std;
const uint TAPE_MAX_LEN = 49152;
//--------------------------------------------------------------------------------------------------
struct action { char write, direction; };
//--------------------------------------------------------------------------------------------------
class tape
{
public:
tape( uint startPos = TAPE_MAX_LEN >> 1 ) : MAX_LEN( TAPE_MAX_LEN ) { _sp = startPos; reset(); }
void reset() { clear( '0' ); headPos = _sp; }
char read(){ return _t[headPos]; }
void input( string a ){ if( a == "" ) return; for( uint s = 0; s < a.length(); s++ ) _t[headPos + s] = a[s]; }
void clear( char c ) { _t.clear(); blk = c; _t.resize( MAX_LEN, blk ); }
void action( const action* a ) { write( a->write ); move( a->direction ); }
void print( int c = 10 )
{
int ml = static_cast<int>( MAX_LEN ), st = static_cast<int>( headPos ) - c, ed = static_cast<int>( headPos ) + c + 1, tx;
for( int x = st; x < ed; x++ )
{ tx = x; if( tx < 0 ) tx += ml; if( tx >= ml ) tx -= ml; cout << _t[tx]; }
cout << endl << setw( c + 1 ) << "^" << endl;
}
private:
void move( char d ) { if( d == 'N' ) return; headPos += d == 'R' ? 1 : -1; if( headPos >= MAX_LEN ) headPos = d == 'R' ? 0 : MAX_LEN - 1; }
void write( char a ) { if( a != 'N' ) { if( a == 'B' ) _t[headPos] = blk; else _t[headPos] = a; } }
string _t; uint headPos, _sp; char blk; const uint MAX_LEN;
};
//--------------------------------------------------------------------------------------------------
class state
{
public:
bool operator ==( const string o ) { return o == name; }
string name, next; char symbol, write, direction;
};
//--------------------------------------------------------------------------------------------------
class actionTable
{
public:
bool loadTable( string file )
{
reset();
ifstream mf; mf.open( file.c_str() ); if( mf.is_open() )
{
string str; state stt;
while( mf.good() )
{
getline( mf, str ); if( str[0] == '\'' ) break;
parseState( str, stt ); states.push_back( stt );
}
while( mf.good() )
{
getline( mf, str ); if( str == "" ) continue;
if( str[0] == '!' ) blank = str.erase( 0, 1 )[0];
if( str[0] == '^' ) curState = str.erase( 0, 1 );
if( str[0] == '>' ) input = str.erase( 0, 1 );
}
mf.close(); return true;
}
cout << "Could not open " << file << endl; return false;
}
bool action( char symbol, action& a )
{
vector<state>::iterator f = states.begin();
while( true )
{
f = find( f, states.end(), curState );
if( f == states.end() ) return false;
if( ( *f ).symbol == '*' || ( *f ).symbol == symbol || ( ( *f ).symbol == 'B' && blank == symbol ) )
{ a.direction = ( *f ).direction; a.write = ( *f ).write; curState = ( *f ).next; break; }
f++;
}
return true;
}
void reset() { states.clear(); blank = '0'; curState = input = ""; }
string getInput() { return input; }
char getBlank() { return blank; }
private:
void parseState( string str, state& stt )
{
string a[5]; int idx = 0;
for( string::iterator si = str.begin(); si != str.end(); si++ )
{ if( ( *si ) == ';' ) idx++; else a[idx].append( &( *si ), 1 ); }
stt.name = a[0]; stt.symbol = a[1][0]; stt.write = a[2][0]; stt.direction = a[3][0]; stt.next = a[4];
}
vector<state> states; char blank; string curState, input;
};
//--------------------------------------------------------------------------------------------------
class utm
{
public:
utm() { files[0] = "incrementer.utm"; files[1] = "busy_beaver.utm"; files[2] = "sort.utm"; }
void start()
{
while( true )
{
reset(); int t = showMenu(); if( t == 0 ) return;
if( !at.loadTable( files[t - 1] ) ) return; startMachine();
}
}
private:
void simulate()
{
char r; action a;
while( true ) { tp.print(); r = tp.read(); if( !( at.action( r, a ) ) ) break; tp.action( &a ); }
cout << endl << endl; system( "pause" );
}
int showMenu()
{
int t = -1;
while( t < 0 || t > 3 )
{
system( "cls" ); cout << "1. Incrementer\n2. Busy beaver\n3. Sort\n\n0. Quit";
cout << endl << endl << "Choose an action "; cin >> t;
}
return t;
}
void reset() { tp.reset(); at.reset(); }
void startMachine() { system( "cls" ); tp.clear( at.getBlank() ); tp.input( at.getInput() ); simulate(); }
tape tp; actionTable at; string files[7];
};
//--------------------------------------------------------------------------------------------------
int main( int a, char* args[] ){ utm mm; mm.start(); return 0; }
//--------------------------------------------------------------------------------------------------
|
http://rosettacode.org/wiki/Totient_function | Totient function | The totient function is also known as:
Euler's totient function
Euler's phi totient function
phi totient function
Φ function (uppercase Greek phi)
φ function (lowercase Greek phi)
Definitions (as per number theory)
The totient function:
counts the integers up to a given positive integer n that are relatively prime to n
counts the integers k in the range 1 ≤ k ≤ n for which the greatest common divisor gcd(n,k) is equal to 1
counts numbers ≤ n and prime to n
If the totient number (for N) is one less than N, then N is prime.
Task
Create a totient function and:
Find and display (1 per line) for the 1st 25 integers:
the integer (the index)
the totient number for that integer
indicate if that integer is prime
Find and display the count of the primes up to 100
Find and display the count of the primes up to 1,000
Find and display the count of the primes up to 10,000
Find and display the count of the primes up to 100,000 (optional)
Show all output here.
Related task
Perfect totient numbers
Also see
Wikipedia: Euler's totient function.
MathWorld: totient function.
OEIS: Euler totient function phi(n).
| #ALGOL_68 | ALGOL 68 | BEGIN
# returns the number of integers k where 1 <= k <= n that are mutually prime to n #
PROC totient = ( INT n )INT:
IF n < 3 THEN 1
ELIF n = 3 THEN 2
ELSE
INT result := n;
INT v := n;
INT i := 2;
WHILE i * i <= v DO
IF v MOD i = 0 THEN
WHILE v MOD i = 0 DO v OVERAB i OD;
result -:= result OVER i
FI;
IF i = 2 THEN
i := 1
FI;
i +:= 2
OD;
IF v > 1 THEN result -:= result OVER v FI;
result
FI # totient # ;
# show the totient function values for the first 25 integers #
print( ( " n phi(n) remarks", newline ) );
FOR n TO 25 DO
INT tn = totient( n );
print( ( whole( n, -2 ), ": ", whole( tn, -5 ), IF tn = n - 1 AND tn /= 0 THEN " n is prime" ELSE "" FI, newline ) )
OD;
# use the totient function to count primes #
INT n100 := 0, n1000 := 0, n10000 := 0, n100000 := 0;
FOR n TO 100 000 DO
IF totient( n ) = n - 1 THEN
IF n <= 100 THEN n100 +:= 1 FI;
IF n <= 1 000 THEN n1000 +:= 1 FI;
IF n <= 10 000 THEN n10000 +:= 1 FI;
IF n <= 100 000 THEN n100000 +:= 1 FI
FI
OD;
print( ( "There are ", whole( n100, -6 ), " primes below 100", newline ) );
print( ( "There are ", whole( n1000, -6 ), " primes below 1 000", newline ) );
print( ( "There are ", whole( n10000, -6 ), " primes below 10 000", newline ) );
print( ( "There are ", whole( n100000, -6 ), " primes below 100 000", newline ) )
END |
http://rosettacode.org/wiki/Topswops | Topswops | Topswops is a card game created by John Conway in the 1970's.
Assume you have a particular permutation of a set of n cards numbered 1..n on both of their faces, for example the arrangement of four cards given by [2, 4, 1, 3] where the leftmost card is on top.
A round is composed of reversing the first m cards where m is the value of the topmost card.
Rounds are repeated until the topmost card is the number 1 and the number of swaps is recorded.
For our example the swaps produce:
[2, 4, 1, 3] # Initial shuffle
[4, 2, 1, 3]
[3, 1, 2, 4]
[2, 1, 3, 4]
[1, 2, 3, 4]
For a total of four swaps from the initial ordering to produce the terminating case where 1 is on top.
For a particular number n of cards, topswops(n) is the maximum swaps needed for any starting permutation of the n cards.
Task
The task is to generate and show here a table of n vs topswops(n) for n in the range 1..10 inclusive.
Note
Topswops is also known as Fannkuch from the German word Pfannkuchen meaning pancake.
Related tasks
Number reversal game
Sorting algorithms/Pancake sort
| #C.2B.2B | C++ |
#include <iostream>
#include <vector>
#include <numeric>
#include <algorithm>
int topswops(int n) {
std::vector<int> list(n);
std::iota(std::begin(list), std::end(list), 1);
int max_steps = 0;
do {
auto temp_list = list;
for (int steps = 1; temp_list[0] != 1; ++steps) {
std::reverse(std::begin(temp_list), std::begin(temp_list) + temp_list[0]);
if (steps > max_steps) max_steps = steps;
}
} while (std::next_permutation(std::begin(list), std::end(list)));
return max_steps;
}
int main() {
for (int i = 1; i <= 10; ++i) {
std::cout << i << ": " << topswops(i) << std::endl;
}
return 0;
} |
http://rosettacode.org/wiki/Trigonometric_functions | Trigonometric functions | Task
If your language has a library or built-in functions for trigonometry, show examples of:
sine
cosine
tangent
inverses (of the above)
using the same angle in radians and degrees.
For the non-inverse functions, each radian/degree pair should use arguments that evaluate to the same angle (that is, it's not necessary to use the same angle for all three regular functions as long as the two sine calls use the same angle).
For the inverse functions, use the same number and convert its answer to radians and degrees.
If your language does not have trigonometric functions available or only has some available, write functions to calculate the functions based on any known approximation or identity.
| #Asymptote | Asymptote | real pi = 4 * atan(1);
real radian = pi / 4.0;
real angulo = 45.0 * pi / 180;
write("Radians : ", radian);
write("Degrees : ", angulo / pi * 180);
write();
write("Sine : ", sin(radian), sin(angulo));
write("Cosine : ", cos(radian), cos(angulo));
write("Tangent : ", tan(radian), tan(angulo));
write();
real temp = asin(sin(radian));
write("Arc Sine : ", temp, temp * 180 / pi);
temp = acos(cos(radian));
write("Arc Cosine : ", temp, temp * 180 / pi);
temp = atan(tan(radian));
write("Arc Tangent : ", temp, temp * 180 / pi); |
http://rosettacode.org/wiki/Trabb_Pardo%E2%80%93Knuth_algorithm | Trabb Pardo–Knuth algorithm | The TPK algorithm is an early example of a programming chrestomathy.
It was used in Donald Knuth and Luis Trabb Pardo's Stanford tech report The Early Development of Programming Languages.
The report traces the early history of work in developing computer languages in the 1940s and 1950s, giving several translations of the algorithm.
From the wikipedia entry:
ask for 11 numbers to be read into a sequence S
reverse sequence S
for each item in sequence S
result := call a function to do an operation
if result overflows
alert user
else
print result
The task is to implement the algorithm:
Use the function:
f
(
x
)
=
|
x
|
0.5
+
5
x
3
{\displaystyle f(x)=|x|^{0.5}+5x^{3}}
The overflow condition is an answer of greater than 400.
The 'user alert' should not stop processing of other items of the sequence.
Print a prompt before accepting eleven, textual, numeric inputs.
You may optionally print the item as well as its associated result, but the results must be in reverse order of input.
The sequence S may be 'implied' and so not shown explicitly.
Print and show the program in action from a typical run here. (If the output is graphical rather than text then either add a screendump or describe textually what is displayed).
| #AWK | AWK |
# syntax: GAWK -f TRABB_PARDO-KNUTH_ALGORITHM.AWK
BEGIN {
printf("enter 11 numbers: ")
getline S
n = split(S,arr," ")
if (n != 11) {
printf("%d numbers entered; S/B 11\n",n)
exit(1)
}
for (i=n; i>0; i--) {
x = f(arr[i])
printf("f(%s) = %s\n",arr[i],(x>400) ? "too large" : x)
}
exit(0)
}
function abs(x) { if (x >= 0) { return x } else { return -x } }
function f(x) { return sqrt(abs(x)) + 5 * x ^ 3 }
|
http://rosettacode.org/wiki/Twelve_statements | Twelve statements | This puzzle is borrowed from math-frolic.blogspot.
Given the following twelve statements, which of them are true?
1. This is a numbered list of twelve statements.
2. Exactly 3 of the last 6 statements are true.
3. Exactly 2 of the even-numbered statements are true.
4. If statement 5 is true, then statements 6 and 7 are both true.
5. The 3 preceding statements are all false.
6. Exactly 4 of the odd-numbered statements are true.
7. Either statement 2 or 3 is true, but not both.
8. If statement 7 is true, then 5 and 6 are both true.
9. Exactly 3 of the first 6 statements are true.
10. The next two statements are both true.
11. Exactly 1 of statements 7, 8 and 9 are true.
12. Exactly 4 of the preceding statements are true.
Task
When you get tired of trying to figure it out in your head,
write a program to solve it, and print the correct answer or answers.
Extra credit
Print out a table of near misses, that is, solutions that are contradicted by only a single statement.
| #Kotlin | Kotlin | // version 1.1.3
typealias Predicate = (String) -> Boolean
val predicates = listOf<Predicate>(
{ it.length == 13 }, // indexing starts at 0 but first bit ignored
{ (7..12).count { i -> it[i] == '1' } == 3 },
{ (2..12 step 2).count { i -> it[i] == '1' } == 2 },
{ it[5] == '0' || (it[6] == '1' && it[7] == '1') },
{ it[2] == '0' && it[3] == '0' && it[4] == '0' },
{ (1..11 step 2).count { i -> it[i] == '1' } == 4 },
{ (it[2] == '1') xor (it[3] == '1') },
{ it[7] == '0' || (it[5] == '1' && it[6] == '1') },
{ (1..6).count { i -> it[i] == '1' } == 3 },
{ it[11] == '1' && it[12] == '1' },
{ (7..9).count { i -> it[i] == '1' } == 1 },
{ (1..11).count { i -> it[i] == '1' } == 4 }
)
fun show(s: String, indent: Boolean) {
if (indent) print(" ")
for (i in s.indices) if (s[i] == '1') print("$i ")
println()
}
fun main(args: Array<String>) {
println("Exact hits:")
for (i in 0..4095) {
val s = i.toString(2).padStart(13, '0')
var j = 1
if (predicates.all { it(s) == (s[j++] == '1') }) show(s, true)
}
println("\nNear misses:")
for (i in 0..4095) {
val s = i.toString(2).padStart(13, '0')
var j = 1
if (predicates.count { it(s) == (s[j++] == '1') } == 11) {
var k = 1
val iof = predicates.indexOfFirst { it(s) != (s[k++] == '1') } + 1
print(" (Fails at statement ${"%2d".format(iof)}) ")
show(s, false)
}
}
} |
http://rosettacode.org/wiki/Truth_table | Truth table | A truth table is a display of the inputs to, and the output of a Boolean function organized as a table where each row gives one combination of input values and the corresponding value of the function.
Task
Input a Boolean function from the user as a string then calculate and print a formatted truth table for the given function.
(One can assume that the user input is correct).
Print and show output for Boolean functions of two and three input variables, but any program should not be limited to that many variables in the function.
Either reverse-polish or infix notation expressions are allowed.
Related tasks
Boolean values
Ternary logic
See also
Wolfram MathWorld entry on truth tables.
some "truth table" examples from Google.
| #Julia | Julia | module TruthTable
using Printf
using MacroTools
isvariablename(::Any) = false
isvariablename(s::Symbol) = all(x -> isletter(x) || x == '_', string(s))
function table(expr)
if !isvariablename(expr) && !Meta.isexpr(expr, :call)
throw(ArgumentError("expr must be a boolean expression"))
end
exprstr = string(expr)
# Collect variable names
symset = Set{Symbol}()
MacroTools.prewalk(expr) do node
isvariablename(node) && push!(symset, node)
return node
end
symlist = collect(symset)
# Create assignment assertions + evaluate
blocks = Vector{Expr}(undef, 2 ^ length(symlist) + 1)
blocks[1] = quote
println(join(lpad.($(symlist), 6), " | "), " || ", $exprstr)
end
for (i, tup) in enumerate(Iterators.product(Iterators.repeated((false, true), length(symlist))...))
blocks[i + 1] = quote
let $(Expr(:(=), Expr(:tuple, symlist...), Expr(:tuple, tup...)))
println(join(lpad.($(Expr(:tuple, symlist...)), 6), " | "), " || ", lpad($expr, $(length(exprstr))))
end
end
end
return esc(Expr(:block, blocks...))
end
macro table(expr)
return table(expr)
end
end # module TruthTable |
http://rosettacode.org/wiki/Ulam_spiral_(for_primes) | Ulam spiral (for primes) | An Ulam spiral (of primes) is a method of visualizing primes when expressed in a (normally counter-clockwise) outward spiral (usually starting at 1), constructed on a square grid, starting at the "center".
An Ulam spiral is also known as a prime spiral.
The first grid (green) is shown with sequential integers, starting at 1.
In an Ulam spiral of primes, only the primes are shown (usually indicated by some glyph such as a dot or asterisk), and all non-primes as shown as a blank (or some other whitespace).
Of course, the grid and border are not to be displayed (but they are displayed here when using these Wiki HTML tables).
Normally, the spiral starts in the "center", and the 2nd number is to the viewer's right and the number spiral starts from there in a counter-clockwise direction.
There are other geometric shapes that are used as well, including clock-wise spirals.
Also, some spirals (for the 2nd number) is viewed upwards from the 1st number instead of to the right, but that is just a matter of orientation.
Sometimes, the starting number can be specified to show more visual striking patterns (of prime densities).
[A larger than necessary grid (numbers wise) is shown here to illustrate the pattern of numbers on the diagonals (which may be used by the method to orientate the direction of spiral-construction algorithm within the example computer programs)].
Then, in the next phase in the transformation of the Ulam prime spiral, the non-primes are translated to blanks.
In the orange grid below, the primes are left intact, and all non-primes are changed to blanks.
Then, in the final transformation of the Ulam spiral (the yellow grid), translate the primes to a glyph such as a • or some other suitable glyph.
65
64
63
62
61
60
59
58
57
66
37
36
35
34
33
32
31
56
67
38
17
16
15
14
13
30
55
68
39
18
5
4
3
12
29
54
69
40
19
6
1
2
11
28
53
70
41
20
7
8
9
10
27
52
71
42
21
22
23
24
25
26
51
72
43
44
45
46
47
48
49
50
73
74
75
76
77
78
79
80
81
61
59
37
31
67
17
13
5
3
29
19
2
11
53
41
7
71
23
43
47
73
79
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
The Ulam spiral becomes more visually obvious as the grid increases in size.
Task
For any sized N × N grid, construct and show an Ulam spiral (counter-clockwise) of primes starting at some specified initial number (the default would be 1), with some suitably dotty (glyph) representation to indicate primes, and the absence of dots to indicate non-primes.
You should demonstrate the generator by showing at Ulam prime spiral large enough to (almost) fill your terminal screen.
Related tasks
Spiral matrix
Zig-zag matrix
Identity matrix
Sequence of primes by Trial Division
See also
Wikipedia entry: Ulam spiral
MathWorld™ entry: Prime Spiral
| #Kotlin | Kotlin | object Ulam {
fun generate(n: Int, i: Int = 1, c: Char = '*') {
require(n > 1)
val s = Array(n) { Array(n, { "" }) }
var dir = Direction.RIGHT
var y = n / 2
var x = if (n % 2 == 0) y - 1 else y // shift left for even n's
for (j in i..n * n - 1 + i) {
s[y][x] = if (isPrime(j)) if (c.isDigit()) "%4d".format(j) else " $c " else " ---"
when (dir) {
Direction.RIGHT -> if (x <= n - 1 && s[y - 1][x].none() && j > i) dir = Direction.UP
Direction.UP -> if (s[y][x - 1].none()) dir = Direction.LEFT
Direction.LEFT -> if (x == 0 || s[y + 1][x].none()) dir = Direction.DOWN
Direction.DOWN -> if (s[y][x + 1].none()) dir = Direction.RIGHT
}
when (dir) {
Direction.RIGHT -> x++
Direction.UP -> y--
Direction.LEFT -> x--
Direction.DOWN -> y++
}
}
for (row in s) println("[" + row.joinToString("") + ']')
println()
}
private enum class Direction { RIGHT, UP, LEFT, DOWN }
private fun isPrime(a: Int): Boolean {
when {
a == 2 -> return true
a <= 1 || a % 2 == 0 -> return false
else -> {
val max = Math.sqrt(a.toDouble()).toInt()
for (n in 3..max step 2)
if (a % n == 0) return false
return true
}
}
}
}
fun main(args: Array<String>) {
Ulam.generate(9, c = '0')
Ulam.generate(9)
} |
http://rosettacode.org/wiki/Truncatable_primes | Truncatable primes | A truncatable prime is a prime number that when you successively remove digits from one end of the prime, you are left with a new prime number.
Examples
The number 997 is called a left-truncatable prime as the numbers 997, 97, and 7 are all prime.
The number 7393 is a right-truncatable prime as the numbers 7393, 739, 73, and 7 formed by removing digits from its right are also prime.
No zeroes are allowed in truncatable primes.
Task
The task is to find the largest left-truncatable and right-truncatable primes less than one million (base 10 is implied).
Related tasks
Find largest left truncatable prime in a given base
Sieve of Eratosthenes
See also
Truncatable Prime from MathWorld.]
| #Eiffel | Eiffel |
class
APPLICATION
create
make
feature
make
do
io.put_string ("Largest right truncatable prime: " + find_right_truncatable_primes.out)
io.new_line
io.put_string ("Largest left truncatable prime: " + find_left_truncatable_primes.out)
end
find_right_truncatable_primes: INTEGER
-- Largest right truncatable prime below 1000000.
local
i, maybe_prime: INTEGER
found, is_one: BOOLEAN
do
from
i := 999999
until
found
loop
is_one := True
from
maybe_prime := i
until
not is_one or maybe_prime.out.count = 1
loop
if maybe_prime.out.has ('0') or maybe_prime.out.has ('2') or maybe_prime.out.has ('4') or maybe_prime.out.has ('6') or maybe_prime.out.has ('8') then
is_one := False
else
if not is_prime (maybe_prime) then
is_one := False
elseif is_prime (maybe_prime) and maybe_prime.out.count > 1 then
maybe_prime := truncate_right (maybe_prime)
end
end
end
if is_one then
found := True
Result := i
end
i := i - 2
end
ensure
Result_is_smaller: Result < 1000000
end
find_left_truncatable_primes: INTEGER
-- Largest left truncatable prime below 1000000.
local
i, maybe_prime: INTEGER
found, is_one: BOOLEAN
do
from
i := 999999
until
found
loop
is_one := True
from
maybe_prime := i
until
not is_one or maybe_prime.out.count = 1
loop
if not is_prime (maybe_prime) then
is_one := False
elseif is_prime (maybe_prime) and maybe_prime.out.count > 1 then
if maybe_prime.out.at (2) = '0' then
is_one := False
else
maybe_prime := truncate_left (maybe_prime)
end
end
end
if is_one then
found := True
Result := i
end
i := i - 2
end
ensure
Result_is_smaller: Result < 1000000
end
feature {NONE}
is_prime (n: INTEGER): BOOLEAN
--Is 'n' a prime number?
require
positiv_input: n > 0
local
i: INTEGER
max: REAL_64
math: DOUBLE_MATH
do
create math
if n = 2 then
Result := True
elseif n <= 1 or n \\ 2 = 0 then
Result := False
else
Result := True
max := math.sqrt (n)
from
i := 3
until
i > max
loop
if n \\ i = 0 then
Result := False
end
i := i + 2
end
end
end
truncate_left (n: INTEGER): INTEGER
-- 'n' truncated by one digit from the left side.
require
truncatable: n.out.count > 1
local
st: STRING
do
st := n.out
st.remove_head (1)
Result := st.to_integer
ensure
Result_truncated: Result.out.count = n.out.count - 1
end
truncate_right (n: INTEGER): INTEGER
-- 'n' truncated by one digit from the right side.
require
truncatable: n.out.count > 1
local
st: STRING
do
st := n.out
st.remove_tail (1)
Result := st.to_integer
ensure
Result_truncated: Result.out.count = n.out.count - 1
end
end
|
http://rosettacode.org/wiki/Truncate_a_file | Truncate a file | Task
Truncate a file to a specific length. This should be implemented as a routine that takes two parameters: the filename and the required file length (in bytes).
Truncation can be achieved using system or library calls intended for such a task, if such methods exist, or by creating a temporary file of a reduced size and renaming it, after first deleting the original file, if no other method is available. The file may contain non human readable binary data in an unspecified format, so the routine should be "binary safe", leaving the contents of the untruncated part of the file unchanged.
If the specified filename does not exist, or the provided length is not less than the current file length, then the routine should raise an appropriate error condition.
On some systems, the provided file truncation facilities might not change the file or may extend the file, if the specified length is greater than the current length of the file.
This task permits the use of such facilities. However, such behaviour should be noted, or optionally a warning message relating to an non change or increase in file size may be implemented.
| #Standard_ML | Standard ML | local
open Posix.FileSys
val perm = S.flags [S.irusr, S.iwusr, S.irgrp, S.iwgrp, S.iroth, S.iwoth]
in
fun truncate (path, len) =
let
val fd = createf (path, O_WRONLY, O.noctty, perm)
in
ftruncate (fd, len); Posix.IO.close fd
end
end |
http://rosettacode.org/wiki/Truncate_a_file | Truncate a file | Task
Truncate a file to a specific length. This should be implemented as a routine that takes two parameters: the filename and the required file length (in bytes).
Truncation can be achieved using system or library calls intended for such a task, if such methods exist, or by creating a temporary file of a reduced size and renaming it, after first deleting the original file, if no other method is available. The file may contain non human readable binary data in an unspecified format, so the routine should be "binary safe", leaving the contents of the untruncated part of the file unchanged.
If the specified filename does not exist, or the provided length is not less than the current file length, then the routine should raise an appropriate error condition.
On some systems, the provided file truncation facilities might not change the file or may extend the file, if the specified length is greater than the current length of the file.
This task permits the use of such facilities. However, such behaviour should be noted, or optionally a warning message relating to an non change or increase in file size may be implemented.
| #Tcl | Tcl | package require Tcl 8.5
set f [open "file" r+]; # Truncation is done on channels
chan truncate $f 1234; # Truncate at a particular length (in bytes)
close $f |
http://rosettacode.org/wiki/Truncate_a_file | Truncate a file | Task
Truncate a file to a specific length. This should be implemented as a routine that takes two parameters: the filename and the required file length (in bytes).
Truncation can be achieved using system or library calls intended for such a task, if such methods exist, or by creating a temporary file of a reduced size and renaming it, after first deleting the original file, if no other method is available. The file may contain non human readable binary data in an unspecified format, so the routine should be "binary safe", leaving the contents of the untruncated part of the file unchanged.
If the specified filename does not exist, or the provided length is not less than the current file length, then the routine should raise an appropriate error condition.
On some systems, the provided file truncation facilities might not change the file or may extend the file, if the specified length is greater than the current length of the file.
This task permits the use of such facilities. However, such behaviour should be noted, or optionally a warning message relating to an non change or increase in file size may be implemented.
| #UNIX_Shell | UNIX Shell | # Truncate a file named "myfile" to 1440 kilobytes.
ls myfile >/dev/null &&
dd if=/dev/null of=myfile bs=1 seek=1440k |
http://rosettacode.org/wiki/Tree_traversal | Tree traversal | Task
Implement a binary tree where each node carries an integer, and implement:
pre-order,
in-order,
post-order, and
level-order traversal.
Use those traversals to output the following tree:
1
/ \
/ \
/ \
2 3
/ \ /
4 5 6
/ / \
7 8 9
The correct output should look like this:
preorder: 1 2 4 7 5 3 6 8 9
inorder: 7 4 2 5 1 8 6 9 3
postorder: 7 4 5 2 8 9 6 3 1
level-order: 1 2 3 4 5 6 7 8 9
See also
Wikipedia article: Tree traversal.
| #11l | 11l | T Node
Int data
Node? left
Node? right
F (data, Node? left = N, Node? right = N)
.data = data
.left = left
.right = right
F preorder(visitor) -> N
visitor(.data)
I .left != N
.left.preorder(visitor)
I .right != N
.right.preorder(visitor)
F inorder(visitor) -> N
I .left != N
.left.inorder(visitor)
visitor(.data)
I .right != N
.right.inorder(visitor)
F postorder(visitor) -> N
I .left != N
.left.postorder(visitor)
I .right != N
.right.postorder(visitor)
visitor(.data)
F preorder2(&d, level = 0) -> N
d[level].append(.data)
I .left != N
.left.preorder2(d, level + 1)
I .right != N
.right.preorder2(d, level + 1)
F levelorder(visitor)
DefaultDict[Int, [Int]] d
.preorder2(&d)
L(k) sorted(d.keys())
L(v) d[k]
visitor(v)
V tree = Node(1,
Node(2,
Node(4,
Node(7, N, N),
N),
Node(5, N, N)),
Node(3,
Node(6,
Node(8, N, N),
Node(9, N, N)),
N))
F printwithspace(Int i)
print(‘#. ’.format(i), end' ‘’)
print(‘ preorder: ’, end' ‘’)
tree.preorder(printwithspace)
print()
print(‘ inorder: ’, end' ‘’)
tree.inorder(printwithspace)
print()
print(‘ postorder: ’, end' ‘’)
tree.postorder(printwithspace)
print()
print(‘levelorder: ’, end' ‘’)
tree.levelorder(printwithspace)
print() |
http://rosettacode.org/wiki/Topic_variable | Topic variable | Several programming languages offer syntax shortcuts to deal with the notion of "current" or "topic" variable.
A topic variable is a special variable with a very short name which can also often be omitted.
Demonstrate the utilization and behaviour of the topic variable within the language and explain or demonstrate how the topic variable behaves under different levels of nesting or scope, if this applies, within the language.
For instance you can (but you don't have to) show how the topic variable can be used by assigning the number
3
{\displaystyle 3}
to it and then computing its square and square root.
| #AppleScript | AppleScript | on run
1 + 2
ap({square, squareRoot}, {result})
--> {9, 1.732050807569}
end run
-- square :: Num a => a -> a
on square(x)
x * x
end square
-- squareRoot :: Num a, Float b => a -> b
on squareRoot(x)
x ^ 0.5
end squareRoot
-- GENERIC FUNCTIONS ----------------------------------------------------------
-- A list of functions applied to a list of arguments
-- (<*> | ap) :: [(a -> b)] -> [a] -> [b]
on ap(fs, xs)
set lst to {}
repeat with f in fs
tell mReturn(contents of f)
repeat with x in xs
set end of lst to |λ|(contents of x)
end repeat
end tell
end repeat
return lst
end ap
-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: Handler -> Script
on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn |
http://rosettacode.org/wiki/Universal_Turing_machine | Universal Turing machine | One of the foundational mathematical constructs behind computer science
is the universal Turing Machine.
(Alan Turing introduced the idea of such a machine in 1936–1937.)
Indeed one way to definitively prove that a language
is turing-complete
is to implement a universal Turing machine in it.
Task
Simulate such a machine capable
of taking the definition of any other Turing machine and executing it.
Of course, you will not have an infinite tape,
but you should emulate this as much as is possible.
The three permissible actions on the tape are "left", "right" and "stay".
To test your universal Turing machine (and prove your programming language
is Turing complete!), you should execute the following two Turing machines
based on the following definitions.
Simple incrementer
States: q0, qf
Initial state: q0
Terminating states: qf
Permissible symbols: B, 1
Blank symbol: B
Rules:
(q0, 1, 1, right, q0)
(q0, B, 1, stay, qf)
The input for this machine should be a tape of 1 1 1
Three-state busy beaver
States: a, b, c, halt
Initial state: a
Terminating states: halt
Permissible symbols: 0, 1
Blank symbol: 0
Rules:
(a, 0, 1, right, b)
(a, 1, 1, left, c)
(b, 0, 1, left, a)
(b, 1, 1, right, b)
(c, 0, 1, left, b)
(c, 1, 1, stay, halt)
The input for this machine should be an empty tape.
Bonus:
5-state, 2-symbol probable Busy Beaver machine from Wikipedia
States: A, B, C, D, E, H
Initial state: A
Terminating states: H
Permissible symbols: 0, 1
Blank symbol: 0
Rules:
(A, 0, 1, right, B)
(A, 1, 1, left, C)
(B, 0, 1, right, C)
(B, 1, 1, right, B)
(C, 0, 1, right, D)
(C, 1, 0, left, E)
(D, 0, 1, left, A)
(D, 1, 1, left, D)
(E, 0, 1, stay, H)
(E, 1, 0, left, A)
The input for this machine should be an empty tape.
This machine runs for more than 47 millions steps.
| #Clojure | Clojure |
(defn tape
"Creates a new tape with given blank character and tape contents"
([blank] (tape () blank () blank))
([right blank] (tape () (first right) (rest right) blank))
([left head right blank] [(reverse left) (or head blank) (into () right) blank]))
; Tape operations
(defn- left [[[l & ls] _ rs b] c] [ls (or l b) (conj rs c) b])
(defn- right [[ls _ [r & rs] b] c] [(conj ls c) (or r b) rs b])
(defn- stay [[ls _ rs b] c] [ls c rs b])
(defn- head [[_ c _ b]] (or c b))
(defn- pretty [[ls c rs b]] (concat (reverse ls) [[(or c b)]] rs))
(defn new-machine
"Returns a function that takes a tape as input, and returns the tape
after running the machine specified in `machine`."
[machine]
(let [rules (into {} (for [[s c c' a s'] (:rules machine)]
[[s c] [c' (-> a name symbol resolve) s']]))
finished? (into #{} (:terminating machine))]
(fn [input-tape]
(loop [state (:initial machine) tape input-tape]
(if (finished? state)
(pretty tape)
(let [[out action new-state] (get rules [state (head tape)])]
(recur new-state (action tape out))))))))
|
http://rosettacode.org/wiki/Totient_function | Totient function | The totient function is also known as:
Euler's totient function
Euler's phi totient function
phi totient function
Φ function (uppercase Greek phi)
φ function (lowercase Greek phi)
Definitions (as per number theory)
The totient function:
counts the integers up to a given positive integer n that are relatively prime to n
counts the integers k in the range 1 ≤ k ≤ n for which the greatest common divisor gcd(n,k) is equal to 1
counts numbers ≤ n and prime to n
If the totient number (for N) is one less than N, then N is prime.
Task
Create a totient function and:
Find and display (1 per line) for the 1st 25 integers:
the integer (the index)
the totient number for that integer
indicate if that integer is prime
Find and display the count of the primes up to 100
Find and display the count of the primes up to 1,000
Find and display the count of the primes up to 10,000
Find and display the count of the primes up to 100,000 (optional)
Show all output here.
Related task
Perfect totient numbers
Also see
Wikipedia: Euler's totient function.
MathWorld: totient function.
OEIS: Euler totient function phi(n).
| #ALGOL-M | ALGOL-M |
BEGIN
% RETURN P MOD Q %
INTEGER FUNCTION MOD (P, Q);
INTEGER P, Q;
BEGIN
MOD := P - Q * (P / Q);
END;
% RETURN GREATEST COMMON DIVISOR OF X AND Y %
INTEGER FUNCTION GCD (X, Y);
INTEGER X, Y;
BEGIN
INTEGER R;
IF X < Y THEN
BEGIN
INTEGER TEMP;
TEMP := X;
X := Y;
Y := TEMP;
END;
WHILE (R := MOD(X, Y)) <> 0 DO
BEGIN
X := Y;
Y := R;
END;
GCD := Y;
END;
% RETURN PHI (ALSO CALLED TOTIENT) OF N %
INTEGER FUNCTION PHI(N);
INTEGER N;
BEGIN
INTEGER I, COUNT;
COUNT := 1;
FOR I := 2 STEP 1 UNTIL N DO
BEGIN
IF GCD(N,I) = 1 THEN COUNT := COUNT + 1;
END;
PHI := COUNT;
END;
COMMENT - EXERCISE THE FUNCTION;
INTEGER N, TOTIENT, COUNT;
WRITE(" N PHI(N) PRIME?");
FOR N := 1 STEP 1 UNTIL 25 DO
BEGIN
WRITE(N, (TOTIENT := PHI(N)));
WRITEON(IF TOTIENT = (N-1) THEN " YES" ELSE " NO");
END;
COMMENT - AND USE IT TO COUNT PRIMES;
WRITE("");
COUNT := 0;
FOR N := 1 STEP 1 UNTIL 1000 DO
BEGIN
IF PHI(N) = (N-1) THEN COUNT := COUNT + 1;
IF N = 100 THEN
WRITE("PRIMES UP TO 100 =", COUNT)
ELSE IF N = 1000 THEN
WRITE("PRIMES UP TO 1000 =", COUNT);
END;
END
|
http://rosettacode.org/wiki/Topswops | Topswops | Topswops is a card game created by John Conway in the 1970's.
Assume you have a particular permutation of a set of n cards numbered 1..n on both of their faces, for example the arrangement of four cards given by [2, 4, 1, 3] where the leftmost card is on top.
A round is composed of reversing the first m cards where m is the value of the topmost card.
Rounds are repeated until the topmost card is the number 1 and the number of swaps is recorded.
For our example the swaps produce:
[2, 4, 1, 3] # Initial shuffle
[4, 2, 1, 3]
[3, 1, 2, 4]
[2, 1, 3, 4]
[1, 2, 3, 4]
For a total of four swaps from the initial ordering to produce the terminating case where 1 is on top.
For a particular number n of cards, topswops(n) is the maximum swaps needed for any starting permutation of the n cards.
Task
The task is to generate and show here a table of n vs topswops(n) for n in the range 1..10 inclusive.
Note
Topswops is also known as Fannkuch from the German word Pfannkuchen meaning pancake.
Related tasks
Number reversal game
Sorting algorithms/Pancake sort
| #D | D | import std.stdio, std.algorithm, std.range, permutations2;
int topswops(in int n) pure @safe {
static int flip(int[] xa) pure nothrow @safe @nogc {
if (!xa[0]) return 0;
xa[0 .. xa[0] + 1].reverse();
return 1 + flip(xa);
}
return n.iota.array.permutations.map!flip.reduce!max;
}
void main() {
foreach (immutable i; 1 .. 11)
writeln(i, ": ", i.topswops);
} |
http://rosettacode.org/wiki/Trigonometric_functions | Trigonometric functions | Task
If your language has a library or built-in functions for trigonometry, show examples of:
sine
cosine
tangent
inverses (of the above)
using the same angle in radians and degrees.
For the non-inverse functions, each radian/degree pair should use arguments that evaluate to the same angle (that is, it's not necessary to use the same angle for all three regular functions as long as the two sine calls use the same angle).
For the inverse functions, use the same number and convert its answer to radians and degrees.
If your language does not have trigonometric functions available or only has some available, write functions to calculate the functions based on any known approximation or identity.
| #AutoHotkey | AutoHotkey | pi := 4 * atan(1)
radians := pi / 4
degrees := 45.0
result .= "`n" . sin(radians) . " " . sin(degrees * pi / 180)
result .= "`n" . cos(radians) . " " . cos(degrees * pi / 180)
result .= "`n" . tan(radians) . " " . tan(degrees * pi / 180)
temp := asin(sin(radians))
result .= "`n" . temp . " " . temp * 180 / pi
temp := acos(cos(radians))
result .= "`n" . temp . " " . temp * 180 / pi
temp := atan(tan(radians))
result .= "`n" . temp . " " . temp * 180 / pi
msgbox % result
/* output
---------------------------
trig.ahk
---------------------------
0.707107 0.707107
0.707107 0.707107
1.000000 1.000000
0.785398 45.000000
0.785398 45.000000
0.785398 45.000000
*/ |
http://rosettacode.org/wiki/Trabb_Pardo%E2%80%93Knuth_algorithm | Trabb Pardo–Knuth algorithm | The TPK algorithm is an early example of a programming chrestomathy.
It was used in Donald Knuth and Luis Trabb Pardo's Stanford tech report The Early Development of Programming Languages.
The report traces the early history of work in developing computer languages in the 1940s and 1950s, giving several translations of the algorithm.
From the wikipedia entry:
ask for 11 numbers to be read into a sequence S
reverse sequence S
for each item in sequence S
result := call a function to do an operation
if result overflows
alert user
else
print result
The task is to implement the algorithm:
Use the function:
f
(
x
)
=
|
x
|
0.5
+
5
x
3
{\displaystyle f(x)=|x|^{0.5}+5x^{3}}
The overflow condition is an answer of greater than 400.
The 'user alert' should not stop processing of other items of the sequence.
Print a prompt before accepting eleven, textual, numeric inputs.
You may optionally print the item as well as its associated result, but the results must be in reverse order of input.
The sequence S may be 'implied' and so not shown explicitly.
Print and show the program in action from a typical run here. (If the output is graphical rather than text then either add a screendump or describe textually what is displayed).
| #BASIC | BASIC |
10 REM Trabb Pardo-Knuth algorithm
20 REM Used "magic numbers" because of strict specification
30 REM of the algorithm.
40 DEF FNF(N) = SQR(ABS(N))+5*N*N*N
50 DIM S(10)
60 PRINT "Enter 11 numbers."
70 FOR I = 0 TO 10
80 PRINT I+1; "- Enter number";
90 INPUT S(I)
100 NEXT I
110 PRINT
120 REM Reverse
130 FOR I = 0 TO 10/2
140 LET T = S(I)
150 LET S(I) = S(10-I)
160 LET S(10-I) = T
170 NEXT I
180 REM Results
190 PRINT "num", "f(num)"
200 FOR I = 0 TO 10
210 LET R = FNF(S(I))
220 IF R>400 THEN 250
230 PRINT S(I), R
240 GOTO 260
250 PRINT S(I), " overflow"
260 NEXT I
270 END
|
http://rosettacode.org/wiki/Twelve_statements | Twelve statements | This puzzle is borrowed from math-frolic.blogspot.
Given the following twelve statements, which of them are true?
1. This is a numbered list of twelve statements.
2. Exactly 3 of the last 6 statements are true.
3. Exactly 2 of the even-numbered statements are true.
4. If statement 5 is true, then statements 6 and 7 are both true.
5. The 3 preceding statements are all false.
6. Exactly 4 of the odd-numbered statements are true.
7. Either statement 2 or 3 is true, but not both.
8. If statement 7 is true, then 5 and 6 are both true.
9. Exactly 3 of the first 6 statements are true.
10. The next two statements are both true.
11. Exactly 1 of statements 7, 8 and 9 are true.
12. Exactly 4 of the preceding statements are true.
Task
When you get tired of trying to figure it out in your head,
write a program to solve it, and print the correct answer or answers.
Extra credit
Print out a table of near misses, that is, solutions that are contradicted by only a single statement.
| #Mathematica.2FWolfram_Language | Mathematica/Wolfram Language | Print["Answer:\n", Column@Cases[#, {s_, 0} :> s], "\nNear misses:\n",
Column@Cases[#, {s_, 1} :> s]] &[{#,
Count[Boole /@ {Length@# == 12, Total@#[[7 ;;]] == 3,
Total@#[[2 ;; 12 ;; 2]] == 2, #[[5]] (#[[6]] + #[[7]] - 2) ==
0, Total@#[[2 ;; 4]] == 0,
Total@#[[1 ;; 11 ;; 2]] == 4, #[[2]] + #[[3]] ==
1, #[[7]] (#[[5]] + #[[6]] - 2) == 0,
Total@#[[;; 6]] == 3, #[[11]] + #[[12]] == 2,
Total@#[[7 ;; 9]] == 1, Total@#[[;; 11]] == 4} - #,
Except[0]]} & /@ Tuples[{1, 0}, 12]] |
http://rosettacode.org/wiki/Truth_table | Truth table | A truth table is a display of the inputs to, and the output of a Boolean function organized as a table where each row gives one combination of input values and the corresponding value of the function.
Task
Input a Boolean function from the user as a string then calculate and print a formatted truth table for the given function.
(One can assume that the user input is correct).
Print and show output for Boolean functions of two and three input variables, but any program should not be limited to that many variables in the function.
Either reverse-polish or infix notation expressions are allowed.
Related tasks
Boolean values
Ternary logic
See also
Wolfram MathWorld entry on truth tables.
some "truth table" examples from Google.
| #Kotlin | Kotlin | // Version 1.2.31
import java.util.Stack
class Variable(val name: Char, var value: Boolean = false)
lateinit var expr: String
var variables = mutableListOf<Variable>()
fun Char.isOperator() = this in "&|!^"
fun Char.isVariable() = this in variables.map { it.name }
fun evalExpression(): Boolean {
val stack = Stack<Boolean>()
for (e in expr) {
stack.push(
if (e == 'T')
true
else if (e == 'F')
false
else if (e.isVariable())
variables.single { it.name == e }.value
else when (e) {
'&' -> stack.pop() and stack.pop()
'|' -> stack.pop() or stack.pop()
'!' -> !stack.pop()
'^' -> stack.pop() xor stack.pop()
else -> throw RuntimeException("Non-conformant character '$e' in expression")
}
)
}
require(stack.size == 1)
return stack.peek()
}
fun setVariables(pos: Int) {
require(pos <= variables.size)
if (pos == variables.size) {
val vs = variables.map { if (it.value) "T" else "F" }.joinToString(" ")
val es = if (evalExpression()) "T" else "F"
return println("$vs $es")
}
variables[pos].value = false
setVariables(pos + 1)
variables[pos].value = true
setVariables(pos + 1)
}
fun main(args: Array<String>) {
println("Accepts single-character variables (except for 'T' and 'F',")
println("which specify explicit true or false values), postfix, with")
println("&|!^ for and, or, not, xor, respectively; optionally")
println("seperated by spaces or tabs. Just enter nothing to quit.")
while (true) {
print("\nBoolean expression: ")
expr = readLine()!!.toUpperCase().replace(" ", "").replace("\t", "")
if (expr == "") return
variables.clear()
for (e in expr) {
if (!e.isOperator() && e !in "TF" && !e.isVariable()) variables.add(Variable(e))
}
if (variables.isEmpty()) return
val vs = variables.map { it.name }.joinToString(" ")
println("\n$vs $expr")
val h = vs.length + expr.length + 2
repeat(h) { print("=") }
println("\n")
setVariables(0)
}
} |
http://rosettacode.org/wiki/Ulam_spiral_(for_primes) | Ulam spiral (for primes) | An Ulam spiral (of primes) is a method of visualizing primes when expressed in a (normally counter-clockwise) outward spiral (usually starting at 1), constructed on a square grid, starting at the "center".
An Ulam spiral is also known as a prime spiral.
The first grid (green) is shown with sequential integers, starting at 1.
In an Ulam spiral of primes, only the primes are shown (usually indicated by some glyph such as a dot or asterisk), and all non-primes as shown as a blank (or some other whitespace).
Of course, the grid and border are not to be displayed (but they are displayed here when using these Wiki HTML tables).
Normally, the spiral starts in the "center", and the 2nd number is to the viewer's right and the number spiral starts from there in a counter-clockwise direction.
There are other geometric shapes that are used as well, including clock-wise spirals.
Also, some spirals (for the 2nd number) is viewed upwards from the 1st number instead of to the right, but that is just a matter of orientation.
Sometimes, the starting number can be specified to show more visual striking patterns (of prime densities).
[A larger than necessary grid (numbers wise) is shown here to illustrate the pattern of numbers on the diagonals (which may be used by the method to orientate the direction of spiral-construction algorithm within the example computer programs)].
Then, in the next phase in the transformation of the Ulam prime spiral, the non-primes are translated to blanks.
In the orange grid below, the primes are left intact, and all non-primes are changed to blanks.
Then, in the final transformation of the Ulam spiral (the yellow grid), translate the primes to a glyph such as a • or some other suitable glyph.
65
64
63
62
61
60
59
58
57
66
37
36
35
34
33
32
31
56
67
38
17
16
15
14
13
30
55
68
39
18
5
4
3
12
29
54
69
40
19
6
1
2
11
28
53
70
41
20
7
8
9
10
27
52
71
42
21
22
23
24
25
26
51
72
43
44
45
46
47
48
49
50
73
74
75
76
77
78
79
80
81
61
59
37
31
67
17
13
5
3
29
19
2
11
53
41
7
71
23
43
47
73
79
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
The Ulam spiral becomes more visually obvious as the grid increases in size.
Task
For any sized N × N grid, construct and show an Ulam spiral (counter-clockwise) of primes starting at some specified initial number (the default would be 1), with some suitably dotty (glyph) representation to indicate primes, and the absence of dots to indicate non-primes.
You should demonstrate the generator by showing at Ulam prime spiral large enough to (almost) fill your terminal screen.
Related tasks
Spiral matrix
Zig-zag matrix
Identity matrix
Sequence of primes by Trial Division
See also
Wikipedia entry: Ulam spiral
MathWorld™ entry: Prime Spiral
| #Lua | Lua | local function ulamspiral(n, f)
print("n = " .. n)
local function isprime(p)
if p < 2 then return false end
if p % 2 == 0 then return p==2 end
if p % 3 == 0 then return p==3 end
local limit = math.sqrt(p)
for f = 5, limit, 6 do
if p % f == 0 or p % (f+2) == 0 then return false end
end
return true
end
local function spiral(x, y)
if n%2==1 then x, y = n-1-x, n-1-y end
local m = math.min(x, y, n-1-x, n-1-y)
return x<y and (n-2*m-2)^2+(x-m)+(y-m) or (n-2*m)^2-(x-m)-(y-m)
end
for y = 0, n-1 do
for x = 0, n-1 do
io.write(f(isprime(spiral(x,y))))
end
print()
end
print()
end
-- filling a 132 column terminal (with a 2-wide glyph to better preserve aspect ratio)
ulamspiral(132/2, function(b) return b and "██" or " " end) |
http://rosettacode.org/wiki/Truncatable_primes | Truncatable primes | A truncatable prime is a prime number that when you successively remove digits from one end of the prime, you are left with a new prime number.
Examples
The number 997 is called a left-truncatable prime as the numbers 997, 97, and 7 are all prime.
The number 7393 is a right-truncatable prime as the numbers 7393, 739, 73, and 7 formed by removing digits from its right are also prime.
No zeroes are allowed in truncatable primes.
Task
The task is to find the largest left-truncatable and right-truncatable primes less than one million (base 10 is implied).
Related tasks
Find largest left truncatable prime in a given base
Sieve of Eratosthenes
See also
Truncatable Prime from MathWorld.]
| #Elena | Elena | import extensions;
const MAXN = 1000000;
extension mathOp
{
isPrime()
{
int n := cast int(self);
if (n < 2) { ^ false };
if (n < 4) { ^ true };
if (n.mod:2 == 0) { ^ false };
if (n < 9) { ^ true };
if (n.mod:3 == 0) { ^ false };
int r := n.sqrt();
int f := 5;
while (f <= r)
{
if ((n.mod(f) == 0) || (n.mod(f + 2) == 0))
{ ^ false };
f := f + 6
};
^ true
}
isRightTruncatable()
{
int n := self;
while (n != 0)
{
ifnot (n.isPrime())
{ ^ false };
n := n / 10
};
^ true
}
isLeftTruncatable()
{
int n := self;
int tens := 1;
while (tens < n)
{ tens := tens * 10 };
while (n != 0)
{
ifnot (n.isPrime())
{ ^ false };
tens := tens / 10;
n := n - (n / tens * tens)
};
^ true
}
}
public program()
{
var n := MAXN;
var max_lt := 0;
var max_rt := 0;
while (max_lt == 0 || max_rt == 0)
{
if(n.toString().indexOf("0") == -1)
{
if ((max_lt == 0) && (n.isLeftTruncatable()))
{
max_lt := n
};
if ((max_rt == 0) && (n.isRightTruncatable()))
{
max_rt := n
}
};
n := n - 1
};
console.printLine("Largest truncable left is ",max_lt);
console.printLine("Largest truncable right is ",max_rt);
console.readChar()
} |
http://rosettacode.org/wiki/Truncate_a_file | Truncate a file | Task
Truncate a file to a specific length. This should be implemented as a routine that takes two parameters: the filename and the required file length (in bytes).
Truncation can be achieved using system or library calls intended for such a task, if such methods exist, or by creating a temporary file of a reduced size and renaming it, after first deleting the original file, if no other method is available. The file may contain non human readable binary data in an unspecified format, so the routine should be "binary safe", leaving the contents of the untruncated part of the file unchanged.
If the specified filename does not exist, or the provided length is not less than the current file length, then the routine should raise an appropriate error condition.
On some systems, the provided file truncation facilities might not change the file or may extend the file, if the specified length is greater than the current length of the file.
This task permits the use of such facilities. However, such behaviour should be noted, or optionally a warning message relating to an non change or increase in file size may be implemented.
| #VBScript | VBScript |
Sub truncate(fpath,n)
'Check if file exist
Set objfso = CreateObject("Scripting.FileSystemObject")
If objfso.FileExists(fpath) = False Then
WScript.Echo fpath & " does not exist"
Exit Sub
End If
content = ""
'stream the input file
Set objinstream = CreateObject("Adodb.Stream")
With objinstream
.Type = 1
.Open
.LoadFromFile(fpath)
'check if the specified size is larger than the file content
If n <= .Size Then
content = .Read(n)
Else
WScript.Echo "The specified size is larger than the file content"
Exit Sub
End If
.Close
End With
'write the truncated version
Set objoutstream = CreateObject("Adodb.Stream")
With objoutstream
.Type = 1
.Open
.Write content
.SaveToFile fpath,2
.Close
End With
Set objinstream = Nothing
Set objoutstream = Nothing
Set objfso = Nothing
End Sub
'testing
Call truncate("C:\temp\test.txt",30)
|
http://rosettacode.org/wiki/Truncate_a_file | Truncate a file | Task
Truncate a file to a specific length. This should be implemented as a routine that takes two parameters: the filename and the required file length (in bytes).
Truncation can be achieved using system or library calls intended for such a task, if such methods exist, or by creating a temporary file of a reduced size and renaming it, after first deleting the original file, if no other method is available. The file may contain non human readable binary data in an unspecified format, so the routine should be "binary safe", leaving the contents of the untruncated part of the file unchanged.
If the specified filename does not exist, or the provided length is not less than the current file length, then the routine should raise an appropriate error condition.
On some systems, the provided file truncation facilities might not change the file or may extend the file, if the specified length is greater than the current length of the file.
This task permits the use of such facilities. However, such behaviour should be noted, or optionally a warning message relating to an non change or increase in file size may be implemented.
| #Wren | Wren | import "/ioutil" for FileUtil
var fileName = "temp.txt"
// create a file of length 26 bytes
FileUtil.write(fileName, "abcdefghijklmnopqrstuvwxyz")
System.print("Contents before truncation: %(FileUtil.read(fileName))")
// truncate file to 13 bytes
FileUtil.truncate(fileName, 13)
System.print("Contents after truncation : %(FileUtil.read(fileName))")
// attempt to truncate file to 20 bytes
FileUtil.truncate(fileName, 20)
System.print("Contents are still : %(FileUtil.read(fileName))") |
http://rosettacode.org/wiki/Tree_traversal | Tree traversal | Task
Implement a binary tree where each node carries an integer, and implement:
pre-order,
in-order,
post-order, and
level-order traversal.
Use those traversals to output the following tree:
1
/ \
/ \
/ \
2 3
/ \ /
4 5 6
/ / \
7 8 9
The correct output should look like this:
preorder: 1 2 4 7 5 3 6 8 9
inorder: 7 4 2 5 1 8 6 9 3
postorder: 7 4 5 2 8 9 6 3 1
level-order: 1 2 3 4 5 6 7 8 9
See also
Wikipedia article: Tree traversal.
| #AArch64_Assembly | AArch64 Assembly |
/* ARM assembly AARCH64 Raspberry PI 3B */
/* program deftree64.s */
/*******************************************/
/* Constantes file */
/*******************************************/
/* for this file see task include a file in language AArch64 assembly*/
.include "../includeConstantesARM64.inc"
.equ NBVAL, 9
/*******************************************/
/* Structures */
/********************************************/
/* structure tree */
.struct 0
tree_root: // root pointer
.struct tree_root + 8
tree_size: // number of element of tree
.struct tree_size + 8
tree_fin:
/* structure node tree */
.struct 0
node_left: // left pointer
.struct node_left + 8
node_right: // right pointer
.struct node_right + 8
node_value: // element value
.struct node_value + 8
node_fin:
/* structure queue*/
.struct 0
queue_begin: // next pointer
.struct queue_begin + 8
queue_end: // element value
.struct queue_end + 8
queue_fin:
/* structure node queue */
.struct 0
queue_node_next: // next pointer
.struct queue_node_next + 8
queue_node_value: // element value
.struct queue_node_value + 8
queue_node_fin:
/*******************************************/
/* Initialized data */
/*******************************************/
.data
szMessInOrder: .asciz "inOrder :\n"
szMessPreOrder: .asciz "PreOrder :\n"
szMessPostOrder: .asciz "PostOrder :\n"
szMessLevelOrder: .asciz "LevelOrder :\n"
szCarriageReturn: .asciz "\n"
/* datas error display */
szMessErreur: .asciz "Error detected.\n"
/* datas message display */
szMessResult: .ascii "Element value : @ \n"
/*******************************************/
/* UnInitialized data */
/*******************************************/
.bss
.align 4
sZoneConv: .skip 24
stTree: .skip tree_fin // place to structure tree
stQueue: .skip queue_fin // place to structure queue
/*******************************************/
/* code section */
/*******************************************/
.text
.global main
main:
mov x1,1 // node tree value
1:
ldr x0,qAdrstTree // structure tree address
bl insertElement // add element value x1
cmp x0,-1
beq 99f
add x1,x1,1 // increment value
cmp x1,NBVAL // end ?
ble 1b // no -> loop
ldr x0,qAdrszMessPreOrder
bl affichageMess
ldr x3,qAdrstTree // tree root address (begin structure)
ldr x0,[x3,#tree_root]
ldr x1,qAdrdisplayElement // function to execute
bl preOrder
ldr x0,qAdrszMessInOrder
bl affichageMess
ldr x3,qAdrstTree
ldr x0,[x3,#tree_root]
ldr x1,qAdrdisplayElement // function to execute
bl inOrder
ldr x0,qAdrszMessPostOrder
bl affichageMess
ldr x3,qAdrstTree
ldr x0,[x3,#tree_root]
ldr x1,qAdrdisplayElement // function to execute
bl postOrder
ldr x0,qAdrszMessLevelOrder
bl affichageMess
ldr x3,qAdrstTree
ldr x0,[x3,#tree_root]
ldr x1,qAdrdisplayElement // function to execute
bl levelOrder
b 100f
99: // display error
ldr x0,qAdrszMessErreur
bl affichageMess
100: // standard end of the program
mov x8,EXIT // request to exit program
svc 0 // perform system call
qAdrszMessInOrder: .quad szMessInOrder
qAdrszMessPreOrder: .quad szMessPreOrder
qAdrszMessPostOrder: .quad szMessPostOrder
qAdrszMessLevelOrder: .quad szMessLevelOrder
qAdrszMessErreur: .quad szMessErreur
qAdrszCarriageReturn: .quad szCarriageReturn
qAdrstTree: .quad stTree
qAdrstQueue: .quad stQueue
qAdrdisplayElement: .quad displayElement
/******************************************************************/
/* insert element in the tree */
/******************************************************************/
/* x0 contains the address of the tree structure */
/* x1 contains the value of element */
/* x0 returns address of element or - 1 if error */
insertElement:
stp x1,lr,[sp,-16]! // save registers
mov x4,x0
mov x0,node_fin // reservation place one element
bl allocHeap
cmp x0,-1 // allocation error
beq 100f
mov x5,x0
str x1,[x5,node_value] // store value in address heap
mov x1,0
str x1,[x5,node_left] // init left pointer with zero
str x1,[x5,node_right] // init right pointer with zero
ldr x2,[x4,tree_size] // load tree size
cbnz x2,1f // 0 element ?
str x5,[x4,tree_root] // yes -> store in root
b 6f
1: // else search free address in tree
ldr x3,[x4,tree_root] // start with address root
add x6,x2,1 // increment tree size
clz x7,x6 // compute zeroes left bits
add x7,x7,1 // for sustract the first left bit
lsl x6,x6,x7 // shift number in left
2:
tst x6,1<<63 // test left bit
lsl x6,x6,1 // shift left bit
bne 3f // bit at one
ldr x1,[x3,node_left] // no store node address in left pointer
cbz x1,4f // if equal zero
mov x3,x1 // else loop with next node
b 2b
3: // yes
ldr x1,[x3,node_right] // store node address in right pointer
cbz x1,5f // if equal zero
mov x3,x1 // else loop with next node
b 2b
4:
str x5,[x3,node_left]
b 6f
5:
str x5,[x3,node_right]
6:
add x2,x2,1 // increment tree size
str x2,[x4,tree_size]
100:
ldp x1,lr,[sp],16 // restaur 2 registers
ret // return to address lr x30
/******************************************************************/
/* preOrder */
/******************************************************************/
/* x0 contains the address of the node */
/* x1 function address */
preOrder:
stp x1,lr,[sp,-16]! // save registers
stp x2,x3,[sp,-16]! // save registers
cmp x0,#0
beq 100f
mov x2,x0
blr x1 // call function
ldr x0,[x2,#node_left]
bl preOrder
ldr x0,[x2,#node_right]
bl preOrder
100:
ldp x2,x3,[sp],16 // restaur 2 registers
ldp x1,lr,[sp],16 // restaur 2 registers
ret // return to address lr x30
/******************************************************************/
/* inOrder */
/******************************************************************/
/* x0 contains the address of the node */
/* x1 function address */
inOrder:
stp x1,lr,[sp,-16]! // save registers
stp x2,x3,[sp,-16]! // save registers
cbz x0,100f
mov x3,x0
mov x2,x1
ldr x0,[x3,node_left]
bl inOrder
mov x0,x3
blr x2 // call function
ldr x0,[x3,node_right]
mov x1,x2
bl inOrder
100:
ldp x2,x3,[sp],16 // restaur 2 registers
ldp x1,lr,[sp],16 // restaur 2 registers
ret // return to address lr x30
/******************************************************************/
/* postOrder */
/******************************************************************/
/* x0 contains the address of the node */
/* x1 function address */
postOrder:
stp x1,lr,[sp,-16]! // save registers
stp x2,x3,[sp,-16]! // save registers
cbz x0,100f
mov x3,x0
mov x2,x1
ldr x0,[x3,#node_left]
bl postOrder
ldr x0,[x3,#node_right]
mov x1,x2
bl postOrder
mov x0,x3
blr x2 // call function
100:
ldp x2,x3,[sp],16 // restaur 2 registers
ldp x1,lr,[sp],16 // restaur 2 registers
ret // return to address lr x30
/******************************************************************/
/* levelOrder */
/******************************************************************/
/* x0 contains the address of the node */
/* x1 function address */
levelOrder:
stp x1,lr,[sp,-16]! // save registers
stp x2,x3,[sp,-16]! // save registers
cbz x0,100f
mov x2,x1
mov x1,x0
ldr x0,qAdrstQueue // adresse queue
bl enqueueNode // queue the node
1: // begin loop
ldr x0,qAdrstQueue
bl isEmptyQueue // is queue empty
cbz x0,100f // yes -> end
ldr x0,qAdrstQueue
bl dequeueNode
mov x3,x0 // save node
blr x2 // call function
ldr x14,[x3,#node_left] // left node ok ?
cbz x14,2f
ldr x0,qAdrstQueue // yes -> enqueue
mov x1,x14
bl enqueueNode
2:
ldr x14,[x3,#node_right] // right node ok ?
cbz x14,3f
ldr x0,qAdrstQueue // yes -> enqueue
mov x1,x14
bl enqueueNode
3:
b 1b // and loop
100:
ldp x2,x3,[sp],16 // restaur 2 registers
ldp x1,lr,[sp],16 // restaur 2 registers
ret // return to address lr x30
/******************************************************************/
/* display node */
/******************************************************************/
/* x0 contains node address */
displayElement:
stp x1,lr,[sp,-16]! // save registers
ldr x0,[x0,#node_value]
ldr x1,qAdrsZoneConv
bl conversion10S
ldr x0,qAdrszMessResult
ldr x1,qAdrsZoneConv
bl strInsertAtCharInc // insert result at @ character
bl affichageMess
100:
ldp x1,lr,[sp],16 // restaur 2 registers
ret // return to address lr x30
qAdrszMessResult: .quad szMessResult
qAdrsZoneConv: .quad sZoneConv
/******************************************************************/
/* enqueue node */
/******************************************************************/
/* x0 contains the address of the queue */
/* x1 contains the value of element */
/* x0 returns address of element or - 1 if error */
enqueueNode:
stp x1,lr,[sp,-16]! // save registers
stp x2,x3,[sp,-16]! // save registers
mov x14,x0
mov x0,#queue_node_fin // allocation place heap
bl allocHeap
cmp x0,#-1 // allocation error
beq 100f
mov x15,x0 // save heap address
str x1,[x15,#queue_node_value] // store node value
mov x1,#0
str x1,[x15,#queue_node_next] // init pointer next
ldr x0,[x14,#queue_end]
cbz x0,1f
str x15,[x0,#queue_node_next]
b 2f
1:
str x15,[x14,#queue_begin]
2:
str x15,[x14,#queue_end]
mov x0,#0
100:
ldp x2,x3,[sp],16 // restaur 2 registers
ldp x1,lr,[sp],16 // restaur 2 registers
ret // return to address lr x30
/******************************************************************/
/* dequeue node */
/******************************************************************/
/* x0 contains the address of the queue */
/* x0 returns address of element or - 1 if error */
dequeueNode:
stp x1,lr,[sp,-16]! // save registers
ldr x14,[x0,#queue_begin]
ldr x15,[x14,#queue_node_value]
ldr x16,[x14,#queue_node_next]
str x16,[x0,#queue_begin]
cbnz x16,1f
str x16,[x0,#queue_end]
1:
mov x0,x15
100:
ldp x1,lr,[sp],16 // restaur 2 registers
ret // return to address lr x30
/******************************************************************/
/* dequeue node */
/******************************************************************/
/* x0 contains the address of the queue */
/* x0 returns 0 if empty else 1 */
isEmptyQueue:
ldr x0,[x0,#queue_begin]
cmp x0,#0
cset x0,ne
ret // return
/******************************************************************/
/* memory allocation on the heap */
/******************************************************************/
/* x0 contains the size to allocate */
/* x0 returns address of memory heap or - 1 if error */
/* CAUTION : The size of the allowance must be a multiple of 4 */
allocHeap:
stp x8,lr,[sp,-16]! // save registers
// allocation
mov x16,x0 // save size
mov x0,0 // read address start heap
mov x8,BRK // call system 'brk'
svc 0
mov x15,x0 // save address heap for return
add x0,x0,x16 // reservation place for size
mov x8,BRK // call system 'brk'
svc 0
cmp x0,-1 // allocation error
beq 100f
mov x0,x15 // return address memory heap
100:
ldp x8,lr,[sp],16 // restaur 2 registers
ret // return to address lr x30
/***********************************************/
/* File Include fonctions */
/********************************************************/
/* for this file see task include a file in language AArch64 assembly */
.include "../includeARM64.inc"
|
http://rosettacode.org/wiki/Topic_variable | Topic variable | Several programming languages offer syntax shortcuts to deal with the notion of "current" or "topic" variable.
A topic variable is a special variable with a very short name which can also often be omitted.
Demonstrate the utilization and behaviour of the topic variable within the language and explain or demonstrate how the topic variable behaves under different levels of nesting or scope, if this applies, within the language.
For instance you can (but you don't have to) show how the topic variable can be used by assigning the number
3
{\displaystyle 3}
to it and then computing its square and square root.
| #Axe | Axe | 3
Disp *3▶Dec,i |
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