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http://rosettacode.org/wiki/Polymorphism
Polymorphism
Task Create two classes   Point(x,y)   and   Circle(x,y,r)   with a polymorphic function print, accessors for (x,y,r), copy constructor, assignment and destructor and every possible default constructors
#Java
Java
class Point { protected int x, y; public Point() { this(0); } public Point(int x) { this(x, 0); } public Point(int x, int y) { this.x = x; this.y = y; } public Point(Point p) { this(p.x, p.y); } public int getX() { return this.x; } public int getY() { return this.y; } public void setX(int x) { this.x = x; } public void setY(int y) { this.y = y; } public void print() { System.out.println("Point x: " + this.x + " y: " + this.y); } }   class Circle extends Point { private int r; public Circle(Point p) { this(p, 0); } public Circle(Point p, int r) { super(p); this.r = r; } public Circle() { this(0); } public Circle(int x) { this(x, 0); } public Circle(int x, int y) { this(x, y, 0); } public Circle(int x, int y, int r) { super(x, y); this.r = r; } public Circle(Circle c) { this(c.x, c.y, c.r); } public int getR() { return this.r; } public void setR(int r) { this.r = r; } public void print() { System.out.println("Circle x: " + this.x + " y: " + this.y + " r: " + this.r); } }   public class test { public static void main(String args[]) { Point p = new Point(); Point c = new Circle(); p.print(); c.print(); } }
http://rosettacode.org/wiki/Poker_hand_analyser
Poker hand analyser
Task Create a program to parse a single five card poker hand and rank it according to this list of poker hands. A poker hand is specified as a space separated list of five playing cards. Each input card has two characters indicating face and suit. Example 2d       (two of diamonds). Faces are:    a, 2, 3, 4, 5, 6, 7, 8, 9, 10, j, q, k Suits are:    h (hearts),   d (diamonds),   c (clubs),   and   s (spades),   or alternatively,   the unicode card-suit characters:    ♥ ♦ ♣ ♠ Duplicate cards are illegal. The program should analyze a single hand and produce one of the following outputs: straight-flush four-of-a-kind full-house flush straight three-of-a-kind two-pair one-pair high-card invalid Examples 2♥ 2♦ 2♣ k♣ q♦: three-of-a-kind 2♥ 5♥ 7♦ 8♣ 9♠: high-card a♥ 2♦ 3♣ 4♣ 5♦: straight 2♥ 3♥ 2♦ 3♣ 3♦: full-house 2♥ 7♥ 2♦ 3♣ 3♦: two-pair 2♥ 7♥ 7♦ 7♣ 7♠: four-of-a-kind 10♥ j♥ q♥ k♥ a♥: straight-flush 4♥ 4♠ k♠ 5♦ 10♠: one-pair q♣ 10♣ 7♣ 6♣ q♣: invalid The programs output for the above examples should be displayed here on this page. Extra credit use the playing card characters introduced with Unicode 6.0 (U+1F0A1 - U+1F0DE). allow two jokers use the symbol   joker duplicates would be allowed (for jokers only) five-of-a-kind would then be the highest hand More extra credit examples joker 2♦ 2♠ k♠ q♦: three-of-a-kind joker 5♥ 7♦ 8♠ 9♦: straight joker 2♦ 3♠ 4♠ 5♠: straight joker 3♥ 2♦ 3♠ 3♦: four-of-a-kind joker 7♥ 2♦ 3♠ 3♦: three-of-a-kind joker 7♥ 7♦ 7♠ 7♣: five-of-a-kind joker j♥ q♥ k♥ A♥: straight-flush joker 4♣ k♣ 5♦ 10♠: one-pair joker k♣ 7♣ 6♣ 4♣: flush joker 2♦ joker 4♠ 5♠: straight joker Q♦ joker A♠ 10♠: straight joker Q♦ joker A♦ 10♦: straight-flush joker 2♦ 2♠ joker q♦: four-of-a-kind Related tasks Playing cards Card shuffles Deal cards_for_FreeCell War Card_Game Go Fish
#Python
Python
from collections import namedtuple   class Card(namedtuple('Card', 'face, suit')): def __repr__(self): return ''.join(self)     suit = '♥ ♦ ♣ ♠'.split() # ordered strings of faces faces = '2 3 4 5 6 7 8 9 10 j q k a' lowaces = 'a 2 3 4 5 6 7 8 9 10 j q k' # faces as lists face = faces.split() lowace = lowaces.split()     def straightflush(hand): f,fs = ( (lowace, lowaces) if any(card.face == '2' for card in hand) else (face, faces) ) ordered = sorted(hand, key=lambda card: (f.index(card.face), card.suit)) first, rest = ordered[0], ordered[1:] if ( all(card.suit == first.suit for card in rest) and ' '.join(card.face for card in ordered) in fs ): return 'straight-flush', ordered[-1].face return False   def fourofakind(hand): allfaces = [f for f,s in hand] allftypes = set(allfaces) if len(allftypes) != 2: return False for f in allftypes: if allfaces.count(f) == 4: allftypes.remove(f) return 'four-of-a-kind', [f, allftypes.pop()] else: return False   def fullhouse(hand): allfaces = [f for f,s in hand] allftypes = set(allfaces) if len(allftypes) != 2: return False for f in allftypes: if allfaces.count(f) == 3: allftypes.remove(f) return 'full-house', [f, allftypes.pop()] else: return False   def flush(hand): allstypes = {s for f, s in hand} if len(allstypes) == 1: allfaces = [f for f,s in hand] return 'flush', sorted(allfaces, key=lambda f: face.index(f), reverse=True) return False   def straight(hand): f,fs = ( (lowace, lowaces) if any(card.face == '2' for card in hand) else (face, faces) ) ordered = sorted(hand, key=lambda card: (f.index(card.face), card.suit)) first, rest = ordered[0], ordered[1:] if ' '.join(card.face for card in ordered) in fs: return 'straight', ordered[-1].face return False   def threeofakind(hand): allfaces = [f for f,s in hand] allftypes = set(allfaces) if len(allftypes) <= 2: return False for f in allftypes: if allfaces.count(f) == 3: allftypes.remove(f) return ('three-of-a-kind', [f] + sorted(allftypes, key=lambda f: face.index(f), reverse=True)) else: return False   def twopair(hand): allfaces = [f for f,s in hand] allftypes = set(allfaces) pairs = [f for f in allftypes if allfaces.count(f) == 2] if len(pairs) != 2: return False p0, p1 = pairs other = [(allftypes - set(pairs)).pop()] return 'two-pair', pairs + other if face.index(p0) > face.index(p1) else pairs[::-1] + other   def onepair(hand): allfaces = [f for f,s in hand] allftypes = set(allfaces) pairs = [f for f in allftypes if allfaces.count(f) == 2] if len(pairs) != 1: return False allftypes.remove(pairs[0]) return 'one-pair', pairs + sorted(allftypes, key=lambda f: face.index(f), reverse=True)   def highcard(hand): allfaces = [f for f,s in hand] return 'high-card', sorted(allfaces, key=lambda f: face.index(f), reverse=True)   handrankorder = (straightflush, fourofakind, fullhouse, flush, straight, threeofakind, twopair, onepair, highcard)   def rank(cards): hand = handy(cards) for ranker in handrankorder: rank = ranker(hand) if rank: break assert rank, "Invalid: Failed to rank cards: %r" % cards return rank   def handy(cards='2♥ 2♦ 2♣ k♣ q♦'): hand = [] for card in cards.split(): f, s = card[:-1], card[-1] assert f in face, "Invalid: Don't understand card face %r" % f assert s in suit, "Invalid: Don't understand card suit %r" % s hand.append(Card(f, s)) assert len(hand) == 5, "Invalid: Must be 5 cards in a hand, not %i" % len(hand) assert len(set(hand)) == 5, "Invalid: All cards in the hand must be unique %r" % cards return hand     if __name__ == '__main__': hands = ["2♥ 2♦ 2♣ k♣ q♦", "2♥ 5♥ 7♦ 8♣ 9♠", "a♥ 2♦ 3♣ 4♣ 5♦", "2♥ 3♥ 2♦ 3♣ 3♦", "2♥ 7♥ 2♦ 3♣ 3♦", "2♥ 7♥ 7♦ 7♣ 7♠", "10♥ j♥ q♥ k♥ a♥"] + [ "4♥ 4♠ k♠ 5♦ 10♠", "q♣ 10♣ 7♣ 6♣ 4♣", ] print("%-18s %-15s %s" % ("HAND", "CATEGORY", "TIE-BREAKER")) for cards in hands: r = rank(cards) print("%-18r %-15s %r" % (cards, r[0], r[1]))
http://rosettacode.org/wiki/Population_count
Population count
Population count You are encouraged to solve this task according to the task description, using any language you may know. The   population count   is the number of   1s   (ones)   in the binary representation of a non-negative integer. Population count   is also known as:   pop count   popcount   sideways sum   bit summation   Hamming weight For example,   5   (which is   101   in binary)   has a population count of   2. Evil numbers   are non-negative integers that have an   even   population count. Odious numbers     are  positive integers that have an    odd   population count. Task write a function (or routine) to return the population count of a non-negative integer. all computation of the lists below should start with   0   (zero indexed). display the   pop count   of the   1st   thirty powers of   3       (30,   31,   32,   33,   34,   ∙∙∙   329). display the   1st   thirty     evil     numbers. display the   1st   thirty   odious   numbers. display each list of integers on one line   (which may or may not include a title),   each set of integers being shown should be properly identified. See also The On-Line Encyclopedia of Integer Sequences:   A000120 population count. The On-Line Encyclopedia of Integer Sequences:   A000069 odious numbers. The On-Line Encyclopedia of Integer Sequences:   A001969 evil numbers.
#Fermat
Fermat
Func Popcount(n) = if n = 0 then 0 else if 2*(n\2)=n then Popcount(n\2) else Popcount((n-1)\2)+1 fi fi. Func Odiousness(n) = p:=Popcount(n);if 2*(p\2) = p then 0 else 1 fi.   for n=0 to 29 do !Popcount(3^n);!' ' od e:=0 n:=0 while e<30 do if Odiousness(n)=0 then !n;!' ';e:=e+1 fi; n:=n+1 od e:=0 n:=0 while e<30 do if Odiousness(n)=1 then !n;!' ';e:=e+1 fi; n:=n+1 od
http://rosettacode.org/wiki/Polynomial_long_division
Polynomial long division
This page uses content from Wikipedia. The original article was at Polynomial long division. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance) In algebra, polynomial long division is an algorithm for dividing a polynomial by another polynomial of the same or lower degree. Let us suppose a polynomial is represented by a vector, x {\displaystyle x} (i.e., an ordered collection of coefficients) so that the i {\displaystyle i} th element keeps the coefficient of x i {\displaystyle x^{i}} , and the multiplication by a monomial is a shift of the vector's elements "towards right" (injecting ones from left) followed by a multiplication of each element by the coefficient of the monomial. Then a pseudocode for the polynomial long division using the conventions described above could be: degree(P): return the index of the last non-zero element of P; if all elements are 0, return -∞ polynomial_long_division(N, D) returns (q, r): // N, D, q, r are vectors if degree(D) < 0 then error q ← 0 while degree(N) ≥ degree(D) d ← D shifted right by (degree(N) - degree(D)) q(degree(N) - degree(D)) ← N(degree(N)) / d(degree(d)) // by construction, degree(d) = degree(N) of course d ← d * q(degree(N) - degree(D)) N ← N - d endwhile r ← N return (q, r) Note: vector * scalar multiplies each element of the vector by the scalar; vectorA - vectorB subtracts each element of the vectorB from the element of the vectorA with "the same index". The vectors in the pseudocode are zero-based. Error handling (for allocations or for wrong inputs) is not mandatory. Conventions can be different; in particular, note that if the first coefficient in the vector is the highest power of x for the polynomial represented by the vector, then the algorithm becomes simpler. Example for clarification This example is from Wikipedia, but changed to show how the given pseudocode works. 0 1 2 3 ---------------------- N: -42 0 -12 1 degree = 3 D: -3 1 0 0 degree = 1 d(N) - d(D) = 2, so let's shift D towards right by 2: N: -42 0 -12 1 d: 0 0 -3 1 N(3)/d(3) = 1, so d is unchanged. Now remember that "shifting by 2" is like multiplying by x2, and the final multiplication (here by 1) is the coefficient of this monomial. Let's store this into q: 0 1 2 --------------- q: 0 0 1 now compute N - d, and let it be the "new" N, and let's loop N: -42 0 -9 0 degree = 2 D: -3 1 0 0 degree = 1 d(N) - d(D) = 1, right shift D by 1 and let it be d N: -42 0 -9 0 d: 0 -3 1 0 * -9/1 = -9 q: 0 -9 1 d: 0 27 -9 0 N ← N - d N: -42 -27 0 0 degree = 1 D: -3 1 0 0 degree = 1 looping again... d(N)-d(D)=0, so no shift is needed; we multiply D by -27 (= -27/1) storing the result in d, then q: -27 -9 1 and N: -42 -27 0 0 - d: 81 -27 0 0 = N: -123 0 0 0 (last N) d(N) < d(D), so now r ← N, and the result is: 0 1 2 ------------- q: -27 -9 1 → x2 - 9x - 27 r: -123 0 0 → -123 Related task   Polynomial derivative
#Phix
Phix
-- demo\rosetta\Polynomial_long_division.exw with javascript_semantics function degree(sequence p) for i=length(p) to 1 by -1 do if p[i]!=0 then return i end if end for return -1 end function function poly_div(sequence n, d) d = deep_copy(d) while length(d)<length(n) do d &=0 end while integer dn = degree(n), dd = degree(d) if dd<0 then throw("divide by zero") end if sequence quo = repeat(0,dn), rem = deep_copy(n) while dn>=dd do integer k = dn-dd, qk = rem[dn]/d[dd] sequence d2 = d[1..length(d)-k] quo[k+1] = qk for i=1 to length(d2) do integer mi = -i rem[mi] -= d2[mi]*qk end for dn = degree(rem) end while return {quo,rem} end function function poly(sequence si) -- display helper string r = "" for t=length(si) to 1 by -1 do integer sit = si[t] if sit!=0 then if sit=1 and t>1 then r &= iff(r=""? "":" + ") elsif sit=-1 and t>1 then r &= iff(r=""?"-":" - ") else if r!="" then r &= iff(sit<0?" - ":" + ") sit = abs(sit) end if r &= sprintf("%d",sit) end if r &= iff(t>1?"x"&iff(t>2?sprintf("^%d",t-1):""):"") end if end for if r="" then r="0" end if return r end function constant tests = {{{-42,0,-12,1},{-3,1}}, {{-3,1},{-42,0,-12,1}}, {{-42,0,-12,1},{-3,1,1}}, {{2,3,1},{1,1}}, {{3,5,6,-4,1},{1,2,1}}, {{3,0,7,0,0,0,0,0,3,0,0,1},{1,0,0,5,0,0,0,1}}, {{-56,87,-94,-55,22,-7},{2,0,1}}, } constant fmt = "%40s / %-16s = %25s rem %s\n" for i=1 to length(tests) do sequence {num,den} = tests[i], {quo,rem} = poly_div(num,den) printf(1,fmt,apply({num,den,quo,rem},poly)) end for
http://rosettacode.org/wiki/Polynomial_regression
Polynomial regression
Find an approximating polynomial of known degree for a given data. Example: For input data: x = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; y = {1, 6, 17, 34, 57, 86, 121, 162, 209, 262, 321}; The approximating polynomial is: 3 x2 + 2 x + 1 Here, the polynomial's coefficients are (3, 2, 1). This task is intended as a subtask for Measure relative performance of sorting algorithms implementations.
#OCaml
OCaml
open Base open Stdio   let mean fa = let open Float in (Array.reduce_exn fa ~f:(+)) / (of_int (Array.length fa))   let regression xs ys = let open Float in let xm = mean xs in let ym = mean ys in let x2m = Array.map xs ~f:(fun x -> x * x) |> mean in let x3m = Array.map xs ~f:(fun x -> x * x * x) |> mean in let x4m = Array.map xs ~f:(fun x -> let x2 = x * x in x2 * x2) |> mean in let xzipy = Array.zip_exn xs ys in let xym = Array.map xzipy ~f:(fun (x, y) -> x * y) |> mean in let x2ym = Array.map xzipy ~f:(fun (x, y) -> x * x * y) |> mean in   let sxx = x2m - xm * xm in let sxy = xym - xm * ym in let sxx2 = x3m - xm * x2m in let sx2x2 = x4m - x2m * x2m in let sx2y = x2ym - x2m * ym in   let b = (sxy * sx2x2 - sx2y * sxx2) / (sxx * sx2x2 - sxx2 * sxx2) in let c = (sx2y * sxx - sxy * sxx2) / (sxx * sx2x2 - sxx2 * sxx2) in let a = ym - b * xm - c * x2m in   let abc xx = a + b * xx + c * xx * xx in   printf "y = %.1f + %.1fx + %.1fx^2\n\n" a b c; printf " Input Approximation\n"; printf " x y y1\n"; Array.iter xzipy ~f:(fun (xi, yi) -> printf "%2g %3g  %5.1f\n" xi yi (abc xi) )   let () = let x = Array.init 11 ~f:Float.of_int in let y = [| 1.; 6.; 17.; 34.; 57.; 86.; 121.; 162.; 209.; 262.; 321. |] in regression x y
http://rosettacode.org/wiki/Power_set
Power set
A   set   is a collection (container) of certain values, without any particular order, and no repeated values. It corresponds with a finite set in mathematics. A set can be implemented as an associative array (partial mapping) in which the value of each key-value pair is ignored. Given a set S, the power set (or powerset) of S, written P(S), or 2S, is the set of all subsets of S. Task By using a library or built-in set type, or by defining a set type with necessary operations, write a function with a set S as input that yields the power set 2S of S. For example, the power set of     {1,2,3,4}     is {{}, {1}, {2}, {1,2}, {3}, {1,3}, {2,3}, {1,2,3}, {4}, {1,4}, {2,4}, {1,2,4}, {3,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}. For a set which contains n elements, the corresponding power set has 2n elements, including the edge cases of empty set. The power set of the empty set is the set which contains itself (20 = 1): P {\displaystyle {\mathcal {P}}} ( ∅ {\displaystyle \varnothing } ) = { ∅ {\displaystyle \varnothing } } And the power set of the set which contains only the empty set, has two subsets, the empty set and the set which contains the empty set (21 = 2): P {\displaystyle {\mathcal {P}}} ({ ∅ {\displaystyle \varnothing } }) = { ∅ {\displaystyle \varnothing } , { ∅ {\displaystyle \varnothing } } } Extra credit: Demonstrate that your language supports these last two powersets.
#Go
Go
package main   import ( "fmt" "strconv" "strings" )   // types needed to implement general purpose sets are element and set   // element is an interface, allowing different kinds of elements to be // implemented and stored in sets. type elem interface { // an element must be distinguishable from other elements to satisfy // the mathematical definition of a set. a.eq(b) must give the same // result as b.eq(a). Eq(elem) bool // String result is used only for printable output. Given a, b where // a.eq(b), it is not required that a.String() == b.String(). fmt.Stringer }   // integer type satisfying element interface type Int int   func (i Int) Eq(e elem) bool { j, ok := e.(Int) return ok && i == j }   func (i Int) String() string { return strconv.Itoa(int(i)) }   // a set is a slice of elem's. methods are added to implement // the element interface, to allow nesting. type set []elem   // uniqueness of elements can be ensured by using add method func (s *set) add(e elem) { if !s.has(e) { *s = append(*s, e) } }   func (s *set) has(e elem) bool { for _, ex := range *s { if e.Eq(ex) { return true } } return false }   func (s set) ok() bool { for i, e0 := range s { for _, e1 := range s[i+1:] { if e0.Eq(e1) { return false } } } return true }   // elem.Eq func (s set) Eq(e elem) bool { t, ok := e.(set) if !ok { return false } if len(s) != len(t) { return false } for _, se := range s { if !t.has(se) { return false } } return true }   // elem.String func (s set) String() string { if len(s) == 0 { return "∅" } var buf strings.Builder buf.WriteRune('{') for i, e := range s { if i > 0 { buf.WriteRune(',') } buf.WriteString(e.String()) } buf.WriteRune('}') return buf.String() }   // method required for task func (s set) powerSet() set { r := set{set{}} for _, es := range s { var u set for _, er := range r { er := er.(set) u = append(u, append(er[:len(er):len(er)], es)) } r = append(r, u...) } return r }   func main() { var s set for _, i := range []Int{1, 2, 2, 3, 4, 4, 4} { s.add(i) } fmt.Println(" s:", s, "length:", len(s)) ps := s.powerSet() fmt.Println(" 𝑷(s):", ps, "length:", len(ps))   fmt.Println("\n(extra credit)") var empty set fmt.Println(" empty:", empty, "len:", len(empty)) ps = empty.powerSet() fmt.Println(" 𝑷(∅):", ps, "len:", len(ps)) ps = ps.powerSet() fmt.Println("𝑷(𝑷(∅)):", ps, "len:", len(ps))   fmt.Println("\n(regression test for earlier bug)") s = set{Int(1), Int(2), Int(3), Int(4), Int(5)} fmt.Println(" s:", s, "length:", len(s), "ok:", s.ok()) ps = s.powerSet() fmt.Println(" 𝑷(s):", "length:", len(ps), "ok:", ps.ok()) for _, e := range ps { if !e.(set).ok() { panic("invalid set in ps") } } }
http://rosettacode.org/wiki/Primality_by_trial_division
Primality by trial division
Task Write a boolean function that tells whether a given integer is prime. Remember that   1   and all non-positive numbers are not prime. Use trial division. Even numbers greater than   2   may be eliminated right away. A loop from   3   to   √ n    will suffice,   but other loops are allowed. Related tasks   count in factors   prime decomposition   AKS test for primes   factors of an integer   Sieve of Eratosthenes   factors of a Mersenne number   trial factoring of a Mersenne number   partition an integer X into N primes   sequence of primes by Trial Division
#Crystal
Crystal
Mathematicaly basis of Prime Generators https://www.academia.edu/19786419/PRIMES-UTILS_HANDBOOK https://www.academia.edu/42734410/Improved_Primality_Testing_and_Factorization_in_Ruby_revised
http://rosettacode.org/wiki/Price_fraction
Price fraction
A friend of mine runs a pharmacy.   He has a specialized function in his Dispensary application which receives a decimal value of currency and replaces it to a standard value.   This value is regulated by a government department. Task Given a floating point value between   0.00   and   1.00,   rescale according to the following table: >= 0.00 < 0.06  := 0.10 >= 0.06 < 0.11  := 0.18 >= 0.11 < 0.16  := 0.26 >= 0.16 < 0.21  := 0.32 >= 0.21 < 0.26  := 0.38 >= 0.26 < 0.31  := 0.44 >= 0.31 < 0.36  := 0.50 >= 0.36 < 0.41  := 0.54 >= 0.41 < 0.46  := 0.58 >= 0.46 < 0.51  := 0.62 >= 0.51 < 0.56  := 0.66 >= 0.56 < 0.61  := 0.70 >= 0.61 < 0.66  := 0.74 >= 0.66 < 0.71  := 0.78 >= 0.71 < 0.76  := 0.82 >= 0.76 < 0.81  := 0.86 >= 0.81 < 0.86  := 0.90 >= 0.86 < 0.91  := 0.94 >= 0.91 < 0.96  := 0.98 >= 0.96 < 1.01  := 1.00
#K
K
  le:- 0.96 0.91 0.86 0.81 0.76 0.71 0.66 0.61 0.56 0.51 0.46 0.41 0.36 0.31 0.26 0.21 0.16 0.11 0.06 0.0 out: 1.00 0.98 0.94 0.90 0.86 0.82 0.78 0.74 0.70 0.66 0.62 0.58 0.54 0.50 0.44 0.38 0.32 0.26 0.18 0.1   pf:{out@_bin[le;-x]}'  
http://rosettacode.org/wiki/Price_fraction
Price fraction
A friend of mine runs a pharmacy.   He has a specialized function in his Dispensary application which receives a decimal value of currency and replaces it to a standard value.   This value is regulated by a government department. Task Given a floating point value between   0.00   and   1.00,   rescale according to the following table: >= 0.00 < 0.06  := 0.10 >= 0.06 < 0.11  := 0.18 >= 0.11 < 0.16  := 0.26 >= 0.16 < 0.21  := 0.32 >= 0.21 < 0.26  := 0.38 >= 0.26 < 0.31  := 0.44 >= 0.31 < 0.36  := 0.50 >= 0.36 < 0.41  := 0.54 >= 0.41 < 0.46  := 0.58 >= 0.46 < 0.51  := 0.62 >= 0.51 < 0.56  := 0.66 >= 0.56 < 0.61  := 0.70 >= 0.61 < 0.66  := 0.74 >= 0.66 < 0.71  := 0.78 >= 0.71 < 0.76  := 0.82 >= 0.76 < 0.81  := 0.86 >= 0.81 < 0.86  := 0.90 >= 0.86 < 0.91  := 0.94 >= 0.91 < 0.96  := 0.98 >= 0.96 < 1.01  := 1.00
#Kotlin
Kotlin
// version 1.0.6   fun rescale(price: Double): Double = when { price < 0.06 -> 0.10 price < 0.11 -> 0.18 price < 0.16 -> 0.26 price < 0.21 -> 0.32 price < 0.26 -> 0.38 price < 0.31 -> 0.44 price < 0.36 -> 0.50 price < 0.41 -> 0.54 price < 0.46 -> 0.58 price < 0.51 -> 0.62 price < 0.56 -> 0.66 price < 0.61 -> 0.70 price < 0.66 -> 0.74 price < 0.71 -> 0.78 price < 0.76 -> 0.82 price < 0.81 -> 0.86 price < 0.86 -> 0.90 price < 0.91 -> 0.94 price < 0.96 -> 0.98 else -> 1.00 }   fun main(args: Array<String>) { var d: Double for (i in 1..100) { d = i / 100.0 print(String.format("%4.2f -> %4.2f ", d, rescale(d))) if (i % 5 == 0) println() } }
http://rosettacode.org/wiki/Proper_divisors
Proper divisors
The   proper divisors   of a positive integer N are those numbers, other than N itself, that divide N without remainder. For N > 1 they will always include 1,   but for N == 1 there are no proper divisors. Examples The proper divisors of     6     are   1, 2, and 3. The proper divisors of   100   are   1, 2, 4, 5, 10, 20, 25, and 50. Task Create a routine to generate all the proper divisors of a number. use it to show the proper divisors of the numbers 1 to 10 inclusive. Find a number in the range 1 to 20,000 with the most proper divisors. Show the number and just the count of how many proper divisors it has. Show all output here. Related tasks   Amicable pairs   Abundant, deficient and perfect number classifications   Aliquot sequence classifications   Factors of an integer   Prime decomposition
#Pascal
Pascal
{$IFDEF FPC}{$MODE DELPHI}{$ELSE}{$APPTYPE CONSOLE}{$ENDIF} uses sysutils; const MAXPROPERDIVS = 1920; type tRes = array[0..MAXPROPERDIVS] of LongWord; tPot = record potPrim, potMax :LongWord; end;   tprimeFac = record pfPrims : array[1..10] of tPot; pfCnt, pfNum : LongWord; end; tSmallPrimes = array[0..6541] of longWord;   var SmallPrimes: tSmallPrimes;   procedure InitSmallPrimes; var pr,testPr,j,maxprimidx: Longword; isPrime : boolean; Begin maxprimidx := 0; SmallPrimes[0] := 2; pr := 3; repeat isprime := true; j := 0; repeat testPr := SmallPrimes[j]; IF testPr*testPr > pr then break; If pr mod testPr = 0 then Begin isprime := false; break; end; inc(j); until false;   if isprime then Begin inc(maxprimidx); SmallPrimes[maxprimidx]:= pr; end; inc(pr,2); until pr > 1 shl 16 -1; end;   procedure PrimeFacOut(primeDecomp:tprimeFac); var i : LongWord; begin with primeDecomp do Begin write(pfNum,' = '); For i := 1 to pfCnt-1 do with pfPrims[i] do If potMax = 1 then write(potPrim,'*') else write(potPrim,'^',potMax,'*'); with pfPrims[pfCnt] do If potMax = 1 then write(potPrim) else write(potPrim,'^',potMax); end; end;   procedure PrimeDecomposition(n:LongWord;var res:tprimeFac); var i,pr,cnt,quot{to minimize divisions} : LongWord; Begin res.pfNum := n; res.pfCnt:= 0; i := 0; cnt := 0; repeat pr := SmallPrimes[i]; IF pr*pr>n then Break;   quot := n div pr; IF pr*quot = n then with res do Begin inc(pfCnt); with pfPrims[pfCnt] do Begin potPrim := pr; potMax := 0; repeat n := quot; quot := quot div pr; inc(potMax); until pr*quot <> n; end; end; inc(i); until false; //a big prime left over? IF n <> 1 then with res do Begin inc(pfCnt); with pfPrims[pfCnt] do Begin potPrim := n; potMax := 1; end; end; end;   function CntProperDivs(const primeDecomp:tprimeFac):LongWord; //count of proper divisors var i: LongWord; begin result := 1; with primeDecomp do For i := 1 to pfCnt do result := result*(pfPrims[i].potMax+1); //remove dec(result); end;   function findProperdivs(n:LongWord;var res:TRes):LongWord; //simple trial division to get a sorted list of all proper divisors var i,j: LongWord; Begin result := 0; i := 1; j := n; while j>i do begin j := n DIV i; IF i*j = n then Begin //smaller factor part at the beginning upwards res[result]:= i; IF i <> j then //bigger factor at the end downwards res[MAXPROPERDIVS-result]:= j else //n is square number res[MAXPROPERDIVS-result]:= 0; inc(result); end; inc(i); end;   If result>0 then Begin //move close together i := result; j := MAXPROPERDIVS-result+1; result := 2*result-1; repeat res[i] := res[j]; inc(j); inc(i); until i > result;   if res[result-1] = 0 then dec(result); end; end;   procedure AllFacsOut(n: Longword); var res:TRes; i,k,j:LongInt; Begin j := findProperdivs(n,res); write(n:5,' : '); For k := 0 to j-2 do write(res[k],','); IF j>=1 then write(res[j-1]); writeln; end;   var primeDecomp: tprimeFac; rs : tRes; i,j,max,maxcnt: LongWord; BEGIN InitSmallPrimes; For i := 1 to 10 do AllFacsOut(i); writeln; max := 0; maxCnt := 0; For i := 1 to 20*1000 do Begin PrimeDecomposition(i,primeDecomp); j := CntProperDivs(primeDecomp); IF j> maxCnt then Begin maxcnt := j; max := i; end; end; PrimeDecomposition(max,primeDecomp); j := CntProperDivs(primeDecomp);   PrimeFacOut(primeDecomp);writeln(' ',j:10,' factors'); writeln; //https://en.wikipedia.org/wiki/Highly_composite_number <= HCN //http://wwwhomes.uni-bielefeld.de/achim/highly.txt the first 1200 HCN max := 3491888400; PrimeDecomposition(max,primeDecomp); j := CntProperDivs(primeDecomp); PrimeFacOut(primeDecomp);writeln(' ',j:10,' factors'); writeln; END.
http://rosettacode.org/wiki/Probabilistic_choice
Probabilistic choice
Given a mapping between items and their required probability of occurrence, generate a million items randomly subject to the given probabilities and compare the target probability of occurrence versus the generated values. The total of all the probabilities should equal one. (Because floating point arithmetic is involved, this is subject to rounding errors). aleph 1/5.0 beth 1/6.0 gimel 1/7.0 daleth 1/8.0 he 1/9.0 waw 1/10.0 zayin 1/11.0 heth 1759/27720 # adjusted so that probabilities add to 1 Related task Random number generator (device)
#Ursala
Ursala
#import std #import nat #import flo   outcomes = <'aleph ','beth ','gimel ','daleth','he ','waw ','zayin ','heth '> probabilities = ^lrNCT(~&,minus/1.+ plus:-0) div/*1. float* skip/5 iota12   simulation =   ^(~&rn,div+ float~~rmPlX)^*D/~& iota; ^A(~&h,length)*K2+ * stochasm@p/probabilities !* outcomes   format =   :/' frequency probability'+ * ^lrlrTPT/~&n (printf/'%12.8f')^~/~&m outcomes-$probabilities@n   #show+   results = format simulation 1000000
http://rosettacode.org/wiki/Priority_queue
Priority queue
A priority queue is somewhat similar to a queue, with an important distinction: each item is added to a priority queue with a priority level, and will be later removed from the queue with the highest priority element first. That is, the items are (conceptually) stored in the queue in priority order instead of in insertion order. Task Create a priority queue.   The queue must support at least two operations:   Insertion.   An element is added to the queue with a priority (a numeric value).   Top item removal.   Deletes the element or one of the elements with the current top priority and return it. Optionally, other operations may be defined, such as peeking (find what current top priority/top element is), merging (combining two priority queues into one), etc. To test your implementation, insert a number of elements into the queue, each with some random priority. Then dequeue them sequentially; now the elements should be sorted by priority. You can use the following task/priority items as input data: Priority Task ══════════ ════════════════ 3 Clear drains 4 Feed cat 5 Make tea 1 Solve RC tasks 2 Tax return The implementation should try to be efficient.   A typical implementation has   O(log n)   insertion and extraction time,   where   n   is the number of items in the queue. You may choose to impose certain limits such as small range of allowed priority levels, limited capacity, etc.   If so, discuss the reasons behind it.
#PicoLisp
PicoLisp
# Insert item into priority queue (de insertPQ (Queue Prio Item) (idx Queue (cons Prio Item) T) )   # Remove and return top item from priority queue (de removePQ (Queue) (cdar (idx Queue (peekPQ Queue) NIL)) )   # Find top element in priority queue (de peekPQ (Queue) (let V (val Queue) (while (cadr V) (setq V @) ) (car V) ) )   # Merge second queue into first (de mergePQ (Queue1 Queue2) (balance Queue1 (sort (conc (idx Queue1) (idx Queue2)))) )
http://rosettacode.org/wiki/Prime_decomposition
Prime decomposition
The prime decomposition of a number is defined as a list of prime numbers which when all multiplied together, are equal to that number. Example 12 = 2 × 2 × 3, so its prime decomposition is {2, 2, 3} Task Write a function which returns an array or collection which contains the prime decomposition of a given number   n {\displaystyle n}   greater than   1. If your language does not have an isPrime-like function available, you may assume that you have a function which determines whether a number is prime (note its name before your code). If you would like to test code from this task, you may use code from trial division or the Sieve of Eratosthenes. Note: The program must not be limited by the word size of your computer or some other artificial limit; it should work for any number regardless of size (ignoring the physical limits of RAM etc). Related tasks   count in factors   factors of an integer   Sieve of Eratosthenes   primality by trial division   factors of a Mersenne number   trial factoring of a Mersenne number   partition an integer X into N primes   sequence of primes by Trial Division
#Groovy
Groovy
def factorize = { long target ->   if (target == 1) return [1L]   if (target < 4) return [1L, target]   def targetSqrt = Math.sqrt(target) def lowfactors = (2L..targetSqrt).findAll { (target % it) == 0 } if (lowfactors == []) return [1L, target] def nhalf = lowfactors.size() - ((lowfactors[-1]**2 == target) ? 1 : 0)   [1] + lowfactors + (0..<nhalf).collect { target.intdiv(lowfactors[it]) }.reverse() + [target] }   def decomposePrimes = { target -> def factors = factorize(target) - [1] def primeFactors = [] factors.eachWithIndex { f, i -> if (i==0 || factors[0..<i].every {f % it != 0}) { primeFactors << f def pfPower = f*f while (target % pfPower == 0) { primeFactors << f pfPower *= f } } } primeFactors }
http://rosettacode.org/wiki/Plot_coordinate_pairs
Plot coordinate pairs
Task Plot a function represented as    x,  y    numerical arrays. Post the resulting image for the following input arrays (taken from Python's Example section on Time a function): x = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}; y = {2.7, 2.8, 31.4, 38.1, 58.0, 76.2, 100.5, 130.0, 149.3, 180.0}; This task is intended as a subtask for Measure relative performance of sorting algorithms implementations.
#M2000_Interpreter
M2000 Interpreter
  Module Pairs { \\ written in version 9.5 rev. 13 \\ use Gdi+ antialiasing (not work with Wine in Linux, but we get no error) smooth on Const center=2, right=3, left=1, blue=1, angle=0, dotline=3 Const size9pt=9, size11pt=11 Cls ,0 ' use current background color, set split screen from line 0 Cursor 0,3 Report center, "Coordinate pairs" x = (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) y = (2.7, 2.8, 31.4, 38.1, 58.0, 76.2, 100.5, 130.0, 149.3, 180.0) dx=scale.x/2/len(x) dy=dx 'ratio 1:1 graH=dy*len(x) Basex=scale.x/4 Basey=(scale.y+graH)/2 Move Basex, Basey \\ draw use relative coordinates Draw 0,-graH \\ Step just move graphic cursor Step 0, graH Draw scale.x/2 Step -scale.x/2 \\ scX is 1, not used max=Y#max() \\ Auto scale for Y, using 0 for start of axis Y scY=-graH/((max+5^log(max) ) div 100)/100 \\ make vertical axis using dots with numbers center per dx j=1 For i=basex+dx to basex+dx*x#max() Step dx Move i, basey Step 0, twipsy*10 Legend format$("{0}",array(x,j)), "courier", size9pt, angle, center Width 1, dotline { draw 0, -graH-twipsy*10,7} j++ Next i \\ the same for horizontal axis HalfTextHeight=Size.y("1","courier", size9pt)/2 For i=basey-dy to basey-dy*x#max() Step dy Move basex, i Step -twipsx*10 Width 1, dotline { draw scale.x/2+twipsx*10,,7} Move basex-100, i+HalfTextHeight Legend format$("{0}",(i-basey)/scY), "courier", size9pt, angle, left Next i ex=each(x) : ey=each(y) \\ start from first point. We use Draw to for absolute coordinates Move array(x,0)*dx+Basex, array(y,0)*scy+Basey While ex, ey { Width 2 { Draw to array(ex)*dx+Basex, array(ey)*scy+Basey, blue } } \\ second pass for marks and labels ex=each(x) : ey=each(y) While ex, ey { Move array(ex)*dx+Basex, array(ey)*scy+Basey Step -75, -75 Pen 12 {draw 150: draw 0,150 : draw -150 : draw 0,-150} Pen 13 { Step 200, -200 Legend format$("({0}-{1})",array(ex),array(ey) ), "courier bold", size11pt, angle, right } } \\ screenshot to clipboard Screenshot$="" Move 0,0 Copy scale.x, scale.y to Screenshot$ Clipboard Screenshot$ a$=key$ } Pairs  
http://rosettacode.org/wiki/Polymorphism
Polymorphism
Task Create two classes   Point(x,y)   and   Circle(x,y,r)   with a polymorphic function print, accessors for (x,y,r), copy constructor, assignment and destructor and every possible default constructors
#JavaScript
JavaScript
/* create new Point in one of these ways: * var p = new Point(x,y); * var p = new Point(a_point); * default value for x,y is 0 */ function Point() { var arg1 = arguments[0]; var arg2 = arguments[1];   if (arg1 instanceof Point) { this.x = arg1.x; this.y = arg1.y; } else { this.x = arg1 == null ? 0 : arg1; this.y = arg2 == null ? 0 : arg1; }   this.set_x = function(_x) {this.x = _x;} this.set_y = function(_y) {this.y = _y;} }   Point.prototype.print = function() { var out = "Point(" + this.x + "," + this.y + ")"; print(out); }   /* create new Circle in one of these ways: * var c = new Circle(x,y,r); * var c = new Circle(a_circle); * var c = new Circle(a_point,r); * default value for x,y,r is 0 */ function Circle() { var arg1 = arguments[0]; var arg2 = arguments[1]; var arg3 = arguments[2];   if (arg1 instanceof Circle) { this.x = arg1.x; this.y = arg1.y; this.r = arg1.r; } else if (arg1 instanceof Point) { this.x = arg1.x; this.y = arg1.y; this.r = arg2 == null ? 0 : arg2; } else { this.x = arg1 == null ? 0 : arg1; this.y = arg2 == null ? 0 : arg2; this.r = arg3 == null ? 0 : arg3; }   this.set_x = function(_x) {this.x = _x;} this.set_y = function(_y) {this.y = _y;} this.set_r = function(_r) {this.r = _r;} }   Circle.prototype.print = function() { var out = "Circle(" + this.x + "," + this.y + "," + this.r + ")"; print(out); }
http://rosettacode.org/wiki/Poker_hand_analyser
Poker hand analyser
Task Create a program to parse a single five card poker hand and rank it according to this list of poker hands. A poker hand is specified as a space separated list of five playing cards. Each input card has two characters indicating face and suit. Example 2d       (two of diamonds). Faces are:    a, 2, 3, 4, 5, 6, 7, 8, 9, 10, j, q, k Suits are:    h (hearts),   d (diamonds),   c (clubs),   and   s (spades),   or alternatively,   the unicode card-suit characters:    ♥ ♦ ♣ ♠ Duplicate cards are illegal. The program should analyze a single hand and produce one of the following outputs: straight-flush four-of-a-kind full-house flush straight three-of-a-kind two-pair one-pair high-card invalid Examples 2♥ 2♦ 2♣ k♣ q♦: three-of-a-kind 2♥ 5♥ 7♦ 8♣ 9♠: high-card a♥ 2♦ 3♣ 4♣ 5♦: straight 2♥ 3♥ 2♦ 3♣ 3♦: full-house 2♥ 7♥ 2♦ 3♣ 3♦: two-pair 2♥ 7♥ 7♦ 7♣ 7♠: four-of-a-kind 10♥ j♥ q♥ k♥ a♥: straight-flush 4♥ 4♠ k♠ 5♦ 10♠: one-pair q♣ 10♣ 7♣ 6♣ q♣: invalid The programs output for the above examples should be displayed here on this page. Extra credit use the playing card characters introduced with Unicode 6.0 (U+1F0A1 - U+1F0DE). allow two jokers use the symbol   joker duplicates would be allowed (for jokers only) five-of-a-kind would then be the highest hand More extra credit examples joker 2♦ 2♠ k♠ q♦: three-of-a-kind joker 5♥ 7♦ 8♠ 9♦: straight joker 2♦ 3♠ 4♠ 5♠: straight joker 3♥ 2♦ 3♠ 3♦: four-of-a-kind joker 7♥ 2♦ 3♠ 3♦: three-of-a-kind joker 7♥ 7♦ 7♠ 7♣: five-of-a-kind joker j♥ q♥ k♥ A♥: straight-flush joker 4♣ k♣ 5♦ 10♠: one-pair joker k♣ 7♣ 6♣ 4♣: flush joker 2♦ joker 4♠ 5♠: straight joker Q♦ joker A♠ 10♠: straight joker Q♦ joker A♦ 10♦: straight-flush joker 2♦ 2♠ joker q♦: four-of-a-kind Related tasks Playing cards Card shuffles Deal cards_for_FreeCell War Card_Game Go Fish
#Racket
Racket
#lang racket (require (only-in srfi/1 car+cdr))   ;;; -------------------------------------------------------------------------------------------------- ;;; The analyser is first... the rest of it is prettiness surrounding strings and parsing! ;;; -------------------------------------------------------------------------------------------------- ;; (cons f _) and (cons _ s) appear too frequently in patterns to not factor out (define-match-expander F._ (λ (stx) (syntax-case stx () [(_ f) #'(cons f _)]))) (define-match-expander _.S (λ (stx) (syntax-case stx () [(_ s) #'(cons _ s)])))   ;; Matches are easier when the cards are lined up by face: and I always put the cards in my hand with ;; the highest card on the left (should I be telling this?)... anyway face<? is written to leave high ;; cards on the left. There is no need to sort by suit, flushes are all-or-nothing (define (face-sort hand) (sort hand (match-lambda** [(_ 'joker) #f] [('joker _) #t] [((F._ f1) (F._ f2)) (> f1 f2)])))   ;; even playing poker for money, I never managed to consistently determine what effect jokers were ;; having on my hand... so I'll do an exhaustive search of what's best! ;; ;; scoring hands allows us to choose a best value for joker(s) ;; hand-names provides an order (and therefore a score) for each of the available hands (define hand-names (list 'five-of-a-kind 'straight-flush 'four-of-a-kind 'full-house 'flush 'straight 'three-of-a-kind 'two-pair 'one-pair 'high-card))   (define hand-order# (for/hash ((h hand-names) (i (in-range (add1 (length hand-names)) 0 -1))) (values h i))) ;; The score of a hand is (its order*15^5)+(first tiebreaker*15^4)+(2nd tiebreaker*15^3)... ;; powers of 15 because we have a maxmium face value of 14 (ace) -- even though there are 13 cards ;; in a suit. (define (calculate-score analysis) (define-values (hand-name tiebreakers) (car+cdr analysis)) (for/sum ((n (in-naturals)) (tb (cons (hash-ref hand-order# hand-name -1) tiebreakers))) (* tb (expt 15 (- 5 n)))))   ;; score hand produces an analysis of a hand (which can then be returned to analyse-sorted-hand, ;; and a score that can be maximised by choosing the right jokers. (define (score-hand hand . jokers) ; gives an orderable list of hands with tiebreakers (define analysis (analyse-sorted-hand (face-sort (append jokers hand)))) (cons analysis (calculate-score analysis)))   ;; if we have two (or more) jokers, they will be consumed by the recursive call to ;; analyse-sorted-hand score-hand (define all-cards/promise (delay (for*/list ((f (in-range 2 15)) (s '(h d s c))) (cons f s))))   (define (best-jokered-hand cards) ; we've lost the first joker from cards (define-values (best-hand _bhs) (for*/fold ((best-hand #f) (best-score 0)) ((joker (in-list (force all-cards/promise))) (score (in-value (score-hand cards joker))) #:when (> (cdr score) best-score)) (car+cdr score)))   best-hand)   ;; we can abbreviate 2/3/4/5-of-a-kind 2-pair full-house with 2 and 3 (define-match-expander F*2 (λ (stx) (syntax-case stx () [(_ f) #'(list (F._ f) (F._ f))]))) (define-match-expander F*3 (λ (stx) (syntax-case stx () [(_ f) #'(list (F._ f) (F._ f) (F._ f))])))   ;; note that flush? is cheaper to calculate than straight?, so do it first when we test for ;; straight-flush (define flush? (match-lambda [(and `(,(_.S s) ,(_.S s) ,(_.S s) ,(_.S s) ,(_.S s)) `(,(F._ fs) ...)) `(flush ,@fs)] [_ #f]))   (define straight? (match-lambda  ;; '(straight 5) puts this at the bottom of the pile w.r.t the ordering of straights [`(,(F._ 14) ,(F._ 5) ,(F._ 4) ,(F._ 3) ,(F._ 2)) '(straight 5)] [`(,(F._ f5) ,(F._ f4) ,(F._ f3) ,(F._ f2) ,(F._ f1)) (and (= f1 (- f5 4)) (< f1 f2 f3 f4 f5) `(straight ,f5))]))   (define analyse-sorted-hand (match-lambda [(list 'joker cards ...) (best-jokered-hand cards)] [`(,@(F*3 f) ,@(F*2 f)) `(five-of-a-kind ,f)]  ;; get "top" from the straight. a the top card of the flush when there is a (straight 5) will  ;; be the ace ... putting it in the wrong place for the ordering. [(and (? flush?) (app straight? (list 'straight top _ ...))) `(straight-flush ,top)] [(or `(,@(F*2 f) ,@(F*2 f) ,_) `(,_ ,@(F*2 f) ,@(F*2 f))) `(four-of-a-kind ,f)] [(or `(,@(F*3 fh) ,@(F*2 fl)) `(,@(F*2 fh) ,@(F*3 fl))) `(full-house ,fh, fl)] [(app flush? (and rv (list 'flush _ ...))) rv] [(app straight? (and rv (list 'straight _ ...))) rv]  ;; pairs and threes may be padded to the left, middle and right with tie-breakers; the lists of  ;; which we will call l, m and r, respectively (four and 5-of-a-kind don't need tiebreaking,  ;; they're well hard!) [`(,(F._ l) ... ,@(F*3 f) ,(F._ r) ...) `(three-of-a-kind ,f ,@l ,@r)] [`(,(F._ l) ... ,@(F*2 f1) ,(F._ m) ... ,@(F*2 f2) ,(F._ r) ...) `(two-pair ,f1 ,f2 ,@l ,@m ,@r)] [`(,(F._ l) ... ,@(F*2 f) ,(F._ r) ...) `(one-pair ,f ,@l ,@r)] [`(,(F._ f) ...) `(high-card ,@f)] [hand (error 'invalid-hand hand)]))   (define (analyse-hand/string hand-string) (analyse-sorted-hand (face-sort (string->hand hand-string))))   ;;; -------------------------------------------------------------------------------------------------- ;;; Strings to cards, cards to strings -- that kind of thing ;;; -------------------------------------------------------------------------------------------------- (define suit->unicode (match-lambda ('h "♥") ('d "♦") ('c "♣") ('s "♠") (x x)))   (define unicode->suit (match-lambda ("♥" 'h) ("♦" 'd) ("♣" 'c) ("♠" 's) (x x)))   (define (face->number f) (match (string-upcase f) ["T" 10] ["J" 11] ["Q" 12] ["K" 13] ["A" 14] [(app string->number (? number? n)) n] [else 0]))   (define number->face (match-lambda (10 "T") (11 "J") (12 "Q") (13 "K") (14 "A") ((app ~s x) x)))   (define string->card (match-lambda ("joker" 'joker) ((regexp #px"^(.*)(.)$" (list _ (app face->number num) (app unicode->suit suit))) (cons num suit))))   (define (string->hand str) (map string->card (regexp-split #rx" +" (string-trim str))))   (define card->string (match-lambda ['joker "[]"] [(cons (app number->face f) (app suit->unicode s)) (format "~a~a" f s)]))   (define (hand->string h) (string-join (map card->string h) " "))   ;; used for both testing and output (define e.g.-hands (list " 2♥ 2♦ 2♣ k♣ q♦" " 2♥ 5♥ 7♦ 8♣ 9♠" " a♥ 2♦ 3♣ 4♣ 5♦" "10♥ j♦ q♣ k♣ a♦" " 2♥ 3♥ 2♦ 3♣ 3♦" " 2♥ 7♥ 2♦ 3♣ 3♦" " 2♥ 7♥ 7♦ 7♣ 7♠" "10♥ j♥ q♥ k♥ a♥" " 4♥ 4♠ k♠ 5♦ 10♠" " q♣ 10♣ 7♣ 6♣ 4♣"   " joker 2♦ 2♠ k♠ q♦" " joker 5♥ 7♦ 8♠ 9♦" " joker 2♦ 3♠ 4♠ 5♠" " joker 3♥ 2♦ 3♠ 3♦" " joker 7♥ 2♦ 3♠ 3♦" " joker 7♥ 7♦ 7♠ 7♣" " joker j♥ q♥ k♥ A♥" " joker 4♣ k♣ 5♦ 10♠" " joker k♣ 7♣ 6♣ 4♣" " joker 2♦ joker 4♠ 5♠" " joker Q♦ joker A♠ 10♠" " joker Q♦ joker A♦ 10♦" " joker 2♦ 2♠ joker q♦"))   ;;; -------------------------------------------------------------------------------------------------- ;;; Main and test modules ;;; -------------------------------------------------------------------------------------------------- (module+ main (define scored-hands (for/list ((h (map string->hand e.g.-hands))) (define-values (analysis score) (car+cdr (score-hand h))) (list h analysis score)))   (for ((a.s (sort scored-hands > #:key third))) (match-define (list (app hand->string h) a _) a.s) (printf "~a: ~a ~a" h (~a (first a) #:min-width 15) (number->face (second a))) (when (pair? (cddr a)) (printf " [tiebreak: ~a]" (string-join (map number->face (cddr a)) ", "))) (newline)))   (module+ test (require rackunit) (let ((e.g.-strght-flsh '((14 . h) (13 . h) (12 . h) (11 . h) (10 . h)))) (check-match (straight? e.g.-strght-flsh) '(straight 14)) (check-match (flush? e.g.-strght-flsh) '(flush 14 13 12 11 10)) (check-match e.g.-strght-flsh (and (? flush?) (app straight? (list 'straight top _ ...)))))   (define expected-results '((three-of-a-kind 2 13 12) (high-card 9 8 7 5 2) (straight 5) (straight 14) (full-house 3 2) (two-pair 3 2 7) (four-of-a-kind 7) (straight-flush 14) (one-pair 4 13 10 5) (flush 12 10 7 6 4) (three-of-a-kind 2 13 12) (straight 9) (straight 6) (four-of-a-kind 3) (three-of-a-kind 3 7 2) (five-of-a-kind 7) (straight-flush 14) (one-pair 13 10 5 4) (flush 14 13 7 6 4) (straight 6) (straight 14) (straight-flush 14) (four-of-a-kind 2))) (for ((h e.g.-hands) (r expected-results)) (check-equal? (analyse-hand/string h) r)))
http://rosettacode.org/wiki/Population_count
Population count
Population count You are encouraged to solve this task according to the task description, using any language you may know. The   population count   is the number of   1s   (ones)   in the binary representation of a non-negative integer. Population count   is also known as:   pop count   popcount   sideways sum   bit summation   Hamming weight For example,   5   (which is   101   in binary)   has a population count of   2. Evil numbers   are non-negative integers that have an   even   population count. Odious numbers     are  positive integers that have an    odd   population count. Task write a function (or routine) to return the population count of a non-negative integer. all computation of the lists below should start with   0   (zero indexed). display the   pop count   of the   1st   thirty powers of   3       (30,   31,   32,   33,   34,   ∙∙∙   329). display the   1st   thirty     evil     numbers. display the   1st   thirty   odious   numbers. display each list of integers on one line   (which may or may not include a title),   each set of integers being shown should be properly identified. See also The On-Line Encyclopedia of Integer Sequences:   A000120 population count. The On-Line Encyclopedia of Integer Sequences:   A000069 odious numbers. The On-Line Encyclopedia of Integer Sequences:   A001969 evil numbers.
#Forth
Forth
: popcnt ( n -- u) 0 swap BEGIN dup WHILE tuck 1 AND + swap 1 rshift REPEAT DROP ; : odious? ( n -- t|f) popcnt 1 AND ; : evil? ( n -- t|f) odious? 0= ;   CREATE A 30 , : task1 1 0 ." 3**i popcnt: " BEGIN dup A @ < WHILE over popcnt . 1+ swap 3 * swap REPEAT DROP DROP CR ; : task2 0 0 ." evil  : " BEGIN dup A @ < WHILE over evil? IF over . 1+ THEN swap 1+ swap REPEAT DROP DROP CR ; : task3 0 0 ." odious  : " BEGIN dup A @ < WHILE over odious? IF over . 1+ THEN swap 1+ swap REPEAT DROP DROP CR ; task1 task2 task3 BYE
http://rosettacode.org/wiki/Polynomial_long_division
Polynomial long division
This page uses content from Wikipedia. The original article was at Polynomial long division. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance) In algebra, polynomial long division is an algorithm for dividing a polynomial by another polynomial of the same or lower degree. Let us suppose a polynomial is represented by a vector, x {\displaystyle x} (i.e., an ordered collection of coefficients) so that the i {\displaystyle i} th element keeps the coefficient of x i {\displaystyle x^{i}} , and the multiplication by a monomial is a shift of the vector's elements "towards right" (injecting ones from left) followed by a multiplication of each element by the coefficient of the monomial. Then a pseudocode for the polynomial long division using the conventions described above could be: degree(P): return the index of the last non-zero element of P; if all elements are 0, return -∞ polynomial_long_division(N, D) returns (q, r): // N, D, q, r are vectors if degree(D) < 0 then error q ← 0 while degree(N) ≥ degree(D) d ← D shifted right by (degree(N) - degree(D)) q(degree(N) - degree(D)) ← N(degree(N)) / d(degree(d)) // by construction, degree(d) = degree(N) of course d ← d * q(degree(N) - degree(D)) N ← N - d endwhile r ← N return (q, r) Note: vector * scalar multiplies each element of the vector by the scalar; vectorA - vectorB subtracts each element of the vectorB from the element of the vectorA with "the same index". The vectors in the pseudocode are zero-based. Error handling (for allocations or for wrong inputs) is not mandatory. Conventions can be different; in particular, note that if the first coefficient in the vector is the highest power of x for the polynomial represented by the vector, then the algorithm becomes simpler. Example for clarification This example is from Wikipedia, but changed to show how the given pseudocode works. 0 1 2 3 ---------------------- N: -42 0 -12 1 degree = 3 D: -3 1 0 0 degree = 1 d(N) - d(D) = 2, so let's shift D towards right by 2: N: -42 0 -12 1 d: 0 0 -3 1 N(3)/d(3) = 1, so d is unchanged. Now remember that "shifting by 2" is like multiplying by x2, and the final multiplication (here by 1) is the coefficient of this monomial. Let's store this into q: 0 1 2 --------------- q: 0 0 1 now compute N - d, and let it be the "new" N, and let's loop N: -42 0 -9 0 degree = 2 D: -3 1 0 0 degree = 1 d(N) - d(D) = 1, right shift D by 1 and let it be d N: -42 0 -9 0 d: 0 -3 1 0 * -9/1 = -9 q: 0 -9 1 d: 0 27 -9 0 N ← N - d N: -42 -27 0 0 degree = 1 D: -3 1 0 0 degree = 1 looping again... d(N)-d(D)=0, so no shift is needed; we multiply D by -27 (= -27/1) storing the result in d, then q: -27 -9 1 and N: -42 -27 0 0 - d: 81 -27 0 0 = N: -123 0 0 0 (last N) d(N) < d(D), so now r ← N, and the result is: 0 1 2 ------------- q: -27 -9 1 → x2 - 9x - 27 r: -123 0 0 → -123 Related task   Polynomial derivative
#PicoLisp
PicoLisp
(de degree (P) (let I NIL (for (N . C) P (or (=0 C) (setq I N)) ) (dec I) ) )   (de divPoly (N D) (if (lt0 (degree D)) (quit "Div/0" D) (let (Q NIL Diff) (while (ge0 (setq Diff (- (degree N) (degree D)))) (setq Q (need (- -1 Diff) Q 0)) (let E D (do Diff (push 'E 0)) (let F (/ (get N (inc (degree N))) (get E (inc (degree E)))) (set (nth Q (inc Diff)) F) (setq N (mapcar '((N E) (- N (* E F))) N E)) ) ) ) (list Q N) ) ) )
http://rosettacode.org/wiki/Polynomial_long_division
Polynomial long division
This page uses content from Wikipedia. The original article was at Polynomial long division. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance) In algebra, polynomial long division is an algorithm for dividing a polynomial by another polynomial of the same or lower degree. Let us suppose a polynomial is represented by a vector, x {\displaystyle x} (i.e., an ordered collection of coefficients) so that the i {\displaystyle i} th element keeps the coefficient of x i {\displaystyle x^{i}} , and the multiplication by a monomial is a shift of the vector's elements "towards right" (injecting ones from left) followed by a multiplication of each element by the coefficient of the monomial. Then a pseudocode for the polynomial long division using the conventions described above could be: degree(P): return the index of the last non-zero element of P; if all elements are 0, return -∞ polynomial_long_division(N, D) returns (q, r): // N, D, q, r are vectors if degree(D) < 0 then error q ← 0 while degree(N) ≥ degree(D) d ← D shifted right by (degree(N) - degree(D)) q(degree(N) - degree(D)) ← N(degree(N)) / d(degree(d)) // by construction, degree(d) = degree(N) of course d ← d * q(degree(N) - degree(D)) N ← N - d endwhile r ← N return (q, r) Note: vector * scalar multiplies each element of the vector by the scalar; vectorA - vectorB subtracts each element of the vectorB from the element of the vectorA with "the same index". The vectors in the pseudocode are zero-based. Error handling (for allocations or for wrong inputs) is not mandatory. Conventions can be different; in particular, note that if the first coefficient in the vector is the highest power of x for the polynomial represented by the vector, then the algorithm becomes simpler. Example for clarification This example is from Wikipedia, but changed to show how the given pseudocode works. 0 1 2 3 ---------------------- N: -42 0 -12 1 degree = 3 D: -3 1 0 0 degree = 1 d(N) - d(D) = 2, so let's shift D towards right by 2: N: -42 0 -12 1 d: 0 0 -3 1 N(3)/d(3) = 1, so d is unchanged. Now remember that "shifting by 2" is like multiplying by x2, and the final multiplication (here by 1) is the coefficient of this monomial. Let's store this into q: 0 1 2 --------------- q: 0 0 1 now compute N - d, and let it be the "new" N, and let's loop N: -42 0 -9 0 degree = 2 D: -3 1 0 0 degree = 1 d(N) - d(D) = 1, right shift D by 1 and let it be d N: -42 0 -9 0 d: 0 -3 1 0 * -9/1 = -9 q: 0 -9 1 d: 0 27 -9 0 N ← N - d N: -42 -27 0 0 degree = 1 D: -3 1 0 0 degree = 1 looping again... d(N)-d(D)=0, so no shift is needed; we multiply D by -27 (= -27/1) storing the result in d, then q: -27 -9 1 and N: -42 -27 0 0 - d: 81 -27 0 0 = N: -123 0 0 0 (last N) d(N) < d(D), so now r ← N, and the result is: 0 1 2 ------------- q: -27 -9 1 → x2 - 9x - 27 r: -123 0 0 → -123 Related task   Polynomial derivative
#Python
Python
# -*- coding: utf-8 -*-   from itertools import izip   def degree(poly): while poly and poly[-1] == 0: poly.pop() # normalize return len(poly)-1   def poly_div(N, D): dD = degree(D) dN = degree(N) if dD < 0: raise ZeroDivisionError if dN >= dD: q = [0] * dN while dN >= dD: d = [0]*(dN - dD) + D mult = q[dN - dD] = N[-1] / float(d[-1]) d = [coeff*mult for coeff in d] N = [coeffN - coeffd for coeffN, coeffd in izip(N, d)] dN = degree(N) r = N else: q = [0] r = N return q, r   if __name__ == '__main__': print "POLYNOMIAL LONG DIVISION" N = [-42, 0, -12, 1] D = [-3, 1, 0, 0] print "  %s / %s =" % (N,D), print " %s remainder %s" % poly_div(N, D)
http://rosettacode.org/wiki/Polynomial_regression
Polynomial regression
Find an approximating polynomial of known degree for a given data. Example: For input data: x = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; y = {1, 6, 17, 34, 57, 86, 121, 162, 209, 262, 321}; The approximating polynomial is: 3 x2 + 2 x + 1 Here, the polynomial's coefficients are (3, 2, 1). This task is intended as a subtask for Measure relative performance of sorting algorithms implementations.
#Octave
Octave
x = [0:10]; y = [1, 6, 17, 34, 57, 86, 121, 162, 209, 262, 321]; coeffs = polyfit(x, y, 2)
http://rosettacode.org/wiki/Power_set
Power set
A   set   is a collection (container) of certain values, without any particular order, and no repeated values. It corresponds with a finite set in mathematics. A set can be implemented as an associative array (partial mapping) in which the value of each key-value pair is ignored. Given a set S, the power set (or powerset) of S, written P(S), or 2S, is the set of all subsets of S. Task By using a library or built-in set type, or by defining a set type with necessary operations, write a function with a set S as input that yields the power set 2S of S. For example, the power set of     {1,2,3,4}     is {{}, {1}, {2}, {1,2}, {3}, {1,3}, {2,3}, {1,2,3}, {4}, {1,4}, {2,4}, {1,2,4}, {3,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}. For a set which contains n elements, the corresponding power set has 2n elements, including the edge cases of empty set. The power set of the empty set is the set which contains itself (20 = 1): P {\displaystyle {\mathcal {P}}} ( ∅ {\displaystyle \varnothing } ) = { ∅ {\displaystyle \varnothing } } And the power set of the set which contains only the empty set, has two subsets, the empty set and the set which contains the empty set (21 = 2): P {\displaystyle {\mathcal {P}}} ({ ∅ {\displaystyle \varnothing } }) = { ∅ {\displaystyle \varnothing } , { ∅ {\displaystyle \varnothing } } } Extra credit: Demonstrate that your language supports these last two powersets.
#Groovy
Groovy
  def powerSetRec(head, tail) { if (!tail) return [head] powerSetRec(head, tail.tail()) + powerSetRec(head + [tail.head()], tail.tail()) }   def powerSet(set) { powerSetRec([], set as List) as Set}  
http://rosettacode.org/wiki/Primality_by_trial_division
Primality by trial division
Task Write a boolean function that tells whether a given integer is prime. Remember that   1   and all non-positive numbers are not prime. Use trial division. Even numbers greater than   2   may be eliminated right away. A loop from   3   to   √ n    will suffice,   but other loops are allowed. Related tasks   count in factors   prime decomposition   AKS test for primes   factors of an integer   Sieve of Eratosthenes   factors of a Mersenne number   trial factoring of a Mersenne number   partition an integer X into N primes   sequence of primes by Trial Division
#D
D
import std.stdio, std.algorithm, std.range, std.math;   bool isPrime1(in int n) pure nothrow { if (n == 2) return true; if (n <= 1 || (n & 1) == 0) return false;   for(int i = 3; i <= real(n).sqrt; i += 2) if (n % i == 0) return false; return true; }   void main() { iota(2, 40).filter!isPrime1.writeln; }
http://rosettacode.org/wiki/Price_fraction
Price fraction
A friend of mine runs a pharmacy.   He has a specialized function in his Dispensary application which receives a decimal value of currency and replaces it to a standard value.   This value is regulated by a government department. Task Given a floating point value between   0.00   and   1.00,   rescale according to the following table: >= 0.00 < 0.06  := 0.10 >= 0.06 < 0.11  := 0.18 >= 0.11 < 0.16  := 0.26 >= 0.16 < 0.21  := 0.32 >= 0.21 < 0.26  := 0.38 >= 0.26 < 0.31  := 0.44 >= 0.31 < 0.36  := 0.50 >= 0.36 < 0.41  := 0.54 >= 0.41 < 0.46  := 0.58 >= 0.46 < 0.51  := 0.62 >= 0.51 < 0.56  := 0.66 >= 0.56 < 0.61  := 0.70 >= 0.61 < 0.66  := 0.74 >= 0.66 < 0.71  := 0.78 >= 0.71 < 0.76  := 0.82 >= 0.76 < 0.81  := 0.86 >= 0.81 < 0.86  := 0.90 >= 0.86 < 0.91  := 0.94 >= 0.91 < 0.96  := 0.98 >= 0.96 < 1.01  := 1.00
#langur
langur
# using an implied parameter .f ... val .pricefrac = f given .f { case >= 0.00, < 0.06: 0.10 case >= 0.06, < 0.11: 0.18 case >= 0.11, < 0.16: 0.26 case >= 0.16, < 0.21: 0.32 case >= 0.21, < 0.26: 0.38 case >= 0.26, < 0.31: 0.44 case >= 0.31, < 0.36: 0.50 case >= 0.36, < 0.41: 0.54 case >= 0.41, < 0.46: 0.58 case >= 0.46, < 0.51: 0.62 case >= 0.51, < 0.56: 0.66 case >= 0.56, < 0.61: 0.70 case >= 0.61, < 0.66: 0.74 case >= 0.66, < 0.71: 0.78 case >= 0.71, < 0.76: 0.82 case >= 0.76, < 0.81: 0.86 case >= 0.81, < 0.86: 0.90 case >= 0.86, < 0.91: 0.94 case >= 0.91, < 0.96: 0.98 case >= 0.96, <= 1.00: 1.00 default: throw "bad data"   # The default operator between test cases is "and". # That is, writing "case" without a logical operator is the same as writing "case and". # To make a given case act as a switch case does in other languages, use "case or". }   writeln .pricefrac(0.17) writeln .pricefrac(0.71)
http://rosettacode.org/wiki/Proper_divisors
Proper divisors
The   proper divisors   of a positive integer N are those numbers, other than N itself, that divide N without remainder. For N > 1 they will always include 1,   but for N == 1 there are no proper divisors. Examples The proper divisors of     6     are   1, 2, and 3. The proper divisors of   100   are   1, 2, 4, 5, 10, 20, 25, and 50. Task Create a routine to generate all the proper divisors of a number. use it to show the proper divisors of the numbers 1 to 10 inclusive. Find a number in the range 1 to 20,000 with the most proper divisors. Show the number and just the count of how many proper divisors it has. Show all output here. Related tasks   Amicable pairs   Abundant, deficient and perfect number classifications   Aliquot sequence classifications   Factors of an integer   Prime decomposition
#Perl
Perl
use ntheory qw/divisors/; sub proper_divisors { my $n = shift; # Like Pari/GP, divisors(0) = (0,1) and divisors(1) = () return 1 if $n == 0; my @d = divisors($n); pop @d; # divisors are in sorted order, so last entry is the input @d; } say "$_: ", join " ", proper_divisors($_) for 1..10; # 1. For the max, we can do a traditional loop. my($max,$ind) = (0,0); for (1..20000) { my $nd = scalar proper_divisors($_); ($max,$ind) = ($nd,$_) if $nd > $max; } say "$max $ind"; # 2. Or we can use List::Util's max with decoration (this exploits its implementation) { use List::Util qw/max/; no warnings 'numeric'; say max(map { scalar(proper_divisors($_)) . " $_" } 1..20000); }
http://rosettacode.org/wiki/Probabilistic_choice
Probabilistic choice
Given a mapping between items and their required probability of occurrence, generate a million items randomly subject to the given probabilities and compare the target probability of occurrence versus the generated values. The total of all the probabilities should equal one. (Because floating point arithmetic is involved, this is subject to rounding errors). aleph 1/5.0 beth 1/6.0 gimel 1/7.0 daleth 1/8.0 he 1/9.0 waw 1/10.0 zayin 1/11.0 heth 1759/27720 # adjusted so that probabilities add to 1 Related task Random number generator (device)
#VBScript
VBScript
  item = Array("aleph","beth","gimel","daleth","he","waw","zayin","heth") prob = Array(1/5.0, 1/6.0, 1/7.0, 1/8.0, 1/9.0, 1/10.0, 1/11.0, 1759/27720) Dim cnt(7)   'Terminate script if sum of probabilities <> 1. sum = 0 For i = 0 To UBound(prob) sum = sum + prob(i) Next   If sum <> 1 Then WScript.Quit End If   For trial = 1 To 1000000 r = Rnd(1) p = 0 For i = 0 To UBound(prob) p = p + prob(i) If r < p Then cnt(i) = cnt(i) + 1 Exit For End If Next Next   WScript.StdOut.Write "item" & vbTab & "actual" & vbTab & vbTab & "theoretical" WScript.StdOut.WriteLine For i = 0 To UBound(item) WScript.StdOut.Write item(i) & vbTab & FormatNumber(cnt(i)/1000000,6) & vbTab & FormatNumber(prob(i),6) WScript.StdOut.WriteLine Next
http://rosettacode.org/wiki/Priority_queue
Priority queue
A priority queue is somewhat similar to a queue, with an important distinction: each item is added to a priority queue with a priority level, and will be later removed from the queue with the highest priority element first. That is, the items are (conceptually) stored in the queue in priority order instead of in insertion order. Task Create a priority queue.   The queue must support at least two operations:   Insertion.   An element is added to the queue with a priority (a numeric value).   Top item removal.   Deletes the element or one of the elements with the current top priority and return it. Optionally, other operations may be defined, such as peeking (find what current top priority/top element is), merging (combining two priority queues into one), etc. To test your implementation, insert a number of elements into the queue, each with some random priority. Then dequeue them sequentially; now the elements should be sorted by priority. You can use the following task/priority items as input data: Priority Task ══════════ ════════════════ 3 Clear drains 4 Feed cat 5 Make tea 1 Solve RC tasks 2 Tax return The implementation should try to be efficient.   A typical implementation has   O(log n)   insertion and extraction time,   where   n   is the number of items in the queue. You may choose to impose certain limits such as small range of allowed priority levels, limited capacity, etc.   If so, discuss the reasons behind it.
#Prolog
Prolog
priority-queue :- TL0 = [3-'Clear drains', 4-'Feed cat'],   % we can create a priority queue from a list list_to_heap(TL0, Heap0),   % alternatively we can start from an empty queue % get from empty_heap/1.   % now we add the other elements add_to_heap(Heap0, 5, 'Make tea', Heap1), add_to_heap(Heap1, 1, 'Solve RC tasks', Heap2), add_to_heap(Heap2, 2, 'Tax return', Heap3),   % we list the content of the heap: heap_to_list(Heap3, TL1), writeln('Content of the queue'), maplist(writeln, TL1), nl,   % now we retrieve the minimum-priority pair get_from_heap(Heap3, Priority, Key, Heap4), format('Retrieve top of the queue : Priority ~w, Element ~w~n', [Priority, Key]), nl,   % we list the content of the heap: heap_to_list(Heap4, TL2), writeln('Content of the queue'), maplist(writeln, TL2).  
http://rosettacode.org/wiki/Prime_decomposition
Prime decomposition
The prime decomposition of a number is defined as a list of prime numbers which when all multiplied together, are equal to that number. Example 12 = 2 × 2 × 3, so its prime decomposition is {2, 2, 3} Task Write a function which returns an array or collection which contains the prime decomposition of a given number   n {\displaystyle n}   greater than   1. If your language does not have an isPrime-like function available, you may assume that you have a function which determines whether a number is prime (note its name before your code). If you would like to test code from this task, you may use code from trial division or the Sieve of Eratosthenes. Note: The program must not be limited by the word size of your computer or some other artificial limit; it should work for any number regardless of size (ignoring the physical limits of RAM etc). Related tasks   count in factors   factors of an integer   Sieve of Eratosthenes   primality by trial division   factors of a Mersenne number   trial factoring of a Mersenne number   partition an integer X into N primes   sequence of primes by Trial Division
#Haskell
Haskell
factorize n = [ d | p <- [2..n], isPrime p, d <- divs n p ] -- [2..n] >>= (\p-> [p|isPrime p]) >>= divs n where divs n p | rem n p == 0 = p : divs (quot n p) p | otherwise = []
http://rosettacode.org/wiki/Plot_coordinate_pairs
Plot coordinate pairs
Task Plot a function represented as    x,  y    numerical arrays. Post the resulting image for the following input arrays (taken from Python's Example section on Time a function): x = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}; y = {2.7, 2.8, 31.4, 38.1, 58.0, 76.2, 100.5, 130.0, 149.3, 180.0}; This task is intended as a subtask for Measure relative performance of sorting algorithms implementations.
#Maple
Maple
x := Vector([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]): y := Vector([2.7, 2.8, 31.4, 38.1, 58.0, 76.2, 100.5, 130.0, 149.3, 180.0]): plot(x,y,style="point");
http://rosettacode.org/wiki/Plot_coordinate_pairs
Plot coordinate pairs
Task Plot a function represented as    x,  y    numerical arrays. Post the resulting image for the following input arrays (taken from Python's Example section on Time a function): x = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}; y = {2.7, 2.8, 31.4, 38.1, 58.0, 76.2, 100.5, 130.0, 149.3, 180.0}; This task is intended as a subtask for Measure relative performance of sorting algorithms implementations.
#Mathematica.2FWolfram_Language
Mathematica/Wolfram Language
x={0,1,2,3,4,5,6,7,8,9}; y={2.7,2.8,31.4,38.1,58.0,76.2,100.5,130.0,149.3,180.0}; ListPlot[{x, y} // Transpose]
http://rosettacode.org/wiki/Polymorphism
Polymorphism
Task Create two classes   Point(x,y)   and   Circle(x,y,r)   with a polymorphic function print, accessors for (x,y,r), copy constructor, assignment and destructor and every possible default constructors
#jq
jq
  def Point(x;y): {"type": "Point", "x": x, "y": y}; def Point(x): Point(x;0); def Point: Point(0);   def Circle(x;y;r): {"type": "Circle", "x": x, "y": y, "r": r}; def Circle(x;y): Circle(x;y;0); def Circle(x): Circle(x;0); def Circle: Circle(0);   def print: if .type == "Circle" then "\(.type)(\(.x); \(.y); \(.r))" elif .type == "Point" then "\(.type)(\(.x); \(.y))" else empty end;  
http://rosettacode.org/wiki/Polymorphism
Polymorphism
Task Create two classes   Point(x,y)   and   Circle(x,y,r)   with a polymorphic function print, accessors for (x,y,r), copy constructor, assignment and destructor and every possible default constructors
#Julia
Julia
mutable struct Point x::Float64 y::Float64 end   Base.show(io::IO, p::Point) = print(io, "Point($(p.x), $(p.y))")   getx(p::Point) = p.x gety(p::Point) = p.y   setx(p::Point, x) = (p.x = x) sety(p::Point, y) = (p.y = y)   mutable struct Circle x::Float64 y::Float64 r::Float64 end   getx(c::Circle) = c.x gety(c::Circle) = c.y getr(c::Circle) = c.r   setx(c::Circle, x) = (c.x = x) sety(c::Circle, y) = (c.y = y) setr(c::Circle, r) = (c.r = r)   Base.show(io::IO, c::Circle) = print(io, "Circle($(c.x), $(c.y), $(c.r))")
http://rosettacode.org/wiki/Poker_hand_analyser
Poker hand analyser
Task Create a program to parse a single five card poker hand and rank it according to this list of poker hands. A poker hand is specified as a space separated list of five playing cards. Each input card has two characters indicating face and suit. Example 2d       (two of diamonds). Faces are:    a, 2, 3, 4, 5, 6, 7, 8, 9, 10, j, q, k Suits are:    h (hearts),   d (diamonds),   c (clubs),   and   s (spades),   or alternatively,   the unicode card-suit characters:    ♥ ♦ ♣ ♠ Duplicate cards are illegal. The program should analyze a single hand and produce one of the following outputs: straight-flush four-of-a-kind full-house flush straight three-of-a-kind two-pair one-pair high-card invalid Examples 2♥ 2♦ 2♣ k♣ q♦: three-of-a-kind 2♥ 5♥ 7♦ 8♣ 9♠: high-card a♥ 2♦ 3♣ 4♣ 5♦: straight 2♥ 3♥ 2♦ 3♣ 3♦: full-house 2♥ 7♥ 2♦ 3♣ 3♦: two-pair 2♥ 7♥ 7♦ 7♣ 7♠: four-of-a-kind 10♥ j♥ q♥ k♥ a♥: straight-flush 4♥ 4♠ k♠ 5♦ 10♠: one-pair q♣ 10♣ 7♣ 6♣ q♣: invalid The programs output for the above examples should be displayed here on this page. Extra credit use the playing card characters introduced with Unicode 6.0 (U+1F0A1 - U+1F0DE). allow two jokers use the symbol   joker duplicates would be allowed (for jokers only) five-of-a-kind would then be the highest hand More extra credit examples joker 2♦ 2♠ k♠ q♦: three-of-a-kind joker 5♥ 7♦ 8♠ 9♦: straight joker 2♦ 3♠ 4♠ 5♠: straight joker 3♥ 2♦ 3♠ 3♦: four-of-a-kind joker 7♥ 2♦ 3♠ 3♦: three-of-a-kind joker 7♥ 7♦ 7♠ 7♣: five-of-a-kind joker j♥ q♥ k♥ A♥: straight-flush joker 4♣ k♣ 5♦ 10♠: one-pair joker k♣ 7♣ 6♣ 4♣: flush joker 2♦ joker 4♠ 5♠: straight joker Q♦ joker A♠ 10♠: straight joker Q♦ joker A♦ 10♦: straight-flush joker 2♦ 2♠ joker q♦: four-of-a-kind Related tasks Playing cards Card shuffles Deal cards_for_FreeCell War Card_Game Go Fish
#Raku
Raku
use v6;   grammar PokerHand {   # Raku Grammar to parse and rank 5-card poker hands # E.g. PokerHand.parse("2♥ 3♥ 2♦ 3♣ 3♦"); # 2013-12-21: handle 'joker' wildcards; maximum of two   rule TOP { :my %*PLAYED; { %*PLAYED = () } [ <face-card> | <joker> ]**5 }   token face-card {<face><suit> <?{ my $card = ~$/.lc; # disallow duplicates ++%*PLAYED{$card} <= 1; }> }   token joker {:i 'joker' <?{ my $card = ~$/.lc; # allow two jokers in a hand ++%*PLAYED{$card} <= 2; }> }   token face {:i <[2..9 jqka]> | 10 } token suit {<[♥ ♦ ♣ ♠]>} }   class PokerHand::Actions { method TOP($/) { my UInt @n = n-of-a-kind($/); my $flush = 'flush' if flush($/); my $straight = 'straight' if straight($/); make rank(@n[0], @n[1], $flush, $straight); } multi sub rank(5,*@) { 'five-of-a-kind' } multi sub rank($,$,'flush','straight') { 'straight-flush' } multi sub rank(4,*@) { 'four-of-a-kind' } multi sub rank($,$,'flush',$) { 'flush' } multi sub rank($,$,$,'straight') { 'straight' } multi sub rank(3,2,*@) { 'full-house' } multi sub rank(3,*@) { 'three-of-a-kind' } multi sub rank(2,2,*@) { 'two-pair' } multi sub rank(2,*@) { 'one-pair' } multi sub rank(*@) { 'high-card' }   sub n-of-a-kind($/) { my %faces := bag @<face-card>.map: -> $/ {~$<face>.lc}; my @counts = %faces.values.sort.reverse; @counts[0] += @<joker>; return @counts; }   sub flush($/) { my @suits = unique @<face-card>.map: -> $/ {~$<suit>}; return +@suits == 1; }   sub straight($/) { # allow both ace-low and ace-high straights constant @Faces = [ "a 2 3 4 5 6 7 8 9 10 j q k a".split: ' ' ]; constant @Possible-Straights = [ (4 ..^ @Faces).map: { set @Faces[$_-4 .. $_] } ];   my $faces = set @<face-card>.map: -> $/ {~$<face>.lc}; my $jokers = +@<joker>;   return ?( @Possible-Straights.first: { +($faces ∩ $_) + $jokers == 5 } ); } }   my PokerHand::Actions $actions .= new;   for ("2♥ 2♦ 2♣ k♣ q♦", # three-of-a-kind "2♥ 5♥ 7♦ 8♣ 9♠", # high-card "a♥ 2♦ 3♣ 4♣ 5♦", # straight "2♥ 3♥ 2♦ 3♣ 3♦", # full-house "2♥ 7♥ 2♦ 3♣ 3♦", # two-pair "2♥ 7♥ 7♦ 7♣ 7♠", # four-of-a-kind "10♥ j♥ q♥ k♥ a♥", # straight-flush "4♥ 4♠ k♠ 5♦ 10♠", # one-pair "q♣ 10♣ 7♣ 6♣ 4♣", # flush "a♥ a♥ 3♣ 4♣ 5♦", # invalid ## EXTRA CREDIT ## "joker 2♦ 2♠ k♠ q♦", # three-of-a-kind "joker 5♥ 7♦ 8♠ 9♦", # straight "joker 2♦ 3♠ 4♠ 5♠", # straight "joker 3♥ 2♦ 3♠ 3♦", # four-of-a-kind "joker 7♥ 2♦ 3♠ 3♦", # three-of-a-kind "joker 7♥ 7♦ 7♠ 7♣", # five-of-a-kind "joker j♥ q♥ k♥ A♥", # straight-flush "joker 4♣ k♣ 5♦ 10♠", # one-pair "joker k♣ 7♣ 6♣ 4♣", # flush "joker 2♦ joker 4♠ 5♠", # straight "joker Q♦ joker A♠ 10♠", # straight "joker Q♦ joker A♦ 10♦", # straight-flush "joker 2♦ 2♠ joker q♦", # four of a kind ) { my $rank = do with PokerHand.parse($_, :$actions) { .ast; } else { 'invalid'; } say "$_: $rank"; }  
http://rosettacode.org/wiki/Population_count
Population count
Population count You are encouraged to solve this task according to the task description, using any language you may know. The   population count   is the number of   1s   (ones)   in the binary representation of a non-negative integer. Population count   is also known as:   pop count   popcount   sideways sum   bit summation   Hamming weight For example,   5   (which is   101   in binary)   has a population count of   2. Evil numbers   are non-negative integers that have an   even   population count. Odious numbers     are  positive integers that have an    odd   population count. Task write a function (or routine) to return the population count of a non-negative integer. all computation of the lists below should start with   0   (zero indexed). display the   pop count   of the   1st   thirty powers of   3       (30,   31,   32,   33,   34,   ∙∙∙   329). display the   1st   thirty     evil     numbers. display the   1st   thirty   odious   numbers. display each list of integers on one line   (which may or may not include a title),   each set of integers being shown should be properly identified. See also The On-Line Encyclopedia of Integer Sequences:   A000120 population count. The On-Line Encyclopedia of Integer Sequences:   A000069 odious numbers. The On-Line Encyclopedia of Integer Sequences:   A001969 evil numbers.
#Fortran
Fortran
program population_count implicit none   integer, parameter :: i64 = selected_int_kind(18) integer(i64) :: x integer :: i, n   x = 1 write(*, "(a8)", advance = "no") "3**i :" do i = 1, 30 write(*, "(i3)", advance = "no") popcnt(x) x = x * 3 end do   write(*,*) write(*, "(a8)", advance = "no") "Evil :" n = 0 x = 0 do while(n < 30) if(mod(popcnt(x), 2) == 0) then n = n + 1 write(*, "(i3)", advance = "no") x end if x = x + 1 end do   write(*,*) write(*, "(a8)", advance = "no") "Odious :" n = 0 x = 0 do while(n < 30) if(mod(popcnt(x), 2) /= 0) then n = n + 1 write(*, "(i3)", advance = "no") x end if x = x + 1 end do   contains   integer function popcnt(x) integer(i64), intent(in) :: x integer :: i   popcnt = 0 do i = 0, 63 if(btest(x, i)) popcnt = popcnt + 1 end do   end function end program
http://rosettacode.org/wiki/Polynomial_long_division
Polynomial long division
This page uses content from Wikipedia. The original article was at Polynomial long division. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance) In algebra, polynomial long division is an algorithm for dividing a polynomial by another polynomial of the same or lower degree. Let us suppose a polynomial is represented by a vector, x {\displaystyle x} (i.e., an ordered collection of coefficients) so that the i {\displaystyle i} th element keeps the coefficient of x i {\displaystyle x^{i}} , and the multiplication by a monomial is a shift of the vector's elements "towards right" (injecting ones from left) followed by a multiplication of each element by the coefficient of the monomial. Then a pseudocode for the polynomial long division using the conventions described above could be: degree(P): return the index of the last non-zero element of P; if all elements are 0, return -∞ polynomial_long_division(N, D) returns (q, r): // N, D, q, r are vectors if degree(D) < 0 then error q ← 0 while degree(N) ≥ degree(D) d ← D shifted right by (degree(N) - degree(D)) q(degree(N) - degree(D)) ← N(degree(N)) / d(degree(d)) // by construction, degree(d) = degree(N) of course d ← d * q(degree(N) - degree(D)) N ← N - d endwhile r ← N return (q, r) Note: vector * scalar multiplies each element of the vector by the scalar; vectorA - vectorB subtracts each element of the vectorB from the element of the vectorA with "the same index". The vectors in the pseudocode are zero-based. Error handling (for allocations or for wrong inputs) is not mandatory. Conventions can be different; in particular, note that if the first coefficient in the vector is the highest power of x for the polynomial represented by the vector, then the algorithm becomes simpler. Example for clarification This example is from Wikipedia, but changed to show how the given pseudocode works. 0 1 2 3 ---------------------- N: -42 0 -12 1 degree = 3 D: -3 1 0 0 degree = 1 d(N) - d(D) = 2, so let's shift D towards right by 2: N: -42 0 -12 1 d: 0 0 -3 1 N(3)/d(3) = 1, so d is unchanged. Now remember that "shifting by 2" is like multiplying by x2, and the final multiplication (here by 1) is the coefficient of this monomial. Let's store this into q: 0 1 2 --------------- q: 0 0 1 now compute N - d, and let it be the "new" N, and let's loop N: -42 0 -9 0 degree = 2 D: -3 1 0 0 degree = 1 d(N) - d(D) = 1, right shift D by 1 and let it be d N: -42 0 -9 0 d: 0 -3 1 0 * -9/1 = -9 q: 0 -9 1 d: 0 27 -9 0 N ← N - d N: -42 -27 0 0 degree = 1 D: -3 1 0 0 degree = 1 looping again... d(N)-d(D)=0, so no shift is needed; we multiply D by -27 (= -27/1) storing the result in d, then q: -27 -9 1 and N: -42 -27 0 0 - d: 81 -27 0 0 = N: -123 0 0 0 (last N) d(N) < d(D), so now r ← N, and the result is: 0 1 2 ------------- q: -27 -9 1 → x2 - 9x - 27 r: -123 0 0 → -123 Related task   Polynomial derivative
#R
R
polylongdiv <- function(n,d) { gd <- length(d) pv <- vector("numeric", length(n)) pv[1:gd] <- d if ( length(n) >= gd ) { q <- c() while ( length(n) >= gd ) { q <- c(q, n[1]/pv[1]) n <- n - pv * (n[1]/pv[1]) n <- n[2:length(n)] pv <- pv[1:(length(pv)-1)] } list(q=q, r=n) } else { list(q=c(0), r=n) } }   # an utility function to print polynomial print.polynomial <- function(p) { i <- length(p)-1 for(a in p) { if ( i == 0 ) { cat(a, "\n") } else { cat(a, "x^", i, " + ", sep="") } i <- i - 1 } }   r <- polylongdiv(c(1,-12,0,-42), c(1,-3)) print.polynomial(r$q) print.polynomial(r$r)
http://rosettacode.org/wiki/Polynomial_regression
Polynomial regression
Find an approximating polynomial of known degree for a given data. Example: For input data: x = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; y = {1, 6, 17, 34, 57, 86, 121, 162, 209, 262, 321}; The approximating polynomial is: 3 x2 + 2 x + 1 Here, the polynomial's coefficients are (3, 2, 1). This task is intended as a subtask for Measure relative performance of sorting algorithms implementations.
#PARI.2FGP
PARI/GP
polinterpolate([0,1,2,3,4,5,6,7,8,9,10],[1,6,17,34,57,86,121,162,209,262,321])
http://rosettacode.org/wiki/Power_set
Power set
A   set   is a collection (container) of certain values, without any particular order, and no repeated values. It corresponds with a finite set in mathematics. A set can be implemented as an associative array (partial mapping) in which the value of each key-value pair is ignored. Given a set S, the power set (or powerset) of S, written P(S), or 2S, is the set of all subsets of S. Task By using a library or built-in set type, or by defining a set type with necessary operations, write a function with a set S as input that yields the power set 2S of S. For example, the power set of     {1,2,3,4}     is {{}, {1}, {2}, {1,2}, {3}, {1,3}, {2,3}, {1,2,3}, {4}, {1,4}, {2,4}, {1,2,4}, {3,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}. For a set which contains n elements, the corresponding power set has 2n elements, including the edge cases of empty set. The power set of the empty set is the set which contains itself (20 = 1): P {\displaystyle {\mathcal {P}}} ( ∅ {\displaystyle \varnothing } ) = { ∅ {\displaystyle \varnothing } } And the power set of the set which contains only the empty set, has two subsets, the empty set and the set which contains the empty set (21 = 2): P {\displaystyle {\mathcal {P}}} ({ ∅ {\displaystyle \varnothing } }) = { ∅ {\displaystyle \varnothing } , { ∅ {\displaystyle \varnothing } } } Extra credit: Demonstrate that your language supports these last two powersets.
#Haskell
Haskell
import Data.Set import Control.Monad   powerset :: Ord a => Set a -> Set (Set a) powerset = fromList . fmap fromList . listPowerset . toList   listPowerset :: [a] -> [[a]] listPowerset = filterM (const [True, False])
http://rosettacode.org/wiki/Primality_by_trial_division
Primality by trial division
Task Write a boolean function that tells whether a given integer is prime. Remember that   1   and all non-positive numbers are not prime. Use trial division. Even numbers greater than   2   may be eliminated right away. A loop from   3   to   √ n    will suffice,   but other loops are allowed. Related tasks   count in factors   prime decomposition   AKS test for primes   factors of an integer   Sieve of Eratosthenes   factors of a Mersenne number   trial factoring of a Mersenne number   partition an integer X into N primes   sequence of primes by Trial Division
#Delphi
Delphi
function IsPrime(aNumber: Integer): Boolean; var I: Integer; begin Result:= True; if(aNumber = 2) then Exit;   Result:= not ((aNumber mod 2 = 0) or (aNumber <= 1)); if not Result then Exit;   for I:=3 to Trunc(Sqrt(aNumber)) do if(aNumber mod I = 0) then begin Result:= False; Break; end; end;
http://rosettacode.org/wiki/Price_fraction
Price fraction
A friend of mine runs a pharmacy.   He has a specialized function in his Dispensary application which receives a decimal value of currency and replaces it to a standard value.   This value is regulated by a government department. Task Given a floating point value between   0.00   and   1.00,   rescale according to the following table: >= 0.00 < 0.06  := 0.10 >= 0.06 < 0.11  := 0.18 >= 0.11 < 0.16  := 0.26 >= 0.16 < 0.21  := 0.32 >= 0.21 < 0.26  := 0.38 >= 0.26 < 0.31  := 0.44 >= 0.31 < 0.36  := 0.50 >= 0.36 < 0.41  := 0.54 >= 0.41 < 0.46  := 0.58 >= 0.46 < 0.51  := 0.62 >= 0.51 < 0.56  := 0.66 >= 0.56 < 0.61  := 0.70 >= 0.61 < 0.66  := 0.74 >= 0.66 < 0.71  := 0.78 >= 0.71 < 0.76  := 0.82 >= 0.76 < 0.81  := 0.86 >= 0.81 < 0.86  := 0.90 >= 0.86 < 0.91  := 0.94 >= 0.91 < 0.96  := 0.98 >= 0.96 < 1.01  := 1.00
#Liberty_BASIC
Liberty BASIC
  dim DR(38) 'decimal range dim PF(38) 'corresponding price fraction range$="0.06 0.11 0.16 0.21 0.26 0.31 0.36 0.41 0.46 0.51 0.56 0.61 0.66 0.71 0.76 0.81 0.86 0.91 0.96 0.01" frac$="0.10 0.18 0.26 0.32 0.38 0.44 0.50 0.54 0.58 0.62 0.66 0.70 0.74 0.78 0.82 0.86 0.90 0.94 0.98 1.00" for i = 1 to 38 DR(i)=val(word$(range$,i)) PF(i)=val(word$(frac$,i)) next   for i = 0 to .99 step 0.03 print i;" -> ";PriceFraction(i) next end   Function PriceFraction(n) PriceFraction=n 'return original if outside test bounds for i = 1 to 38 if n<=DR(i) then PriceFraction=PF(i) exit for end if next end function  
http://rosettacode.org/wiki/Proper_divisors
Proper divisors
The   proper divisors   of a positive integer N are those numbers, other than N itself, that divide N without remainder. For N > 1 they will always include 1,   but for N == 1 there are no proper divisors. Examples The proper divisors of     6     are   1, 2, and 3. The proper divisors of   100   are   1, 2, 4, 5, 10, 20, 25, and 50. Task Create a routine to generate all the proper divisors of a number. use it to show the proper divisors of the numbers 1 to 10 inclusive. Find a number in the range 1 to 20,000 with the most proper divisors. Show the number and just the count of how many proper divisors it has. Show all output here. Related tasks   Amicable pairs   Abundant, deficient and perfect number classifications   Aliquot sequence classifications   Factors of an integer   Prime decomposition
#Phix
Phix
global function factors(atom n, integer include1=0) -- -- returns a list of all integer factors of n -- if include1 is 0 (the default), result does not contain either 1 or n -- if include1 is 1 the result contains 1 and n -- if include1 is -1 the result contains 1 but not n -- if n=0 then return {} end if check_limits(n,"factors") sequence lfactors = {}, hfactors = {} atom hfactor integer p = 2, lim = floor(sqrt(n)) if include1!=0 then lfactors = {1} if n!=1 and include1=1 then hfactors = {n} end if end if while p<=lim do if remainder(n,p)=0 then lfactors = append(lfactors,p) hfactor = n/p if hfactor=p then exit end if hfactors = prepend(hfactors,hfactor) end if p += 1 end while return lfactors & hfactors end function
http://rosettacode.org/wiki/Probabilistic_choice
Probabilistic choice
Given a mapping between items and their required probability of occurrence, generate a million items randomly subject to the given probabilities and compare the target probability of occurrence versus the generated values. The total of all the probabilities should equal one. (Because floating point arithmetic is involved, this is subject to rounding errors). aleph 1/5.0 beth 1/6.0 gimel 1/7.0 daleth 1/8.0 he 1/9.0 waw 1/10.0 zayin 1/11.0 heth 1759/27720 # adjusted so that probabilities add to 1 Related task Random number generator (device)
#Wren
Wren
import "random" for Random import "/fmt" for Fmt   var letters = ["aleph", "beth", "gimel", "daleth", "he", "waw", "zayin", "heth"] var actual = [0] * 8 var probs = [1/5, 1/6, 1/7, 1/8, 1/9, 1/10, 1/11, 0] var cumProbs = [0] * 8   cumProbs[0] = probs[0] for (i in 1..6) cumProbs[i] = cumProbs[i-1] + probs[i] cumProbs[7] = 1 probs[7] = 1 - cumProbs[6] var n = 1e6 var rand = Random.new() (1..n).each { |i| var r = rand.float() var index = (r <= cumProbs[0]) ? 0 : (r <= cumProbs[1]) ? 1 : (r <= cumProbs[2]) ? 2 : (r <= cumProbs[3]) ? 3 : (r <= cumProbs[4]) ? 4 : (r <= cumProbs[5]) ? 5 : (r <= cumProbs[6]) ? 6 : 7 actual[index] = actual[index] + 1 }   var sumActual = 0 System.print("Letter\t Actual Expected") System.print("------\t-------- --------") for (i in 0..7) { var generated = actual[i]/n Fmt.print("$s\t$8.6f $8.6f", letters[i], generated, probs[i]) sumActual = sumActual + generated } System.print("\t-------- --------") Fmt.print("\t$8.6f 1.000000", sumActual)
http://rosettacode.org/wiki/Priority_queue
Priority queue
A priority queue is somewhat similar to a queue, with an important distinction: each item is added to a priority queue with a priority level, and will be later removed from the queue with the highest priority element first. That is, the items are (conceptually) stored in the queue in priority order instead of in insertion order. Task Create a priority queue.   The queue must support at least two operations:   Insertion.   An element is added to the queue with a priority (a numeric value).   Top item removal.   Deletes the element or one of the elements with the current top priority and return it. Optionally, other operations may be defined, such as peeking (find what current top priority/top element is), merging (combining two priority queues into one), etc. To test your implementation, insert a number of elements into the queue, each with some random priority. Then dequeue them sequentially; now the elements should be sorted by priority. You can use the following task/priority items as input data: Priority Task ══════════ ════════════════ 3 Clear drains 4 Feed cat 5 Make tea 1 Solve RC tasks 2 Tax return The implementation should try to be efficient.   A typical implementation has   O(log n)   insertion and extraction time,   where   n   is the number of items in the queue. You may choose to impose certain limits such as small range of allowed priority levels, limited capacity, etc.   If so, discuss the reasons behind it.
#PureBasic
PureBasic
Structure taskList List description.s() ;implements FIFO queue EndStructure   Structure task *tl.tList ;pointer to a list of task descriptions Priority.i ;tasks priority, lower value has more priority EndStructure   Structure priorityQueue maxHeapSize.i ;increases as needed heapItemCount.i ;number of elements currently in heap Array heap.task(0) ;elements hold FIFO queues ordered by priorities, lowest first map heapMap.taskList() ;holds lists of tasks with the same priority that are FIFO queues EndStructure   Procedure insertPQ(*PQ.priorityQueue, description.s, p) If FindMapElement(*PQ\heapMap(), Str(p)) LastElement(*PQ\heapMap()\description()) AddElement(*PQ\heapMap()\description()) *PQ\heapMap()\description() = description Else Protected *tl.taskList = AddMapElement(*PQ\heapMap(), Str(p)) AddElement(*tl\description()) *tl\description() = description   Protected pos = *PQ\heapItemCount   *PQ\heapItemCount + 1 If *PQ\heapItemCount > *PQ\maxHeapSize Select *PQ\maxHeapSize Case 0 *PQ\maxHeapSize = 128 Default *PQ\maxHeapSize * 2 EndSelect Redim *PQ\heap.task(*PQ\maxHeapSize) EndIf   While pos > 0 And p < *PQ\heap((pos - 1) / 2)\Priority *PQ\heap(pos) = *PQ\heap((pos - 1) / 2) pos = (pos - 1) / 2 Wend   *PQ\heap(pos)\tl = *tl *PQ\heap(pos)\Priority = p EndIf EndProcedure   Procedure.s removePQ(*PQ.priorityQueue) Protected *tl.taskList = *PQ\heap(0)\tl, description.s FirstElement(*tl\description()) description = *tl\description() If ListSize(*tl\description()) > 1 DeleteElement(*tl\description()) Else DeleteMapElement(*PQ\heapMap(), Str(*PQ\heap(0)\Priority))   *PQ\heapItemCount - 1 *PQ\heap(0) = *PQ\heap(*PQ\heapItemCount)   Protected pos Repeat Protected child1 = 2 * pos + 1 Protected child2 = 2 * pos + 2 If child1 >= *PQ\heapItemCount Break EndIf   Protected smallestChild If child2 >= *PQ\heapItemCount smallestChild = child1 ElseIf *PQ\heap(child1)\Priority <= *PQ\heap(child2)\Priority smallestChild = child1 Else smallestChild = child2 EndIf   If (*PQ\heap(smallestChild)\Priority >= *PQ\heap(pos)\Priority) Break EndIf Swap *PQ\heap(pos)\tl, *PQ\heap(smallestChild)\tl Swap *PQ\heap(pos)\Priority, *PQ\heap(smallestChild)\Priority pos = smallestChild ForEver EndIf   ProcedureReturn description EndProcedure   Procedure isEmptyPQ(*PQ.priorityQueue) ;returns 1 if empty, otherwise returns 0 If *PQ\heapItemCount ProcedureReturn 0 EndIf ProcedureReturn 1 EndProcedure   If OpenConsole() Define PQ.priorityQueue insertPQ(PQ, "Clear drains", 3) insertPQ(PQ, "Answer Phone 1", 8) insertPQ(PQ, "Feed cat", 4) insertPQ(PQ, "Answer Phone 2", 8) insertPQ(PQ, "Make tea", 5) insertPQ(PQ, "Sleep", 9) insertPQ(PQ, "Check email", 3) insertPQ(PQ, "Solve RC tasks", 1) insertPQ(PQ, "Answer Phone 3", 8) insertPQ(PQ, "Exercise", 9) insertPQ(PQ, "Answer Phone 4", 8) insertPQ(PQ, "Tax return", 2)   While Not isEmptyPQ(PQ) PrintN(removePQ(PQ)) Wend   Print(#CRLF$ + #CRLF$ + "Press ENTER to exit"): Input() CloseConsole() EndIf
http://rosettacode.org/wiki/Prime_decomposition
Prime decomposition
The prime decomposition of a number is defined as a list of prime numbers which when all multiplied together, are equal to that number. Example 12 = 2 × 2 × 3, so its prime decomposition is {2, 2, 3} Task Write a function which returns an array or collection which contains the prime decomposition of a given number   n {\displaystyle n}   greater than   1. If your language does not have an isPrime-like function available, you may assume that you have a function which determines whether a number is prime (note its name before your code). If you would like to test code from this task, you may use code from trial division or the Sieve of Eratosthenes. Note: The program must not be limited by the word size of your computer or some other artificial limit; it should work for any number regardless of size (ignoring the physical limits of RAM etc). Related tasks   count in factors   factors of an integer   Sieve of Eratosthenes   primality by trial division   factors of a Mersenne number   trial factoring of a Mersenne number   partition an integer X into N primes   sequence of primes by Trial Division
#Icon_and_Unicon
Icon and Unicon
procedure main() factors := primedecomp(2^43-1) # a big int end   procedure primedecomp(n) #: return a list of factors local F,o,x F := []   every writes(o,n|(x := genfactors(n))) do { \o := "*" /o := "=" put(F,x) # build a list of factors to satisfy the task } write() return F end   link factors
http://rosettacode.org/wiki/Plot_coordinate_pairs
Plot coordinate pairs
Task Plot a function represented as    x,  y    numerical arrays. Post the resulting image for the following input arrays (taken from Python's Example section on Time a function): x = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}; y = {2.7, 2.8, 31.4, 38.1, 58.0, 76.2, 100.5, 130.0, 149.3, 180.0}; This task is intended as a subtask for Measure relative performance of sorting algorithms implementations.
#MATLAB
MATLAB
>> x = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]; >> y = [2.7, 2.8, 31.4, 38.1, 58.0, 76.2, 100.5, 130.0, 149.3, 180.0]; >> plot(x,y,'.-')
http://rosettacode.org/wiki/Plot_coordinate_pairs
Plot coordinate pairs
Task Plot a function represented as    x,  y    numerical arrays. Post the resulting image for the following input arrays (taken from Python's Example section on Time a function): x = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}; y = {2.7, 2.8, 31.4, 38.1, 58.0, 76.2, 100.5, 130.0, 149.3, 180.0}; This task is intended as a subtask for Measure relative performance of sorting algorithms implementations.
#Maxima
Maxima
(%i1) ".." (m, n) := makelist (i, i, m, n); infix ("..")$ (%i2) x: 0 .. 9$ y:[2.7, 2.8, 31.4, 38.1, 58.0, 76.2, 100.5, 130.0, 149.3, 180.0]$ (%i3) plot2d(['discrete, x, y], [style, [points,5,1,1]], [gnuplot_term, png], [gnuplot_out_file, "qsort-range-10-9.png"])$
http://rosettacode.org/wiki/Polymorphism
Polymorphism
Task Create two classes   Point(x,y)   and   Circle(x,y,r)   with a polymorphic function print, accessors for (x,y,r), copy constructor, assignment and destructor and every possible default constructors
#Kotlin
Kotlin
// version 1.1.2   open class Point(var x: Int, var y: Int) { constructor(): this(0, 0)   constructor(x: Int) : this(x, 0)   constructor(p: Point) : this(p.x, p.y)   open protected fun finalize() = println("Finalizing $this...")   override fun toString() = "Point at ($x, $y)"   open fun print() = println(this) }   class Circle(x: Int, y: Int, var r: Int) : Point(x, y) { constructor(): this(0, 0, 0)   constructor(x: Int) : this(x, 0, 0)   constructor(x: Int, r: Int) : this(x, 0, r)   constructor(c: Circle) : this(c.x, c.y, c.r)   // for simplicity not calling super.finalize() below though this would normally be done in practice override protected fun finalize() = println("Finalizing $this...")   override fun toString() = "Circle at center ($x, $y), radius $r"   override fun print() = println(this) }   fun createObjects() { val points = listOf(Point(), Point(1), Point(2, 3), Point(Point(3, 4))) for (point in points) point.print() val circles = listOf(Circle(), Circle(1), Circle(2, 3), Circle(4, 5, 6), Circle(Circle(7, 8, 9))) for (circle in circles) circle.print() println() }   fun main(args: Array<String>) { createObjects() System.gc() // try and force garbage collection Thread.sleep(2000) // allow time for finalizers to run println() val p = Point(5, 6) p.print() p.y = 7 // change y coordinate p.print() val c = Circle(5, 6, 7) c.print() c.r = 8 c.print() // change radius /* note that finalizers for p and c are not called */ }
http://rosettacode.org/wiki/Poker_hand_analyser
Poker hand analyser
Task Create a program to parse a single five card poker hand and rank it according to this list of poker hands. A poker hand is specified as a space separated list of five playing cards. Each input card has two characters indicating face and suit. Example 2d       (two of diamonds). Faces are:    a, 2, 3, 4, 5, 6, 7, 8, 9, 10, j, q, k Suits are:    h (hearts),   d (diamonds),   c (clubs),   and   s (spades),   or alternatively,   the unicode card-suit characters:    ♥ ♦ ♣ ♠ Duplicate cards are illegal. The program should analyze a single hand and produce one of the following outputs: straight-flush four-of-a-kind full-house flush straight three-of-a-kind two-pair one-pair high-card invalid Examples 2♥ 2♦ 2♣ k♣ q♦: three-of-a-kind 2♥ 5♥ 7♦ 8♣ 9♠: high-card a♥ 2♦ 3♣ 4♣ 5♦: straight 2♥ 3♥ 2♦ 3♣ 3♦: full-house 2♥ 7♥ 2♦ 3♣ 3♦: two-pair 2♥ 7♥ 7♦ 7♣ 7♠: four-of-a-kind 10♥ j♥ q♥ k♥ a♥: straight-flush 4♥ 4♠ k♠ 5♦ 10♠: one-pair q♣ 10♣ 7♣ 6♣ q♣: invalid The programs output for the above examples should be displayed here on this page. Extra credit use the playing card characters introduced with Unicode 6.0 (U+1F0A1 - U+1F0DE). allow two jokers use the symbol   joker duplicates would be allowed (for jokers only) five-of-a-kind would then be the highest hand More extra credit examples joker 2♦ 2♠ k♠ q♦: three-of-a-kind joker 5♥ 7♦ 8♠ 9♦: straight joker 2♦ 3♠ 4♠ 5♠: straight joker 3♥ 2♦ 3♠ 3♦: four-of-a-kind joker 7♥ 2♦ 3♠ 3♦: three-of-a-kind joker 7♥ 7♦ 7♠ 7♣: five-of-a-kind joker j♥ q♥ k♥ A♥: straight-flush joker 4♣ k♣ 5♦ 10♠: one-pair joker k♣ 7♣ 6♣ 4♣: flush joker 2♦ joker 4♠ 5♠: straight joker Q♦ joker A♠ 10♠: straight joker Q♦ joker A♦ 10♦: straight-flush joker 2♦ 2♠ joker q♦: four-of-a-kind Related tasks Playing cards Card shuffles Deal cards_for_FreeCell War Card_Game Go Fish
#REXX
REXX
/* REXX --------------------------------------------------------------- * 10.12.2013 Walter Pachl *--------------------------------------------------------------------*/   d.1='2h 2d 2s ks qd'; x.1='three-of-a-kind' d.2='2h 5h 7d 8s 9d'; x.2='high-card' d.3='ah 2d 3s 4s 5s'; x.3='straight' d.4='2h 3h 2d 3s 3d'; x.4='full-house' d.5='2h 7h 2d 3s 3d'; x.5='two-pair' d.6='2h 7h 7d 7s 7c'; x.6='four-of-a-kind' d.7='th jh qh kh ah'; x.7='straight-flush' d.8='4h 4c kc 5d tc'; x.8='one-pair' d.9='qc tc 7c 6c 4c'; x.9='flush' d.10='ah 2h 3h 4h' d.11='ah 2h 3h 4h 5h 6h' d.12='2h 2h 3h 4h 5h' d.13='xh 2h 3h 4h 5h' d.14='2x 2h 3h 4h 5h' Do ci=1 To 14 Call poker d.ci,x.ci end Exit   poker: Parse Arg deck,expected have.=0 f.=0; fmax=0 s.=0; smax=0 cnt.=0 If words(deck)<5 Then Return err('less than 5 cards') If words(deck)>5 Then Return err('more than 5 cards') Do i=1 To 5 c=word(deck,i) Parse Var c f +1 s If have.f.s=1 Then Return err('duplicate card:' c) have.f.s=1 m=pos(f,'a23456789tjqk') If m=0 Then Return err('invalid face' f 'in' c) cnt.m=cnt.m+1 n=pos(s,'hdcs') If n=0 Then Return err('invalid suit' s 'in' c) f.m=f.m+1; fmax=max(fmax,f.m) s.n=s.n+1; smax=max(smax,s.n) End cntl='' cnt.14=cnt.1 Do i=1 To 14 cntl=cntl||cnt.i End Select When fmax=4 Then res='four-of-a-kind' When fmax=3 Then Do If x_pair() Then res='full-house' Else res='three-of-a-kind' End When fmax=2 Then Do If x_2pair() Then res='two-pair' Else res='one-pair' End When smax=5 Then Do If x_street() Then res='straight-flush' Else res='flush' End When x_street() Then res='straight' Otherwise res='high-card' End Say deck res If res<>expected Then Say copies(' ',14) expected Return   x_pair: Do p=1 To 13 If f.p=2 Then return 1 End Return 0   x_2pair: pp=0 Do p=1 To 13 If f.p=2 Then pp=pp+1 End Return pp=2   x_street: Return pos('11111',cntl)>0   err: Say deck 'Error:' arg(1) Return 0
http://rosettacode.org/wiki/Population_count
Population count
Population count You are encouraged to solve this task according to the task description, using any language you may know. The   population count   is the number of   1s   (ones)   in the binary representation of a non-negative integer. Population count   is also known as:   pop count   popcount   sideways sum   bit summation   Hamming weight For example,   5   (which is   101   in binary)   has a population count of   2. Evil numbers   are non-negative integers that have an   even   population count. Odious numbers     are  positive integers that have an    odd   population count. Task write a function (or routine) to return the population count of a non-negative integer. all computation of the lists below should start with   0   (zero indexed). display the   pop count   of the   1st   thirty powers of   3       (30,   31,   32,   33,   34,   ∙∙∙   329). display the   1st   thirty     evil     numbers. display the   1st   thirty   odious   numbers. display each list of integers on one line   (which may or may not include a title),   each set of integers being shown should be properly identified. See also The On-Line Encyclopedia of Integer Sequences:   A000120 population count. The On-Line Encyclopedia of Integer Sequences:   A000069 odious numbers. The On-Line Encyclopedia of Integer Sequences:   A001969 evil numbers.
#Free_Pascal
Free Pascal
program populationCount(input, output, stdErr); var // general advice: iterator variables are _signed_ iterator: int64; // the variable we’d like to count the set bits in number: qWord; // how many evil numbers we’ve already found evilCount: int64; // odious numbers odiousNumber: array[1..30] of qWord; odiousIterator: int64; begin // population count for powers of three for iterator := 0 to 29 do begin number := round(exp(ln(3) * iterator)); write(popCnt(number):3); end; writeLn();   // evil numbers // (while preserving calculated odious numbers for next sub-task) evilCount := 0; odiousIterator := low(odiousNumber);   // for-loop: because we (pretend to) don’t know, // when and where we’ve found the first 30 numbers of each for iterator := 0 to high(iterator) do begin // implicit typecast: popCnt only accepts _un_-signed integers number := iterator; if odd(popCnt(number)) then begin if odiousIterator <= high(odiousNumber) then begin odiousNumber[odiousIterator] := number; inc(odiousIterator); end; end else begin if evilCount < 30 then begin write(number:20); inc(evilCount); end; end;   if evilCount + odiousIterator > 60 then begin break; end; end; writeLn();   // odious numbers for number in odiousNumber do begin write(number:20); end; writeLn(); end.
http://rosettacode.org/wiki/Polynomial_long_division
Polynomial long division
This page uses content from Wikipedia. The original article was at Polynomial long division. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance) In algebra, polynomial long division is an algorithm for dividing a polynomial by another polynomial of the same or lower degree. Let us suppose a polynomial is represented by a vector, x {\displaystyle x} (i.e., an ordered collection of coefficients) so that the i {\displaystyle i} th element keeps the coefficient of x i {\displaystyle x^{i}} , and the multiplication by a monomial is a shift of the vector's elements "towards right" (injecting ones from left) followed by a multiplication of each element by the coefficient of the monomial. Then a pseudocode for the polynomial long division using the conventions described above could be: degree(P): return the index of the last non-zero element of P; if all elements are 0, return -∞ polynomial_long_division(N, D) returns (q, r): // N, D, q, r are vectors if degree(D) < 0 then error q ← 0 while degree(N) ≥ degree(D) d ← D shifted right by (degree(N) - degree(D)) q(degree(N) - degree(D)) ← N(degree(N)) / d(degree(d)) // by construction, degree(d) = degree(N) of course d ← d * q(degree(N) - degree(D)) N ← N - d endwhile r ← N return (q, r) Note: vector * scalar multiplies each element of the vector by the scalar; vectorA - vectorB subtracts each element of the vectorB from the element of the vectorA with "the same index". The vectors in the pseudocode are zero-based. Error handling (for allocations or for wrong inputs) is not mandatory. Conventions can be different; in particular, note that if the first coefficient in the vector is the highest power of x for the polynomial represented by the vector, then the algorithm becomes simpler. Example for clarification This example is from Wikipedia, but changed to show how the given pseudocode works. 0 1 2 3 ---------------------- N: -42 0 -12 1 degree = 3 D: -3 1 0 0 degree = 1 d(N) - d(D) = 2, so let's shift D towards right by 2: N: -42 0 -12 1 d: 0 0 -3 1 N(3)/d(3) = 1, so d is unchanged. Now remember that "shifting by 2" is like multiplying by x2, and the final multiplication (here by 1) is the coefficient of this monomial. Let's store this into q: 0 1 2 --------------- q: 0 0 1 now compute N - d, and let it be the "new" N, and let's loop N: -42 0 -9 0 degree = 2 D: -3 1 0 0 degree = 1 d(N) - d(D) = 1, right shift D by 1 and let it be d N: -42 0 -9 0 d: 0 -3 1 0 * -9/1 = -9 q: 0 -9 1 d: 0 27 -9 0 N ← N - d N: -42 -27 0 0 degree = 1 D: -3 1 0 0 degree = 1 looping again... d(N)-d(D)=0, so no shift is needed; we multiply D by -27 (= -27/1) storing the result in d, then q: -27 -9 1 and N: -42 -27 0 0 - d: 81 -27 0 0 = N: -123 0 0 0 (last N) d(N) < d(D), so now r ← N, and the result is: 0 1 2 ------------- q: -27 -9 1 → x2 - 9x - 27 r: -123 0 0 → -123 Related task   Polynomial derivative
#Racket
Racket
  #lang racket (define (deg p) (for/fold ([d -inf.0]) ([(pi i) (in-indexed p)]) (if (zero? pi) d i))) (define (lead p) (vector-ref p (deg p))) (define (mono c d) (build-vector (+ d 1) (λ(i) (if (= i d) c 0)))) (define (poly*cx^n c n p) (vector-append (make-vector n 0) (for/vector ([pi p]) (* c pi)))) (define (poly+ p q) (poly/lin 1 p 1 q)) (define (poly- p q) (poly/lin 1 p -1 q)) (define (poly/lin a p b q) (cond [(< (deg p) 0) q] [(< (deg q) 0) p] [(< (deg p) (deg q)) (poly/lin b q a p)] [else (define ap+bq (for/vector #:length (+ (deg p) 1) #:fill 0 ([pi p] [qi q]) (+ (* a pi) (* b qi)))) (for ([i (in-range (+ (deg q) 1) (+ (deg p) 1))]) (vector-set! ap+bq i (* a (vector-ref p i)))) ap+bq]))   (define (poly/ n d) (define N (deg n)) (define D (deg d)) (cond [(< N 0) (error 'poly/ "can't divide by zero")] [(< N D) (values 0 n)] [else (define c (/ (lead n) (lead d))) (define q (mono c (- N D))) (define r (poly- n (poly*cx^n c (- N D) d))) (define-values (q1 r1) (poly/ r d)) (values (poly+ q q1) r1)])) ; Example: (poly/ #(-42 0 -12 1) #(-3 1)) ; Output: '#(-27 -9 1) '#(-123 0)  
http://rosettacode.org/wiki/Polynomial_regression
Polynomial regression
Find an approximating polynomial of known degree for a given data. Example: For input data: x = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; y = {1, 6, 17, 34, 57, 86, 121, 162, 209, 262, 321}; The approximating polynomial is: 3 x2 + 2 x + 1 Here, the polynomial's coefficients are (3, 2, 1). This task is intended as a subtask for Measure relative performance of sorting algorithms implementations.
#Perl
Perl
use strict; use warnings; use Statistics::Regression;   my @x = <0 1 2 3 4 5 6 7 8 9 10>; my @y = <1 6 17 34 57 86 121 162 209 262 321>;   my @model = ('const', 'X', 'X**2');   my $reg = Statistics::Regression->new( '', [@model] ); $reg->include( $y[$_], [ 1.0, $x[$_], $x[$_]**2 ]) for 0..@y-1; my @coeff = $reg->theta();   printf "%-6s %8.3f\n", $model[$_], $coeff[$_] for 0..@model-1;
http://rosettacode.org/wiki/Power_set
Power set
A   set   is a collection (container) of certain values, without any particular order, and no repeated values. It corresponds with a finite set in mathematics. A set can be implemented as an associative array (partial mapping) in which the value of each key-value pair is ignored. Given a set S, the power set (or powerset) of S, written P(S), or 2S, is the set of all subsets of S. Task By using a library or built-in set type, or by defining a set type with necessary operations, write a function with a set S as input that yields the power set 2S of S. For example, the power set of     {1,2,3,4}     is {{}, {1}, {2}, {1,2}, {3}, {1,3}, {2,3}, {1,2,3}, {4}, {1,4}, {2,4}, {1,2,4}, {3,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}. For a set which contains n elements, the corresponding power set has 2n elements, including the edge cases of empty set. The power set of the empty set is the set which contains itself (20 = 1): P {\displaystyle {\mathcal {P}}} ( ∅ {\displaystyle \varnothing } ) = { ∅ {\displaystyle \varnothing } } And the power set of the set which contains only the empty set, has two subsets, the empty set and the set which contains the empty set (21 = 2): P {\displaystyle {\mathcal {P}}} ({ ∅ {\displaystyle \varnothing } }) = { ∅ {\displaystyle \varnothing } , { ∅ {\displaystyle \varnothing } } } Extra credit: Demonstrate that your language supports these last two powersets.
#Icon_and_Unicon
Icon and Unicon
  procedure power_set (s) result := set () if *s = 0 then insert (result, set ()) # empty set else { head := set(?s) # take a random element # and find powerset of remaining part of set tail_pset := power_set (x -- head) result ++:= tail_pset # add powerset of remainder to results every ps := !tail_pset do # and add head to each powerset from the remainder insert (result, ps ++ head) } return result end  
http://rosettacode.org/wiki/Primality_by_trial_division
Primality by trial division
Task Write a boolean function that tells whether a given integer is prime. Remember that   1   and all non-positive numbers are not prime. Use trial division. Even numbers greater than   2   may be eliminated right away. A loop from   3   to   √ n    will suffice,   but other loops are allowed. Related tasks   count in factors   prime decomposition   AKS test for primes   factors of an integer   Sieve of Eratosthenes   factors of a Mersenne number   trial factoring of a Mersenne number   partition an integer X into N primes   sequence of primes by Trial Division
#E
E
def isPrime(n :int) { if (n == 2) { return true } else if (n <= 1 || n %% 2 == 0) { return false } else { def limit := (n :float64).sqrt().ceil() var divisor := 1 while ((divisor += 2) <= limit) { if (n %% divisor == 0) { return false } } return true } }
http://rosettacode.org/wiki/Price_fraction
Price fraction
A friend of mine runs a pharmacy.   He has a specialized function in his Dispensary application which receives a decimal value of currency and replaces it to a standard value.   This value is regulated by a government department. Task Given a floating point value between   0.00   and   1.00,   rescale according to the following table: >= 0.00 < 0.06  := 0.10 >= 0.06 < 0.11  := 0.18 >= 0.11 < 0.16  := 0.26 >= 0.16 < 0.21  := 0.32 >= 0.21 < 0.26  := 0.38 >= 0.26 < 0.31  := 0.44 >= 0.31 < 0.36  := 0.50 >= 0.36 < 0.41  := 0.54 >= 0.41 < 0.46  := 0.58 >= 0.46 < 0.51  := 0.62 >= 0.51 < 0.56  := 0.66 >= 0.56 < 0.61  := 0.70 >= 0.61 < 0.66  := 0.74 >= 0.66 < 0.71  := 0.78 >= 0.71 < 0.76  := 0.82 >= 0.76 < 0.81  := 0.86 >= 0.81 < 0.86  := 0.90 >= 0.86 < 0.91  := 0.94 >= 0.91 < 0.96  := 0.98 >= 0.96 < 1.01  := 1.00
#Lua
Lua
scaleTable = { {0.06, 0.10}, {0.11, 0.18}, {0.16, 0.26}, {0.21, 0.32}, {0.26, 0.38}, {0.31, 0.44}, {0.36, 0.50}, {0.41, 0.54}, {0.46, 0.58}, {0.51, 0.62}, {0.56, 0.66}, {0.61, 0.70}, {0.66, 0.74}, {0.71, 0.78}, {0.76, 0.82}, {0.81, 0.86}, {0.86, 0.90}, {0.91, 0.94}, {0.96, 0.98}, {1.01, 1.00} }   function rescale (price) if price < 0 or price > 1 then return "Out of range!" end for k, v in pairs(scaleTable) do if price < v[1] then return v[2] end end end   math.randomseed(os.time()) for i = 1, 5 do rnd = math.random() print("Random value:", rnd) print("Adjusted price:", rescale(rnd)) print() end
http://rosettacode.org/wiki/Proper_divisors
Proper divisors
The   proper divisors   of a positive integer N are those numbers, other than N itself, that divide N without remainder. For N > 1 they will always include 1,   but for N == 1 there are no proper divisors. Examples The proper divisors of     6     are   1, 2, and 3. The proper divisors of   100   are   1, 2, 4, 5, 10, 20, 25, and 50. Task Create a routine to generate all the proper divisors of a number. use it to show the proper divisors of the numbers 1 to 10 inclusive. Find a number in the range 1 to 20,000 with the most proper divisors. Show the number and just the count of how many proper divisors it has. Show all output here. Related tasks   Amicable pairs   Abundant, deficient and perfect number classifications   Aliquot sequence classifications   Factors of an integer   Prime decomposition
#PHP
PHP
<?php function ProperDivisors($n) { yield 1; $large_divisors = []; for ($i = 2; $i <= sqrt($n); $i++) { if ($n % $i == 0) { yield $i; if ($i*$i != $n) { $large_divisors[] = $n / $i; } } } foreach (array_reverse($large_divisors) as $i) { yield $i; } }   assert([1, 2, 4, 5, 10, 20, 25, 50] == iterator_to_array(ProperDivisors(100)));   foreach (range(1, 10) as $n) { echo "$n =>"; foreach (ProperDivisors($n) as $divisor) { echo " $divisor"; } echo "\n"; }   $divisorsCount = []; for ($i = 1; $i < 20000; $i++) { $divisorsCount[sizeof(iterator_to_array(ProperDivisors($i)))][] = $i; } ksort($divisorsCount);   echo "Numbers with most divisors: ", implode(", ", end($divisorsCount)), ".\n"; echo "They have ", key($divisorsCount), " divisors.\n";    
http://rosettacode.org/wiki/Probabilistic_choice
Probabilistic choice
Given a mapping between items and their required probability of occurrence, generate a million items randomly subject to the given probabilities and compare the target probability of occurrence versus the generated values. The total of all the probabilities should equal one. (Because floating point arithmetic is involved, this is subject to rounding errors). aleph 1/5.0 beth 1/6.0 gimel 1/7.0 daleth 1/8.0 he 1/9.0 waw 1/10.0 zayin 1/11.0 heth 1759/27720 # adjusted so that probabilities add to 1 Related task Random number generator (device)
#XPL0
XPL0
include c:\cxpl\codes; def Size = 10_000_000; int Tbl(12+1); int I, J, N; real X, S0, S1; [for J:= 5 to 12 do Tbl(J):= 0; for I:= 0 to 1_000_000-1 do \generate one million items [N:= Ran(Size); for J:= 5 to 11 do [N:= N - Size/J; if N < 0 then [Tbl(J):= Tbl(J)+1; J:= 100]; ]; if J=12 then Tbl(12):= Tbl(12)+1; ]; S0:= 0.0; S1:= 0.0; for J:= 5 to 11 do [X:= 1.0/float(J); RlOut(0, X); S0:= S0+X; X:= float(Tbl(J)) / 1_000_000.0; RlOut(0, X); S1:= S1+X; CrLf(0); ]; X:= 1759.0 / 27720.0; RlOut(0, X); S0:= S0+X; X:= float(Tbl(12)) / 1_000_000.0; RlOut(0, X); S1:= S1+X; CrLf(0); Text(0, " ------- ------- "); RlOut(0, S0); RlOut(0, S1); ]
http://rosettacode.org/wiki/Probabilistic_choice
Probabilistic choice
Given a mapping between items and their required probability of occurrence, generate a million items randomly subject to the given probabilities and compare the target probability of occurrence versus the generated values. The total of all the probabilities should equal one. (Because floating point arithmetic is involved, this is subject to rounding errors). aleph 1/5.0 beth 1/6.0 gimel 1/7.0 daleth 1/8.0 he 1/9.0 waw 1/10.0 zayin 1/11.0 heth 1759/27720 # adjusted so that probabilities add to 1 Related task Random number generator (device)
#Yabasic
Yabasic
dim letters$(7) data "aleph", "beth", "gimel", "daleth", "he", "waw", "zayin", "heth" letters$(0) = "aleph" letters$(1) = "beth" letters$(2) = "gimel" letters$(3) = "daleth" letters$(4) = "he" letters$(5) = "waw" letters$(6) = "zayin" letters$(7) = "heth" dim actual(7) dim probs(7) probs(0) = 1/5.0 probs(1) = 1/6.0 probs(2) = 1/7.0 probs(3) = 1/8.0 probs(4) = 1/9.0 probs(5) = 1/10.0 probs(6) = 1/11.0 probs(7) = 1759/27720 dim cumProbs(7)   cumProbs(0) = probs(0) for i = 1 to 6 cumProbs(i) = cumProbs(i - 1) + probs(i) next i cumProbs(7) = 1.0   n = 1000000 for test = 1 to n r = ran(1) p = 0.0 for i = 1 to arraysize(probs(),1) p = p + probs(i) if r < p then actual(i) = actual(i) + 1 break end if next i next t   sumActual = 0.0 tab$ = chr$(9) print "Letter Actual Expected" print "------ -------- --------" for i = 0 to 7 print letters$(i), tab$, print actual(i)/n using "#.######", sumActual = sumActual + actual(i)/n print probs(i) using "#.######" next i   print " -------- --------" print " ", sumActual using "#.######", tab$, "1.000000" end
http://rosettacode.org/wiki/Priority_queue
Priority queue
A priority queue is somewhat similar to a queue, with an important distinction: each item is added to a priority queue with a priority level, and will be later removed from the queue with the highest priority element first. That is, the items are (conceptually) stored in the queue in priority order instead of in insertion order. Task Create a priority queue.   The queue must support at least two operations:   Insertion.   An element is added to the queue with a priority (a numeric value).   Top item removal.   Deletes the element or one of the elements with the current top priority and return it. Optionally, other operations may be defined, such as peeking (find what current top priority/top element is), merging (combining two priority queues into one), etc. To test your implementation, insert a number of elements into the queue, each with some random priority. Then dequeue them sequentially; now the elements should be sorted by priority. You can use the following task/priority items as input data: Priority Task ══════════ ════════════════ 3 Clear drains 4 Feed cat 5 Make tea 1 Solve RC tasks 2 Tax return The implementation should try to be efficient.   A typical implementation has   O(log n)   insertion and extraction time,   where   n   is the number of items in the queue. You may choose to impose certain limits such as small range of allowed priority levels, limited capacity, etc.   If so, discuss the reasons behind it.
#Python
Python
>>> import queue >>> pq = queue.PriorityQueue() >>> for item in ((3, "Clear drains"), (4, "Feed cat"), (5, "Make tea"), (1, "Solve RC tasks"), (2, "Tax return")): pq.put(item)     >>> while not pq.empty(): print(pq.get_nowait())     (1, 'Solve RC tasks') (2, 'Tax return') (3, 'Clear drains') (4, 'Feed cat') (5, 'Make tea') >>>
http://rosettacode.org/wiki/Prime_decomposition
Prime decomposition
The prime decomposition of a number is defined as a list of prime numbers which when all multiplied together, are equal to that number. Example 12 = 2 × 2 × 3, so its prime decomposition is {2, 2, 3} Task Write a function which returns an array or collection which contains the prime decomposition of a given number   n {\displaystyle n}   greater than   1. If your language does not have an isPrime-like function available, you may assume that you have a function which determines whether a number is prime (note its name before your code). If you would like to test code from this task, you may use code from trial division or the Sieve of Eratosthenes. Note: The program must not be limited by the word size of your computer or some other artificial limit; it should work for any number regardless of size (ignoring the physical limits of RAM etc). Related tasks   count in factors   factors of an integer   Sieve of Eratosthenes   primality by trial division   factors of a Mersenne number   trial factoring of a Mersenne number   partition an integer X into N primes   sequence of primes by Trial Division
#J
J
q:
http://rosettacode.org/wiki/Plot_coordinate_pairs
Plot coordinate pairs
Task Plot a function represented as    x,  y    numerical arrays. Post the resulting image for the following input arrays (taken from Python's Example section on Time a function): x = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}; y = {2.7, 2.8, 31.4, 38.1, 58.0, 76.2, 100.5, 130.0, 149.3, 180.0}; This task is intended as a subtask for Measure relative performance of sorting algorithms implementations.
#Nim
Nim
import gnuplot   let x = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] y = [2.7, 2.8, 31.4, 38.1, 58.0, 76.2, 100.5, 130.0, 149.3, 180.0]   plot(x, y, "Coordinate pairs") discard stdin.readChar # Needed as when the program exits, “gnuplot” is closed.
http://rosettacode.org/wiki/Plot_coordinate_pairs
Plot coordinate pairs
Task Plot a function represented as    x,  y    numerical arrays. Post the resulting image for the following input arrays (taken from Python's Example section on Time a function): x = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}; y = {2.7, 2.8, 31.4, 38.1, 58.0, 76.2, 100.5, 130.0, 149.3, 180.0}; This task is intended as a subtask for Measure relative performance of sorting algorithms implementations.
#OCaml
OCaml
#load "graphics.cma" open Graphics   let round x = int_of_float (floor(x +. 0.5))   let x = [0; 1; 2; 3; 4; 5; 6; 7; 8; 9] and y = [2.7; 2.8; 31.4; 38.1; 58.0; 76.2; 100.5; 130.0; 149.3; 180.0]   let () = open_graph ""; List.iter2 (fun x y -> (* scale to fit in the window *) let _x = x * 30 and _y = round(y *. 2.0) in plot _x _y) x y; ignore(wait_next_event [Key_pressed]); close_graph(); ;;
http://rosettacode.org/wiki/Polymorphism
Polymorphism
Task Create two classes   Point(x,y)   and   Circle(x,y,r)   with a polymorphic function print, accessors for (x,y,r), copy constructor, assignment and destructor and every possible default constructors
#Lua
Lua
-- Point local Point = {x = 0, y = 0}   function Point:new(o) o = o or {} setmetatable(o, self) self.__index = self return o end   function Point:print() print("Point(" .. self.x .. ", " .. self.y .. ")") end   function Point:copy() return Point:new{x = self.x, y = self.y} end   -- Circle local Circle = Point:new() Circle.r = 0   function Circle:print() print("Circle(" .. self.x .. ", " .. self.y .. ", " .. self.r .. ")") end   function Circle:copy() return Circle:new{x = self.x, y = self.y, r = self.r} end
http://rosettacode.org/wiki/Poker_hand_analyser
Poker hand analyser
Task Create a program to parse a single five card poker hand and rank it according to this list of poker hands. A poker hand is specified as a space separated list of five playing cards. Each input card has two characters indicating face and suit. Example 2d       (two of diamonds). Faces are:    a, 2, 3, 4, 5, 6, 7, 8, 9, 10, j, q, k Suits are:    h (hearts),   d (diamonds),   c (clubs),   and   s (spades),   or alternatively,   the unicode card-suit characters:    ♥ ♦ ♣ ♠ Duplicate cards are illegal. The program should analyze a single hand and produce one of the following outputs: straight-flush four-of-a-kind full-house flush straight three-of-a-kind two-pair one-pair high-card invalid Examples 2♥ 2♦ 2♣ k♣ q♦: three-of-a-kind 2♥ 5♥ 7♦ 8♣ 9♠: high-card a♥ 2♦ 3♣ 4♣ 5♦: straight 2♥ 3♥ 2♦ 3♣ 3♦: full-house 2♥ 7♥ 2♦ 3♣ 3♦: two-pair 2♥ 7♥ 7♦ 7♣ 7♠: four-of-a-kind 10♥ j♥ q♥ k♥ a♥: straight-flush 4♥ 4♠ k♠ 5♦ 10♠: one-pair q♣ 10♣ 7♣ 6♣ q♣: invalid The programs output for the above examples should be displayed here on this page. Extra credit use the playing card characters introduced with Unicode 6.0 (U+1F0A1 - U+1F0DE). allow two jokers use the symbol   joker duplicates would be allowed (for jokers only) five-of-a-kind would then be the highest hand More extra credit examples joker 2♦ 2♠ k♠ q♦: three-of-a-kind joker 5♥ 7♦ 8♠ 9♦: straight joker 2♦ 3♠ 4♠ 5♠: straight joker 3♥ 2♦ 3♠ 3♦: four-of-a-kind joker 7♥ 2♦ 3♠ 3♦: three-of-a-kind joker 7♥ 7♦ 7♠ 7♣: five-of-a-kind joker j♥ q♥ k♥ A♥: straight-flush joker 4♣ k♣ 5♦ 10♠: one-pair joker k♣ 7♣ 6♣ 4♣: flush joker 2♦ joker 4♠ 5♠: straight joker Q♦ joker A♠ 10♠: straight joker Q♦ joker A♦ 10♦: straight-flush joker 2♦ 2♠ joker q♦: four-of-a-kind Related tasks Playing cards Card shuffles Deal cards_for_FreeCell War Card_Game Go Fish
#Ruby
Ruby
class Card include Comparable attr_accessor :ordinal attr_reader :suit, :face   SUITS = %i(♥ ♦ ♣ ♠) FACES = %i(2 3 4 5 6 7 8 9 10 j q k a)   def initialize(str) @face, @suit = parse(str) @ordinal = FACES.index(@face) end   def <=> (other) #used for sorting self.ordinal <=> other.ordinal end   def to_s "#@face#@suit" end   private def parse(str) face, suit = str.chop.to_sym, str[-1].to_sym raise ArgumentError, "invalid card: #{str}" unless FACES.include?(face) && SUITS.include?(suit) [face, suit] end end   class Hand include Comparable attr_reader :cards, :rank   RANKS = %i(high-card one-pair two-pair three-of-a-kind straight flush full-house four-of-a-kind straight-flush five-of-a-kind) WHEEL_FACES = %i(2 3 4 5 a)   def initialize(str_of_cards) @cards = str_of_cards.downcase.tr(',',' ').split.map{|str| Card.new(str)} grouped = @cards.group_by(&:face).values @face_pattern = grouped.map(&:size).sort @rank = categorize @rank_num = RANKS.index(@rank) @tiebreaker = grouped.map{|ar| [ar.size, ar.first.ordinal]}.sort.reverse end   def <=> (other) # used for sorting and comparing self.compare_value <=> other.compare_value end   def to_s @cards.map(&:to_s).join(" ") end   protected # accessible for Hands def compare_value [@rank_num, @tiebreaker] end   private def one_suit? @cards.map(&:suit).uniq.size == 1 end   def consecutive? sort.each_cons(2).all? {|c1,c2| c2.ordinal - c1.ordinal == 1 } end   def sort if @cards.sort.map(&:face) == WHEEL_FACES @cards.detect {|c| c.face == :a}.ordinal = -1 end @cards.sort end   def categorize if consecutive? one_suit? ? :'straight-flush' : :straight elsif one_suit? :flush else case @face_pattern when [1,1,1,1,1] then :'high-card' when [1,1,1,2] then :'one-pair' when [1,2,2] then :'two-pair' when [1,1,3] then :'three-of-a-kind' when [2,3] then :'full-house' when [1,4] then :'four-of-a-kind' when [5] then :'five-of-a-kind' end end end end   # Demo test_hands = <<EOS 2♥ 2♦ 2♣ k♣ q♦ 2♥ 5♥ 7♦ 8♣ 9♠ a♥ 2♦ 3♣ 4♣ 5♦ 2♥ 3♥ 2♦ 3♣ 3♦ 2♥ 7♥ 2♦ 3♣ 3♦ 2♥ 6♥ 2♦ 3♣ 3♦ 10♥ j♥ q♥ k♥ a♥ 4♥ 4♠ k♠ 2♦ 10♠ 4♥ 4♠ k♠ 3♦ 10♠ q♣ 10♣ 7♣ 6♣ 4♣ q♣ 10♣ 7♣ 6♣ 3♣ 9♥ 10♥ q♥ k♥ j♣ 2♥ 3♥ 4♥ 5♥ a♥ 2♥ 2♥ 2♦ 3♣ 3♦ EOS   hands = test_hands.each_line.map{|line| Hand.new(line) } puts "High to low" hands.sort.reverse.each{|hand| puts "#{hand}\t #{hand.rank}" } puts   str = <<EOS joker 2♦ 2♠ k♠ q♦ joker 5♥ 7♦ 8♠ 9♦ joker 2♦ 3♠ 4♠ 5♠ joker 3♥ 2♦ 3♠ 3♦ joker 7♥ 2♦ 3♠ 3♦ joker 7♥ 7♦ 7♠ 7♣ joker j♥ q♥ k♥ A♥ joker 4♣ k♣ 5♦ 10♠ joker k♣ 7♣ 6♣ 4♣ joker 2♦ joker 4♠ 5♠ joker Q♦ joker A♠ 10♠ joker Q♦ joker A♦ 10♦ joker 2♦ 2♠ joker q♦ EOS   # Neither the Card nor the Hand class supports jokers # but since hands are comparable, they are also sortable. # Try every card from a deck for a joker and pick the largest hand:   DECK = Card::FACES.product(Card::SUITS).map(&:join) str.each_line do |line| cards_in_arrays = line.split.map{|c| c == "joker" ? DECK.dup : [c]} #joker is array of all cards all_tries = cards_in_arrays.shift.product(*cards_in_arrays).map{|ar| Hand.new(ar.join" ")} #calculate the Whatshisname product best = all_tries.max puts "#{line.strip}: #{best.rank}" end
http://rosettacode.org/wiki/Population_count
Population count
Population count You are encouraged to solve this task according to the task description, using any language you may know. The   population count   is the number of   1s   (ones)   in the binary representation of a non-negative integer. Population count   is also known as:   pop count   popcount   sideways sum   bit summation   Hamming weight For example,   5   (which is   101   in binary)   has a population count of   2. Evil numbers   are non-negative integers that have an   even   population count. Odious numbers     are  positive integers that have an    odd   population count. Task write a function (or routine) to return the population count of a non-negative integer. all computation of the lists below should start with   0   (zero indexed). display the   pop count   of the   1st   thirty powers of   3       (30,   31,   32,   33,   34,   ∙∙∙   329). display the   1st   thirty     evil     numbers. display the   1st   thirty   odious   numbers. display each list of integers on one line   (which may or may not include a title),   each set of integers being shown should be properly identified. See also The On-Line Encyclopedia of Integer Sequences:   A000120 population count. The On-Line Encyclopedia of Integer Sequences:   A000069 odious numbers. The On-Line Encyclopedia of Integer Sequences:   A001969 evil numbers.
#F.C5.8Drmul.C3.A6
Fōrmulæ
  #define NTERMS 30   function popcount( n as ulongint ) as uinteger if n = 0 then return 0 if n = 1 then return 1 if n mod 2 = 0 then return popcount(n/2) return 1 + popcount( (n - 1)/2 ) end function   dim as ulongint i=0, tp(0 to NTERMS-1), evil(0 to NTERMS-1),_ odious(0 to NTERMS-1), k, ne=0, no=0   while ne < NTERMS or no < NTERMS if i<NTERMS then tp(i) = popcount(3^i) k = popcount(i) if k mod 2 = 0 and ne < NTERMS then evil(ne) = i ne += 1 endif if k mod 2 = 1 and no < NTERMS then odious(no) = i no += 1 end if i += 1 wend   dim as string s_tp = "", s_evil = "", s_odious = ""   for i = 0 to NTERMS-1 s_tp += str(tp(i))+" " s_evil += str(evil(i))+" " s_odious += str(odious(i))+" " next i   print s_tp print s_evil print s_odious
http://rosettacode.org/wiki/Polynomial_long_division
Polynomial long division
This page uses content from Wikipedia. The original article was at Polynomial long division. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance) In algebra, polynomial long division is an algorithm for dividing a polynomial by another polynomial of the same or lower degree. Let us suppose a polynomial is represented by a vector, x {\displaystyle x} (i.e., an ordered collection of coefficients) so that the i {\displaystyle i} th element keeps the coefficient of x i {\displaystyle x^{i}} , and the multiplication by a monomial is a shift of the vector's elements "towards right" (injecting ones from left) followed by a multiplication of each element by the coefficient of the monomial. Then a pseudocode for the polynomial long division using the conventions described above could be: degree(P): return the index of the last non-zero element of P; if all elements are 0, return -∞ polynomial_long_division(N, D) returns (q, r): // N, D, q, r are vectors if degree(D) < 0 then error q ← 0 while degree(N) ≥ degree(D) d ← D shifted right by (degree(N) - degree(D)) q(degree(N) - degree(D)) ← N(degree(N)) / d(degree(d)) // by construction, degree(d) = degree(N) of course d ← d * q(degree(N) - degree(D)) N ← N - d endwhile r ← N return (q, r) Note: vector * scalar multiplies each element of the vector by the scalar; vectorA - vectorB subtracts each element of the vectorB from the element of the vectorA with "the same index". The vectors in the pseudocode are zero-based. Error handling (for allocations or for wrong inputs) is not mandatory. Conventions can be different; in particular, note that if the first coefficient in the vector is the highest power of x for the polynomial represented by the vector, then the algorithm becomes simpler. Example for clarification This example is from Wikipedia, but changed to show how the given pseudocode works. 0 1 2 3 ---------------------- N: -42 0 -12 1 degree = 3 D: -3 1 0 0 degree = 1 d(N) - d(D) = 2, so let's shift D towards right by 2: N: -42 0 -12 1 d: 0 0 -3 1 N(3)/d(3) = 1, so d is unchanged. Now remember that "shifting by 2" is like multiplying by x2, and the final multiplication (here by 1) is the coefficient of this monomial. Let's store this into q: 0 1 2 --------------- q: 0 0 1 now compute N - d, and let it be the "new" N, and let's loop N: -42 0 -9 0 degree = 2 D: -3 1 0 0 degree = 1 d(N) - d(D) = 1, right shift D by 1 and let it be d N: -42 0 -9 0 d: 0 -3 1 0 * -9/1 = -9 q: 0 -9 1 d: 0 27 -9 0 N ← N - d N: -42 -27 0 0 degree = 1 D: -3 1 0 0 degree = 1 looping again... d(N)-d(D)=0, so no shift is needed; we multiply D by -27 (= -27/1) storing the result in d, then q: -27 -9 1 and N: -42 -27 0 0 - d: 81 -27 0 0 = N: -123 0 0 0 (last N) d(N) < d(D), so now r ← N, and the result is: 0 1 2 ------------- q: -27 -9 1 → x2 - 9x - 27 r: -123 0 0 → -123 Related task   Polynomial derivative
#Raku
Raku
sub poly_long_div ( @n is copy, @d ) { return [0], |@n if +@n < +@d;   my @q = gather while +@n >= +@d { @n = @n Z- flat ( ( @d X* take ( @n[0] / @d[0] ) ), 0 xx * ); @n.shift; }   return @q, @n; }   sub xP ( $power ) { $power>1 ?? "x^$power" !! $power==1 ?? 'x' !! '' } sub poly_print ( @c ) { join ' + ', @c.kv.map: { $^v ~ xP( @c.end - $^k ) } }   my @polys = [ [ 1, -12, 0, -42 ], [ 1, -3 ] ], [ [ 1, -12, 0, -42 ], [ 1, 1, -3 ] ], [ [ 1, 3, 2 ], [ 1, 1 ] ], [ [ 1, -4, 6, 5, 3 ], [ 1, 2, 1 ] ];   say '<math>\begin{array}{rr}'; for @polys -> [ @a, @b ] { printf Q"%s , & %s \\\\\n", poly_long_div( @a, @b ).map: { poly_print($_) }; } say '\end{array}</math>';
http://rosettacode.org/wiki/Polynomial_regression
Polynomial regression
Find an approximating polynomial of known degree for a given data. Example: For input data: x = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; y = {1, 6, 17, 34, 57, 86, 121, 162, 209, 262, 321}; The approximating polynomial is: 3 x2 + 2 x + 1 Here, the polynomial's coefficients are (3, 2, 1). This task is intended as a subtask for Measure relative performance of sorting algorithms implementations.
#Phix
Phix
-- demo\rosetta\Polynomial_regression.exw with javascript_semantics constant x = {0,1,2,3,4,5,6,7,8,9,10}, y = {1,6,17,34,57,86,121,162,209,262,321}, n = length(x) function regression() atom xm = 0, ym = 0, x2m = 0, x3m = 0, x4m = 0, xym = 0, x2ym = 0 for i=1 to n do atom xi = x[i], yi = y[i] xm += xi ym += yi x2m += power(xi,2) x3m += power(xi,3) x4m += power(xi,4) xym += xi*yi x2ym += power(xi,2)*yi end for xm /= n ym /= n x2m /= n x3m /= n x4m /= n xym /= n x2ym /= n atom Sxx = x2m-power(xm,2), Sxy = xym-xm*ym, Sxx2 = x3m-xm*x2m, Sx2x2 = x4m-power(x2m,2), Sx2y = x2ym-x2m*ym, B = (Sxy*Sx2x2-Sx2y*Sxx2)/(Sxx*Sx2x2-power(Sxx2,2)), C = (Sx2y*Sxx-Sxy*Sxx2)/(Sxx*Sx2x2-power(Sxx2,2)), A = ym-B*xm-C*x2m return {C,B,A} end function atom {a,b,c} = regression() function f(atom x) return a*x*x+b*x+c end function printf(1,"y=%gx^2+%gx+%g\n",{a,b,c}) printf(1,"\n x y f(x)\n") for i=1 to n do printf(1," %2d %3d  %3g\n",{x[i],y[i],f(x[i])}) end for -- And a simple plot (re-using x,y from above) include pGUI.e include IupGraph.e function get_data(Ihandle graph) integer {w,h} = IupGetIntInt(graph,"DRAWSIZE") IupSetInt(graph,"YTICK",iff(h<240?iff(h<150?80:40):20)) return {{x,y,CD_RED}} end function IupOpen() Ihandle graph = IupGraph(get_data,"RASTERSIZE=640x440") IupSetAttributes(graph,"XTICK=1,XMIN=0,XMAX=10") IupSetAttributes(graph,"YTICK=20,YMIN=0,YMAX=320") Ihandle dlg = IupDialog(graph,`TITLE="simple plot"`) IupSetAttributes(dlg,"MINSIZE=245x150") IupShow(dlg) if platform()!=JS then IupMainLoop() IupClose() end if
http://rosettacode.org/wiki/Power_set
Power set
A   set   is a collection (container) of certain values, without any particular order, and no repeated values. It corresponds with a finite set in mathematics. A set can be implemented as an associative array (partial mapping) in which the value of each key-value pair is ignored. Given a set S, the power set (or powerset) of S, written P(S), or 2S, is the set of all subsets of S. Task By using a library or built-in set type, or by defining a set type with necessary operations, write a function with a set S as input that yields the power set 2S of S. For example, the power set of     {1,2,3,4}     is {{}, {1}, {2}, {1,2}, {3}, {1,3}, {2,3}, {1,2,3}, {4}, {1,4}, {2,4}, {1,2,4}, {3,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}. For a set which contains n elements, the corresponding power set has 2n elements, including the edge cases of empty set. The power set of the empty set is the set which contains itself (20 = 1): P {\displaystyle {\mathcal {P}}} ( ∅ {\displaystyle \varnothing } ) = { ∅ {\displaystyle \varnothing } } And the power set of the set which contains only the empty set, has two subsets, the empty set and the set which contains the empty set (21 = 2): P {\displaystyle {\mathcal {P}}} ({ ∅ {\displaystyle \varnothing } }) = { ∅ {\displaystyle \varnothing } , { ∅ {\displaystyle \varnothing } } } Extra credit: Demonstrate that your language supports these last two powersets.
#J
J
ps =: #~ 2 #:@i.@^ #
http://rosettacode.org/wiki/Primality_by_trial_division
Primality by trial division
Task Write a boolean function that tells whether a given integer is prime. Remember that   1   and all non-positive numbers are not prime. Use trial division. Even numbers greater than   2   may be eliminated right away. A loop from   3   to   √ n    will suffice,   but other loops are allowed. Related tasks   count in factors   prime decomposition   AKS test for primes   factors of an integer   Sieve of Eratosthenes   factors of a Mersenne number   trial factoring of a Mersenne number   partition an integer X into N primes   sequence of primes by Trial Division
#EchoLisp
EchoLisp
(lib 'sequences)   ;; Try divisors iff n = 2k + 1 (define (is-prime? p) (cond [(< p 2) #f] [(zero? (modulo p 2)) (= p 2)] [else (for/and ((d [3 5 .. (1+ (sqrt p))] )) (!zero? (modulo p d)))]))   (filter is-prime? (range 1 100)) → (2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97)   ;; Improve performance , try divisors iff n = 6k+1 or n = 6k+5 (define (is-prime? p) (cond [(< p 2) #f] [(zero? (modulo p 2)) (= p 2)] [(zero? (modulo p 3)) (= p 3)] [(zero? (modulo p 5)) (= p 5)] [else ;; step 6 : try divisors 6n+1 or 6n+5 (for ((d [7 13 .. (1+ (sqrt p))] )) #:break (zero? (modulo p d)) => #f #:break (zero? (modulo p (+ d 4))) => #f #t )]))   (filter is-prime? (range 1 100)) → (2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97)
http://rosettacode.org/wiki/Price_fraction
Price fraction
A friend of mine runs a pharmacy.   He has a specialized function in his Dispensary application which receives a decimal value of currency and replaces it to a standard value.   This value is regulated by a government department. Task Given a floating point value between   0.00   and   1.00,   rescale according to the following table: >= 0.00 < 0.06  := 0.10 >= 0.06 < 0.11  := 0.18 >= 0.11 < 0.16  := 0.26 >= 0.16 < 0.21  := 0.32 >= 0.21 < 0.26  := 0.38 >= 0.26 < 0.31  := 0.44 >= 0.31 < 0.36  := 0.50 >= 0.36 < 0.41  := 0.54 >= 0.41 < 0.46  := 0.58 >= 0.46 < 0.51  := 0.62 >= 0.51 < 0.56  := 0.66 >= 0.56 < 0.61  := 0.70 >= 0.61 < 0.66  := 0.74 >= 0.66 < 0.71  := 0.78 >= 0.71 < 0.76  := 0.82 >= 0.76 < 0.81  := 0.86 >= 0.81 < 0.86  := 0.90 >= 0.86 < 0.91  := 0.94 >= 0.91 < 0.96  := 0.98 >= 0.96 < 1.01  := 1.00
#Maple
Maple
priceFraction := proc(price) local values, standard, newPrice, i; values := [0, 0.06, 0.11, 0.16, 0.21, 0.26, 0.31, 0.36, 0.41, 0.46, 0.51, 0.56, 0.61, 0.66, 0.71, 0.76, 0.81, 0.86, 0.91, 0.96, 1.01]; standard := [0.10, 0.18, 0.26, 0.32, 0.38, 0.44, 0.50, 0.54, 0.58, 0.62, 0.66, 0.70, 0.74, 0.78, 0.82, 0.86, 0.90, 0.94, 0.98, 1.00]; for i to numelems(standard) do if price >= values[i] and price < values[i+1] then newPrice := standard[i]; end if; end do; printf("%f --> %.2f\n", price, newPrice); end proc:   randomize(): for i to 5 do priceFraction (rand(0.0..1.0)()); end do;
http://rosettacode.org/wiki/Price_fraction
Price fraction
A friend of mine runs a pharmacy.   He has a specialized function in his Dispensary application which receives a decimal value of currency and replaces it to a standard value.   This value is regulated by a government department. Task Given a floating point value between   0.00   and   1.00,   rescale according to the following table: >= 0.00 < 0.06  := 0.10 >= 0.06 < 0.11  := 0.18 >= 0.11 < 0.16  := 0.26 >= 0.16 < 0.21  := 0.32 >= 0.21 < 0.26  := 0.38 >= 0.26 < 0.31  := 0.44 >= 0.31 < 0.36  := 0.50 >= 0.36 < 0.41  := 0.54 >= 0.41 < 0.46  := 0.58 >= 0.46 < 0.51  := 0.62 >= 0.51 < 0.56  := 0.66 >= 0.56 < 0.61  := 0.70 >= 0.61 < 0.66  := 0.74 >= 0.66 < 0.71  := 0.78 >= 0.71 < 0.76  := 0.82 >= 0.76 < 0.81  := 0.86 >= 0.81 < 0.86  := 0.90 >= 0.86 < 0.91  := 0.94 >= 0.91 < 0.96  := 0.98 >= 0.96 < 1.01  := 1.00
#Mathematica.2FWolfram_Language
Mathematica/Wolfram Language
PriceFraction[x_]:=Piecewise[{{.1, 0 <= x < 0.06}, {.18, x < .11}, {.26,x < 0.16}, {.32, x < .21}, {.38, x < .26}, {.44, x < 0.31}, {.5, x < .36}, {.54, x < .41}, {.58, x < .46}, {.62, x < .51}, {.66, x < .56}, {.70, x < .61}, {.74, x < .66}, {.78, x < .71}, {.82, x < .76}, {.86, x < .81}, {.90, x < .86}, {.94, x < .91}, {.98, x < .96}}, 1]
http://rosettacode.org/wiki/Proper_divisors
Proper divisors
The   proper divisors   of a positive integer N are those numbers, other than N itself, that divide N without remainder. For N > 1 they will always include 1,   but for N == 1 there are no proper divisors. Examples The proper divisors of     6     are   1, 2, and 3. The proper divisors of   100   are   1, 2, 4, 5, 10, 20, 25, and 50. Task Create a routine to generate all the proper divisors of a number. use it to show the proper divisors of the numbers 1 to 10 inclusive. Find a number in the range 1 to 20,000 with the most proper divisors. Show the number and just the count of how many proper divisors it has. Show all output here. Related tasks   Amicable pairs   Abundant, deficient and perfect number classifications   Aliquot sequence classifications   Factors of an integer   Prime decomposition
#Picat
Picat
go => println(11111=proper_divisors(11111)), nl, foreach(N in 1..10) println(N=proper_divisors(N)) end, nl,   find_most_divisors(20_000), nl.   % Proper divisors of number N proper_divisors(N) = Divisors => Div1 = [ I : I in 1..ceiling(sqrt(N)), N mod I == 0], Divisors = (Div1 ++ [N div I : I in Div1]).sort_remove_dups().delete(N).     % Find the number(s) with the most proper divisors below Limit find_most_divisors(Limit) => MaxN = [], MaxNumDivisors = [], MaxLen = 1,   foreach(N in 1..Limit, not prime(N)) D = proper_divisors(N), Len = D.len,  % Get all numbers with most proper divisors if Len = MaxLen then MaxN := MaxN ++ [N], MaxNumDivisors := MaxNumDivisors ++ [[N=D]] elseif Len > MaxLen then MaxLen := Len, MaxN := [N], MaxNumDivisors := [N=D] end end,   println(maxN=MaxN), println(maxLen=MaxLen), nl.
http://rosettacode.org/wiki/Probabilistic_choice
Probabilistic choice
Given a mapping between items and their required probability of occurrence, generate a million items randomly subject to the given probabilities and compare the target probability of occurrence versus the generated values. The total of all the probabilities should equal one. (Because floating point arithmetic is involved, this is subject to rounding errors). aleph 1/5.0 beth 1/6.0 gimel 1/7.0 daleth 1/8.0 he 1/9.0 waw 1/10.0 zayin 1/11.0 heth 1759/27720 # adjusted so that probabilities add to 1 Related task Random number generator (device)
#zkl
zkl
var names=T("aleph", "beth", "gimel", "daleth", "he", "waw", "zayin", "heth"); var ptable=T(5.0, 6.0, 7.0, 8.0, 9.0, 10.0, 11.0).apply('/.fp(1.0)); ptable=ptable.append(1.0-ptable.sum(0.0)); // add last weight to sum to 1.0 var [const] N=ptable.len();   fcn ridx{ i:=0; s:=(0.0).random(1); while((s-=ptable[i]) > 0) { i+=1 } i }   const M=0d1_000_000; var r=(0).pump(N,List,T(Ref,0)); // list of references to int 0 (0).pump(M,Void,fcn{r[ridx()].inc()}); // 1,000,000 weighted random #s   r=r.apply("value").apply("toFloat"); // (reference to int)-->int-->float   println(" Name Count Ratio Expected"); foreach i in (N){ "%6s%7d %7.4f%% %7.4f%%".fmt(names[i], r[i], r[i]/M*100, ptable[i]*100).println(); }
http://rosettacode.org/wiki/Priority_queue
Priority queue
A priority queue is somewhat similar to a queue, with an important distinction: each item is added to a priority queue with a priority level, and will be later removed from the queue with the highest priority element first. That is, the items are (conceptually) stored in the queue in priority order instead of in insertion order. Task Create a priority queue.   The queue must support at least two operations:   Insertion.   An element is added to the queue with a priority (a numeric value).   Top item removal.   Deletes the element or one of the elements with the current top priority and return it. Optionally, other operations may be defined, such as peeking (find what current top priority/top element is), merging (combining two priority queues into one), etc. To test your implementation, insert a number of elements into the queue, each with some random priority. Then dequeue them sequentially; now the elements should be sorted by priority. You can use the following task/priority items as input data: Priority Task ══════════ ════════════════ 3 Clear drains 4 Feed cat 5 Make tea 1 Solve RC tasks 2 Tax return The implementation should try to be efficient.   A typical implementation has   O(log n)   insertion and extraction time,   where   n   is the number of items in the queue. You may choose to impose certain limits such as small range of allowed priority levels, limited capacity, etc.   If so, discuss the reasons behind it.
#Quackery
Quackery
( ***** priotity queue ***** )   [ stack ] is heap.pq ( --> s )   [ stack ] is comp.pq ( --> s )   [ unpack comp.pq put heap.pq put ] is pq->stacks ( p --> )   [ heap.pq take comp.pq take 2 pack ] is stacks->pq ( --> p )   [ heap.pq share swap peek ] is heapeek ( n --> x )   [ heap.pq take swap poke heap.pq put ] is heapoke ( n x --> )   [ 1 - 1 >> ] is parent ( n --> n )   [ 0 > ] is has-parent ( n --> b )   [ 1 << 1+ ] is child ( n --> n )   [ child heap.pq share size < ] is has-child ( n --> b )   [ 1+ ] is sibling ( n --> n )   [ sibling heap.pq share size < ] is has-sibling ( n --> b )   [ comp.pq share do ] is compare ( x x --> b )   [ 0 peek size ] is pqsize ( p --> n )   [ swap pq->stacks heap.pq take tuck size rot 0 join heap.pq put [ dup has-parent while dup parent rot over heapeek 2dup compare iff [ 2swap unrot heapoke ] again rot 2drop swap ] heapoke stacks->pq ] is topq ( p x --> p )   [ dup heapeek swap [ dup has-child while dup child dup has-sibling if [ dup sibling heapeek over heapeek compare if sibling ] dip over dup heapeek rot dip dup compare iff [ rot heapoke ] again 2drop ] heapoke ] is heapify.pq ( n --> )   [ dup pqsize 0 = if [ $ "Unexpectedly empty priority queue." fail ] pq->stacks heap.pq share behead over [] = iff [ swap heap.pq replace ] else [ swap -1 split swap join heap.pq put 0 heapify.pq ] stacks->pq swap ] is frompq ( p --> p x )   [ ]'[ 2 pack pq->stacks heap.pq share size 1 >> times [ i heapify.pq ] stacks->pq ] is pqwith ( [ --> p )   ( ***** task ***** )   [ 2 pack topq ] is insert ( p n $ --> p )   [ frompq unpack ] is dequeue ( p --> p n $ )   [] pqwith [ 0 peek dip [ 0 peek ] < ]   3 $ "Clear drains" insert 4 $ "Feed cat" insert 5 $ "Make tea" insert 1 $ "Solve RC tasks" insert 2 $ "Tax return" insert   5 times [ dequeue swap echo say ": " echo$ cr ] drop
http://rosettacode.org/wiki/Prime_decomposition
Prime decomposition
The prime decomposition of a number is defined as a list of prime numbers which when all multiplied together, are equal to that number. Example 12 = 2 × 2 × 3, so its prime decomposition is {2, 2, 3} Task Write a function which returns an array or collection which contains the prime decomposition of a given number   n {\displaystyle n}   greater than   1. If your language does not have an isPrime-like function available, you may assume that you have a function which determines whether a number is prime (note its name before your code). If you would like to test code from this task, you may use code from trial division or the Sieve of Eratosthenes. Note: The program must not be limited by the word size of your computer or some other artificial limit; it should work for any number regardless of size (ignoring the physical limits of RAM etc). Related tasks   count in factors   factors of an integer   Sieve of Eratosthenes   primality by trial division   factors of a Mersenne number   trial factoring of a Mersenne number   partition an integer X into N primes   sequence of primes by Trial Division
#Java
Java
public boolean prime(BigInteger i);
http://rosettacode.org/wiki/Plot_coordinate_pairs
Plot coordinate pairs
Task Plot a function represented as    x,  y    numerical arrays. Post the resulting image for the following input arrays (taken from Python's Example section on Time a function): x = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}; y = {2.7, 2.8, 31.4, 38.1, 58.0, 76.2, 100.5, 130.0, 149.3, 180.0}; This task is intended as a subtask for Measure relative performance of sorting algorithms implementations.
#Octave
Octave
x = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]; y = [2.7, 2.8, 31.4, 38.1, 58.0, 76.2, 100.5, 130.0, 149.3, 180.0]; plot(x,y,"o"); pause;
http://rosettacode.org/wiki/Polymorphism
Polymorphism
Task Create two classes   Point(x,y)   and   Circle(x,y,r)   with a polymorphic function print, accessors for (x,y,r), copy constructor, assignment and destructor and every possible default constructors
#M2000_Interpreter
M2000 Interpreter
  \\ block For This {}, or For object [, object2] { }, where object is a group, or a pointer to group, or an item from an array contains a group \\ This is "this context". For This { \\ any new definition here has a temporary use \\ can be nested, but if we use object then we can use dots to access members of it. If we use a second one then we have to use double dots (..x for second object, for access to x member) }  
http://rosettacode.org/wiki/Polymorphism
Polymorphism
Task Create two classes   Point(x,y)   and   Circle(x,y,r)   with a polymorphic function print, accessors for (x,y,r), copy constructor, assignment and destructor and every possible default constructors
#NetRexx
NetRexx
/* NetRexx */   options replace format comments java crossref savelog symbols binary   -- ----------------------------------------------------------------------------- class RCPolymorphism public final   method main(args = String[]) public constant   parry = [Point - Point() - , Point(1.0) - , Point(1.0, 2.0) - , Point(Point(0.3, 0.2)) - , Circle() - , Circle(2.0, 2.0) - , Circle(5.0, 6.0, 7.0) - , Circle(Point(8.0, 9.0)) - , Circle(Point(8.0, 9.0), 4.0) - , Circle(Circle(1.5, 1.4, 1.3)) - ]   loop pp = 0 to parry.length - 1 parry[pp].print end pp   return   -- ----------------------------------------------------------------------------- class RCPolymorphism.Point public binary   properties private x = double y = double className = Point.class.getSimpleName   method Point(x_ = double 0.0, y_ = double 0.0) setX(x_) setY(y_) return   method Point(p = Point) this(p.getX, p.getY) return   method display public returns String hx = '@'Rexx(Integer.toHexString(hashCode())).right(8, 0) str = Rexx(className).left(10)':'hx': (x,y) = (' || - Rexx(getX()).format(null, 3)',' - Rexx(getY()).format(null, 3)')' return str   method getX public returns double return x   method getY public returns double return y   method setX(x_ = double 0.0) inheritable x = x_ return   method setY(y_ = double 0.0) inheritable y = y_ return   method print inheritable say display return   -- ----------------------------------------------------------------------------- class RCPolymorphism.Circle public extends RCPolymorphism.Point binary   properties private r = double className = Circle.class.getSimpleName   method Circle(x_ = double 0.0, y_ = double 0.0, r_ = double 0.0) super(x_, y_) setR(r_) return   method Circle(p_ = RCPolymorphism.Point, r_ = double 0.0) this(p_.getX, p_.getY, r_) return   method Circle(c_ = Circle) this(c_.getX, c_.getY, c_.getR) return   method getR public returns double return r   method setR(r_ = double 0.0) inheritable r = r_ return   method display public returns String hx = '@'Rexx(Integer.toHexString(hashCode())).right(8, 0) str = Rexx(className).left(10)':'hx': (x,y,r) = (' || - Rexx(getX()).format(null, 3)',' - Rexx(getY()).format(null, 3)',' - Rexx(getR()).format(null, 3)')' return str  
http://rosettacode.org/wiki/Poker_hand_analyser
Poker hand analyser
Task Create a program to parse a single five card poker hand and rank it according to this list of poker hands. A poker hand is specified as a space separated list of five playing cards. Each input card has two characters indicating face and suit. Example 2d       (two of diamonds). Faces are:    a, 2, 3, 4, 5, 6, 7, 8, 9, 10, j, q, k Suits are:    h (hearts),   d (diamonds),   c (clubs),   and   s (spades),   or alternatively,   the unicode card-suit characters:    ♥ ♦ ♣ ♠ Duplicate cards are illegal. The program should analyze a single hand and produce one of the following outputs: straight-flush four-of-a-kind full-house flush straight three-of-a-kind two-pair one-pair high-card invalid Examples 2♥ 2♦ 2♣ k♣ q♦: three-of-a-kind 2♥ 5♥ 7♦ 8♣ 9♠: high-card a♥ 2♦ 3♣ 4♣ 5♦: straight 2♥ 3♥ 2♦ 3♣ 3♦: full-house 2♥ 7♥ 2♦ 3♣ 3♦: two-pair 2♥ 7♥ 7♦ 7♣ 7♠: four-of-a-kind 10♥ j♥ q♥ k♥ a♥: straight-flush 4♥ 4♠ k♠ 5♦ 10♠: one-pair q♣ 10♣ 7♣ 6♣ q♣: invalid The programs output for the above examples should be displayed here on this page. Extra credit use the playing card characters introduced with Unicode 6.0 (U+1F0A1 - U+1F0DE). allow two jokers use the symbol   joker duplicates would be allowed (for jokers only) five-of-a-kind would then be the highest hand More extra credit examples joker 2♦ 2♠ k♠ q♦: three-of-a-kind joker 5♥ 7♦ 8♠ 9♦: straight joker 2♦ 3♠ 4♠ 5♠: straight joker 3♥ 2♦ 3♠ 3♦: four-of-a-kind joker 7♥ 2♦ 3♠ 3♦: three-of-a-kind joker 7♥ 7♦ 7♠ 7♣: five-of-a-kind joker j♥ q♥ k♥ A♥: straight-flush joker 4♣ k♣ 5♦ 10♠: one-pair joker k♣ 7♣ 6♣ 4♣: flush joker 2♦ joker 4♠ 5♠: straight joker Q♦ joker A♠ 10♠: straight joker Q♦ joker A♦ 10♦: straight-flush joker 2♦ 2♠ joker q♦: four-of-a-kind Related tasks Playing cards Card shuffles Deal cards_for_FreeCell War Card_Game Go Fish
#Rust
Rust
  fn main() { let hands = vec![ "🂡 🂮 🂭 🂫 🂪", "🃏 🃂 🂢 🂮 🃍", "🃏 🂵 🃇 🂨 🃉", "🃏 🃂 🂣 🂤 🂥", "🃏 🂳 🃂 🂣 🃃", "🃏 🂷 🃂 🂣 🃃", "🃏 🂷 🃇 🂧 🃗", "🃏 🂻 🂽 🂾 🂱", "🃏 🃔 🃞 🃅 🂪", "🃏 🃞 🃗 🃖 🃔", "🃏 🃂 🃟 🂤 🂥", "🃏 🃍 🃟 🂡 🂪", "🃏 🃍 🃟 🃁 🃊", "🃏 🃂 🂢 🃟 🃍", "🃏 🃂 🂢 🃍 🃍", "🃂 🃞 🃍 🃁 🃊", ]; for hand in hands{ println!("{} {}", hand, poker_hand(hand)); } }   fn poker_hand(cards: &str) -> &str { let mut suits = vec![0u8; 4]; let mut faces = vec![0u8; 15]; let mut hand = vec![];   for card in cards.chars(){ if card == ' ' { continue; } let values = get_card_value(card); if values.0 < 14 && hand.contains(&values) { return "invalid"; } hand.push(values); faces[values.0 as usize]+=1; if values.1 >= 0 { suits[values.1 as usize]+=1; } } if hand.len()!=5 { return "invalid"; } faces[13] = faces[0]; //add ace-high count let jokers = faces[14];   //count suits let mut colors = suits.into_iter() .filter(|&x| x > 0).collect::<Vec<_>>(); colors.sort_unstable(); colors[0] += jokers; // add joker suits to the highest one; let is_flush = colors[0] == 5;   //straight let mut is_straight = false; //pointer to optimise some work //avoids looking again at cards that were the start of a sequence //as they cannot be part of another sequence let mut ptr = 14; while ptr>3{ let mut jokers_left = jokers; let mut straight_cards = 0; for i in (0..ptr).rev(){ if faces[i]==0 { if jokers_left == 0 {break;} jokers_left -= 1; } else if i==ptr-1 { ptr-=1; } straight_cards+=1; } ptr-=1; if straight_cards == 5 { is_straight = true; break; } }   //count values let mut values = faces.into_iter().enumerate().take(14).filter(|&x| x.1>0).collect::<Vec<_>>(); //sort by quantity, then by value, high to low values.sort_unstable_by(|a, b| if b.1 == a.1 { (b.0).cmp(&a.0) } else { (b.1).cmp(&a.1)} ); let first_group = values[0].1 + jokers; let second_group = if values.len()>1 {values[1].1} else {0};   match (is_flush, is_straight, first_group, second_group){ (_,_,5,_) => "five-of-a-kind", (true, true, _, _) => if ptr == 8 {"royal-flush"} else {"straight-flush"}, (_,_,4,_) => "four-of-a-kind", (_,_,3,2) => "full-house", (true,_,_,_) => "flush", (_,true,_,_) => "straight", (_,_,3,_) => "three-of-a-kind", (_,_,2,2) => "two-pair", (_,_,2,_) => "one-pair", _ => "high-card" } }   fn get_card_value(card: char) -> (i8,i8) { // transform glyph to face + suit, zero-indexed let base = card as u32 - 0x1F0A1; let mut suit = (base / 16) as i8; let mut face = (base % 16) as i8; if face > 11 && face < 14 { face-=1; } // Unicode has a Knight that we do not want if face == 14 { suit = -1; } //jokers do not have a suit (face, suit) }  
http://rosettacode.org/wiki/Population_count
Population count
Population count You are encouraged to solve this task according to the task description, using any language you may know. The   population count   is the number of   1s   (ones)   in the binary representation of a non-negative integer. Population count   is also known as:   pop count   popcount   sideways sum   bit summation   Hamming weight For example,   5   (which is   101   in binary)   has a population count of   2. Evil numbers   are non-negative integers that have an   even   population count. Odious numbers     are  positive integers that have an    odd   population count. Task write a function (or routine) to return the population count of a non-negative integer. all computation of the lists below should start with   0   (zero indexed). display the   pop count   of the   1st   thirty powers of   3       (30,   31,   32,   33,   34,   ∙∙∙   329). display the   1st   thirty     evil     numbers. display the   1st   thirty   odious   numbers. display each list of integers on one line   (which may or may not include a title),   each set of integers being shown should be properly identified. See also The On-Line Encyclopedia of Integer Sequences:   A000120 population count. The On-Line Encyclopedia of Integer Sequences:   A000069 odious numbers. The On-Line Encyclopedia of Integer Sequences:   A001969 evil numbers.
#FreeBASIC
FreeBASIC
  #define NTERMS 30   function popcount( n as ulongint ) as uinteger if n = 0 then return 0 if n = 1 then return 1 if n mod 2 = 0 then return popcount(n/2) return 1 + popcount( (n - 1)/2 ) end function   dim as ulongint i=0, tp(0 to NTERMS-1), evil(0 to NTERMS-1),_ odious(0 to NTERMS-1), k, ne=0, no=0   while ne < NTERMS or no < NTERMS if i<NTERMS then tp(i) = popcount(3^i) k = popcount(i) if k mod 2 = 0 and ne < NTERMS then evil(ne) = i ne += 1 endif if k mod 2 = 1 and no < NTERMS then odious(no) = i no += 1 end if i += 1 wend   dim as string s_tp = "", s_evil = "", s_odious = ""   for i = 0 to NTERMS-1 s_tp += str(tp(i))+" " s_evil += str(evil(i))+" " s_odious += str(odious(i))+" " next i   print s_tp print s_evil print s_odious
http://rosettacode.org/wiki/Polynomial_long_division
Polynomial long division
This page uses content from Wikipedia. The original article was at Polynomial long division. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance) In algebra, polynomial long division is an algorithm for dividing a polynomial by another polynomial of the same or lower degree. Let us suppose a polynomial is represented by a vector, x {\displaystyle x} (i.e., an ordered collection of coefficients) so that the i {\displaystyle i} th element keeps the coefficient of x i {\displaystyle x^{i}} , and the multiplication by a monomial is a shift of the vector's elements "towards right" (injecting ones from left) followed by a multiplication of each element by the coefficient of the monomial. Then a pseudocode for the polynomial long division using the conventions described above could be: degree(P): return the index of the last non-zero element of P; if all elements are 0, return -∞ polynomial_long_division(N, D) returns (q, r): // N, D, q, r are vectors if degree(D) < 0 then error q ← 0 while degree(N) ≥ degree(D) d ← D shifted right by (degree(N) - degree(D)) q(degree(N) - degree(D)) ← N(degree(N)) / d(degree(d)) // by construction, degree(d) = degree(N) of course d ← d * q(degree(N) - degree(D)) N ← N - d endwhile r ← N return (q, r) Note: vector * scalar multiplies each element of the vector by the scalar; vectorA - vectorB subtracts each element of the vectorB from the element of the vectorA with "the same index". The vectors in the pseudocode are zero-based. Error handling (for allocations or for wrong inputs) is not mandatory. Conventions can be different; in particular, note that if the first coefficient in the vector is the highest power of x for the polynomial represented by the vector, then the algorithm becomes simpler. Example for clarification This example is from Wikipedia, but changed to show how the given pseudocode works. 0 1 2 3 ---------------------- N: -42 0 -12 1 degree = 3 D: -3 1 0 0 degree = 1 d(N) - d(D) = 2, so let's shift D towards right by 2: N: -42 0 -12 1 d: 0 0 -3 1 N(3)/d(3) = 1, so d is unchanged. Now remember that "shifting by 2" is like multiplying by x2, and the final multiplication (here by 1) is the coefficient of this monomial. Let's store this into q: 0 1 2 --------------- q: 0 0 1 now compute N - d, and let it be the "new" N, and let's loop N: -42 0 -9 0 degree = 2 D: -3 1 0 0 degree = 1 d(N) - d(D) = 1, right shift D by 1 and let it be d N: -42 0 -9 0 d: 0 -3 1 0 * -9/1 = -9 q: 0 -9 1 d: 0 27 -9 0 N ← N - d N: -42 -27 0 0 degree = 1 D: -3 1 0 0 degree = 1 looping again... d(N)-d(D)=0, so no shift is needed; we multiply D by -27 (= -27/1) storing the result in d, then q: -27 -9 1 and N: -42 -27 0 0 - d: 81 -27 0 0 = N: -123 0 0 0 (last N) d(N) < d(D), so now r ← N, and the result is: 0 1 2 ------------- q: -27 -9 1 → x2 - 9x - 27 r: -123 0 0 → -123 Related task   Polynomial derivative
#REXX
REXX
/* REXX needed by some... */ z='1 -12 0 -42' /* Numerator */ n='1 -3' /* Denominator */ zx=z nx=n copies('0 ',words(z)-words(n)) qx='' /* Quotient */ Do Until words(zx)<words(n) Parse Value div(zx,nx) With q zx qx=qx q nx=subword(nx,1,words(nx)-1) End Say '('show(z)')/('show(n)')=('show(qx)')' Say 'Remainder:' show(zx) Exit div: Procedure Parse Arg z,n q=word(z,1)/word(n,1) zz='' Do i=1 To words(z) zz=zz word(z,i)-q*word(n,i) End Return q subword(zz,2)   show: Procedure Parse Arg poly d=words(poly)-1 res='' Do i=1 To words(poly) Select When d>1 Then fact='*x**'d When d=1 Then fact='*x' Otherwise fact='' End Select When word(poly,i)=0 Then p='' When word(poly,i)=1 Then p='+'substr(fact,2) When word(poly,i)=-1 Then p='-'substr(fact,2) When word(poly,i)<0 Then p=word(poly,i)||fact Otherwise p='+'word(poly,i)||fact End res=res p d=d-1 End Return strip(space(res,0),'L','+')
http://rosettacode.org/wiki/Polynomial_regression
Polynomial regression
Find an approximating polynomial of known degree for a given data. Example: For input data: x = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; y = {1, 6, 17, 34, 57, 86, 121, 162, 209, 262, 321}; The approximating polynomial is: 3 x2 + 2 x + 1 Here, the polynomial's coefficients are (3, 2, 1). This task is intended as a subtask for Measure relative performance of sorting algorithms implementations.
#PowerShell
PowerShell
  function qr([double[][]]$A) { $m,$n = $A.count, $A[0].count $pm,$pn = ($m-1), ($n-1) [double[][]]$Q = 0..($m-1) | foreach{$row = @(0) * $m; $row[$_] = 1; ,$row} [double[][]]$R = $A | foreach{$row = $_; ,@(0..$pn | foreach{$row[$_]})} foreach ($h in 0..$pn) { [double[]]$u = $R[$h..$pm] | foreach{$_[$h]} [double]$nu = $u | foreach {[double]$sq = 0} {$sq += $_*$_} {[Math]::Sqrt($sq)} $u[0] -= if ($u[0] -lt 1) {$nu} else {-$nu} [double]$nu = $u | foreach {$sq = 0} {$sq += $_*$_} {[Math]::Sqrt($sq)} [double[]]$u = $u | foreach { $_/$nu} [double[][]]$v = 0..($u.Count - 1) | foreach{$i = $_; ,($u | foreach{2*$u[$i]*$_})} [double[][]]$CR = $R | foreach{$row = $_; ,@(0..$pn | foreach{$row[$_]})} [double[][]]$CQ = $Q | foreach{$row = $_; ,@(0..$pm | foreach{$row[$_]})} foreach ($i in $h..$pm) { foreach ($j in $h..$pn) { $R[$i][$j] -= $h..$pm | foreach {[double]$sum = 0} {$sum += $v[$i-$h][$_-$h]*$CR[$_][$j]} {$sum} } } if (0 -eq $h) { foreach ($i in $h..$pm) { foreach ($j in $h..$pm) { $Q[$i][$j] -= $h..$pm | foreach {$sum = 0} {$sum += $v[$i][$_]*$CQ[$_][$j]} {$sum} } } } else { $p = $h-1 foreach ($i in $h..$pm) { foreach ($j in 0..$p) { $Q[$i][$j] -= $h..$pm | foreach {$sum = 0} {$sum += $v[$i-$h][$_-$h]*$CQ[$_][$j]} {$sum} } foreach ($j in $h..$pm) { $Q[$i][$j] -= $h..$pm | foreach {$sum = 0} {$sum += $v[$i-$h][$_-$h]*$CQ[$_][$j]} {$sum} } } } } foreach ($i in 0..$pm) { foreach ($j in $i..$pm) {$Q[$i][$j],$Q[$j][$i] = $Q[$j][$i],$Q[$i][$j]} } [PSCustomObject]@{"Q" = $Q; "R" = $R} }   function leastsquares([Double[][]]$A,[Double[]]$y) { $QR = qr $A [Double[][]]$Q = $QR.Q [Double[][]]$R = $QR.R $m,$n = $A.count, $A[0].count [Double[]]$z = foreach ($j in 0..($m-1)) { 0..($m-1) | foreach {$sum = 0} {$sum += $Q[$_][$j]*$y[$_]} {$sum} } [Double[]]$x = @(0)*$n for ($i = $n-1; $i -ge 0; $i--) { for ($j = $i+1; $j -lt $n; $j++) { $z[$i] -= $x[$j]*$R[$i][$j] } $x[$i] = $z[$i]/$R[$i][$i] } $x }   function polyfit([Double[]]$x,[Double[]]$y,$n) { $m = $x.Count [Double[][]]$A = 0..($m-1) | foreach{$row = @(1) * ($n+1); ,$row} for ($i = 0; $i -lt $m; $i++) { for ($j = $n-1; 0 -le $j; $j--) { $A[$i][$j] = $A[$i][$j+1]*$x[$i] } } leastsquares $A $y }   function show($m) {$m | foreach {write-host "$_"}}   $A = @(@(12,-51,4), @(6,167,-68), @(-4,24,-41)) $x = @(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10) $y = @(1, 6, 17, 34, 57, 86, 121, 162, 209, 262, 321) "polyfit " "X^2 X constant" "$(polyfit $x $y 2)"  
http://rosettacode.org/wiki/Power_set
Power set
A   set   is a collection (container) of certain values, without any particular order, and no repeated values. It corresponds with a finite set in mathematics. A set can be implemented as an associative array (partial mapping) in which the value of each key-value pair is ignored. Given a set S, the power set (or powerset) of S, written P(S), or 2S, is the set of all subsets of S. Task By using a library or built-in set type, or by defining a set type with necessary operations, write a function with a set S as input that yields the power set 2S of S. For example, the power set of     {1,2,3,4}     is {{}, {1}, {2}, {1,2}, {3}, {1,3}, {2,3}, {1,2,3}, {4}, {1,4}, {2,4}, {1,2,4}, {3,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}. For a set which contains n elements, the corresponding power set has 2n elements, including the edge cases of empty set. The power set of the empty set is the set which contains itself (20 = 1): P {\displaystyle {\mathcal {P}}} ( ∅ {\displaystyle \varnothing } ) = { ∅ {\displaystyle \varnothing } } And the power set of the set which contains only the empty set, has two subsets, the empty set and the set which contains the empty set (21 = 2): P {\displaystyle {\mathcal {P}}} ({ ∅ {\displaystyle \varnothing } }) = { ∅ {\displaystyle \varnothing } , { ∅ {\displaystyle \varnothing } } } Extra credit: Demonstrate that your language supports these last two powersets.
#Java
Java
public static ArrayList<String> getpowerset(int a[],int n,ArrayList<String> ps) { if(n<0) { return null; } if(n==0) { if(ps==null) ps=new ArrayList<String>(); ps.add(" "); return ps; } ps=getpowerset(a, n-1, ps); ArrayList<String> tmp=new ArrayList<String>(); for(String s:ps) { if(s.equals(" ")) tmp.add(""+a[n-1]); else tmp.add(s+a[n-1]); } ps.addAll(tmp); return ps; }
http://rosettacode.org/wiki/Primality_by_trial_division
Primality by trial division
Task Write a boolean function that tells whether a given integer is prime. Remember that   1   and all non-positive numbers are not prime. Use trial division. Even numbers greater than   2   may be eliminated right away. A loop from   3   to   √ n    will suffice,   but other loops are allowed. Related tasks   count in factors   prime decomposition   AKS test for primes   factors of an integer   Sieve of Eratosthenes   factors of a Mersenne number   trial factoring of a Mersenne number   partition an integer X into N primes   sequence of primes by Trial Division
#Eiffel
Eiffel
class APPLICATION   create make   feature   make -- Tests the feature is_prime. do io.put_boolean (is_prime (1)) io.new_line io.put_boolean (is_prime (2)) io.new_line io.put_boolean (is_prime (3)) io.new_line io.put_boolean (is_prime (4)) io.new_line io.put_boolean (is_prime (97)) io.new_line io.put_boolean (is_prime (15589)) io.new_line end   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   end
http://rosettacode.org/wiki/Price_fraction
Price fraction
A friend of mine runs a pharmacy.   He has a specialized function in his Dispensary application which receives a decimal value of currency and replaces it to a standard value.   This value is regulated by a government department. Task Given a floating point value between   0.00   and   1.00,   rescale according to the following table: >= 0.00 < 0.06  := 0.10 >= 0.06 < 0.11  := 0.18 >= 0.11 < 0.16  := 0.26 >= 0.16 < 0.21  := 0.32 >= 0.21 < 0.26  := 0.38 >= 0.26 < 0.31  := 0.44 >= 0.31 < 0.36  := 0.50 >= 0.36 < 0.41  := 0.54 >= 0.41 < 0.46  := 0.58 >= 0.46 < 0.51  := 0.62 >= 0.51 < 0.56  := 0.66 >= 0.56 < 0.61  := 0.70 >= 0.61 < 0.66  := 0.74 >= 0.66 < 0.71  := 0.78 >= 0.71 < 0.76  := 0.82 >= 0.76 < 0.81  := 0.86 >= 0.81 < 0.86  := 0.90 >= 0.86 < 0.91  := 0.94 >= 0.91 < 0.96  := 0.98 >= 0.96 < 1.01  := 1.00
#MATLAB_.2F_Octave
MATLAB / Octave
function y = rescale(x)   L = [0,.06:.05:1.02]; V = [.1,.18,.26,.32,.38,.44,.50,.54,.58,.62,.66,.70,.74,.78,.82,.86,.9,.94,.98,1];   y = x; for k=1:numel(x); y(k) = V(sum(L<=x(k))); end; end;   t=0:0.001:1; plot(t,rescale(t));
http://rosettacode.org/wiki/Price_fraction
Price fraction
A friend of mine runs a pharmacy.   He has a specialized function in his Dispensary application which receives a decimal value of currency and replaces it to a standard value.   This value is regulated by a government department. Task Given a floating point value between   0.00   and   1.00,   rescale according to the following table: >= 0.00 < 0.06  := 0.10 >= 0.06 < 0.11  := 0.18 >= 0.11 < 0.16  := 0.26 >= 0.16 < 0.21  := 0.32 >= 0.21 < 0.26  := 0.38 >= 0.26 < 0.31  := 0.44 >= 0.31 < 0.36  := 0.50 >= 0.36 < 0.41  := 0.54 >= 0.41 < 0.46  := 0.58 >= 0.46 < 0.51  := 0.62 >= 0.51 < 0.56  := 0.66 >= 0.56 < 0.61  := 0.70 >= 0.61 < 0.66  := 0.74 >= 0.66 < 0.71  := 0.78 >= 0.71 < 0.76  := 0.82 >= 0.76 < 0.81  := 0.86 >= 0.81 < 0.86  := 0.90 >= 0.86 < 0.91  := 0.94 >= 0.91 < 0.96  := 0.98 >= 0.96 < 1.01  := 1.00
#Mercury
Mercury
:- module price. :- interface. :- import_module int. :- type price == int. :- func standard(price) = price.   :- implementation. :- import_module require, list.   standard(P) = SP :- require(P >= 0, "P must be positive"), Cents = P `mod` 100, P + adjust(Cents) = SP.   :- func adjust(int) = int. adjust(Cents) = adjust(Cents, rules).   :- func adjust(int, list(price_rule)) = int. adjust(_, []) = unexpected("price", "adjust/2", "exhausted rules"). adjust(N, [rule(Low, High, To)|T]) = R :- ( N >= Low, N < High -> To - N = R ; adjust(N, T) = R ).   :- type price_rule ---> rule(int, int, int). :- func rules = list(price_rule). rules = [rule(00, 06, 10), rule(06, 11, 18), rule(11, 16, 26), rule(16, 21, 32), rule(21, 26, 38), rule(26, 31, 44), rule(31, 36, 50), rule(36, 41, 54), rule(41, 46, 58), rule(46, 51, 62), rule(51, 56, 66), rule(56, 61, 70), rule(61, 66, 74), rule(66, 71, 78), rule(71, 76, 82), rule(76, 81, 86), rule(81, 86, 90), rule(86, 91, 94), rule(91, 96, 98), rule(96, 101, 100)].
http://rosettacode.org/wiki/Proper_divisors
Proper divisors
The   proper divisors   of a positive integer N are those numbers, other than N itself, that divide N without remainder. For N > 1 they will always include 1,   but for N == 1 there are no proper divisors. Examples The proper divisors of     6     are   1, 2, and 3. The proper divisors of   100   are   1, 2, 4, 5, 10, 20, 25, and 50. Task Create a routine to generate all the proper divisors of a number. use it to show the proper divisors of the numbers 1 to 10 inclusive. Find a number in the range 1 to 20,000 with the most proper divisors. Show the number and just the count of how many proper divisors it has. Show all output here. Related tasks   Amicable pairs   Abundant, deficient and perfect number classifications   Aliquot sequence classifications   Factors of an integer   Prime decomposition
#PicoLisp
PicoLisp
# Generate all proper divisors. (de propdiv (N) (head -1 (filter '((X) (=0 (% N X))) (range 1 N) )) )   # Obtaining the values from 1 to 10 inclusive. (mapcar propdiv (range 1 10)) # Output: # (NIL (1) (1) (1 2) (1) (1 2 3) (1) (1 2 4) (1 3) (1 2 5))
http://rosettacode.org/wiki/Priority_queue
Priority queue
A priority queue is somewhat similar to a queue, with an important distinction: each item is added to a priority queue with a priority level, and will be later removed from the queue with the highest priority element first. That is, the items are (conceptually) stored in the queue in priority order instead of in insertion order. Task Create a priority queue.   The queue must support at least two operations:   Insertion.   An element is added to the queue with a priority (a numeric value).   Top item removal.   Deletes the element or one of the elements with the current top priority and return it. Optionally, other operations may be defined, such as peeking (find what current top priority/top element is), merging (combining two priority queues into one), etc. To test your implementation, insert a number of elements into the queue, each with some random priority. Then dequeue them sequentially; now the elements should be sorted by priority. You can use the following task/priority items as input data: Priority Task ══════════ ════════════════ 3 Clear drains 4 Feed cat 5 Make tea 1 Solve RC tasks 2 Tax return The implementation should try to be efficient.   A typical implementation has   O(log n)   insertion and extraction time,   where   n   is the number of items in the queue. You may choose to impose certain limits such as small range of allowed priority levels, limited capacity, etc.   If so, discuss the reasons behind it.
#R
R
PriorityQueue <- function() { keys <- values <- NULL insert <- function(key, value) { ord <- findInterval(key, keys) keys <<- append(keys, key, ord) values <<- append(values, value, ord) } pop <- function() { head <- list(key=keys[1],value=values[[1]]) values <<- values[-1] keys <<- keys[-1] return(head) } empty <- function() length(keys) == 0 environment() }   pq <- PriorityQueue() pq$insert(3, "Clear drains") pq$insert(4, "Feed cat") pq$insert(5, "Make tea") pq$insert(1, "Solve RC tasks") pq$insert(2, "Tax return") while(!pq$empty()) { with(pq$pop(), cat(key,":",value,"\n")) }
http://rosettacode.org/wiki/Prime_decomposition
Prime decomposition
The prime decomposition of a number is defined as a list of prime numbers which when all multiplied together, are equal to that number. Example 12 = 2 × 2 × 3, so its prime decomposition is {2, 2, 3} Task Write a function which returns an array or collection which contains the prime decomposition of a given number   n {\displaystyle n}   greater than   1. If your language does not have an isPrime-like function available, you may assume that you have a function which determines whether a number is prime (note its name before your code). If you would like to test code from this task, you may use code from trial division or the Sieve of Eratosthenes. Note: The program must not be limited by the word size of your computer or some other artificial limit; it should work for any number regardless of size (ignoring the physical limits of RAM etc). Related tasks   count in factors   factors of an integer   Sieve of Eratosthenes   primality by trial division   factors of a Mersenne number   trial factoring of a Mersenne number   partition an integer X into N primes   sequence of primes by Trial Division
#JavaScript
JavaScript
function run_factorize(input, output) { var n = new BigInteger(input.value, 10); var TWO = new BigInteger("2", 10); var divisor = new BigInteger("3", 10); var prod = false;   if (n.compareTo(TWO) < 0) return;   output.value = "";   while (true) { var qr = n.divideAndRemainder(TWO); if (qr[1].equals(BigInteger.ZERO)) { if (prod) output.value += "*"; else prod = true; output.value += "2"; n = qr[0]; } else break; }   while (!n.equals(BigInteger.ONE)) { var qr = n.divideAndRemainder(divisor); if (qr[1].equals(BigInteger.ZERO)) { if (prod) output.value += "*"; else prod = true; output.value += divisor; n = qr[0]; } else divisor = divisor.add(TWO); } }
http://rosettacode.org/wiki/Plot_coordinate_pairs
Plot coordinate pairs
Task Plot a function represented as    x,  y    numerical arrays. Post the resulting image for the following input arrays (taken from Python's Example section on Time a function): x = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}; y = {2.7, 2.8, 31.4, 38.1, 58.0, 76.2, 100.5, 130.0, 149.3, 180.0}; This task is intended as a subtask for Measure relative performance of sorting algorithms implementations.
#Ol
Ol
  ; define input arrays (define x '(0 1 2 3 4 5 6 7 8 9)) (define y '(2.7 2.8 31.4 38.1 58.0 76.2 100.5 130.0 149.3 180.0))   ; render (import (lib gl2)) (glOrtho 0 10 0 200 0 1)   (gl:set-renderer (lambda (mouse) (glClear GL_COLOR_BUFFER_BIT) (glColor3f 0 1 0) (glBegin GL_LINE_STRIP) (map glVertex2f x y) (glEnd)))  
http://rosettacode.org/wiki/Polymorphism
Polymorphism
Task Create two classes   Point(x,y)   and   Circle(x,y,r)   with a polymorphic function print, accessors for (x,y,r), copy constructor, assignment and destructor and every possible default constructors
#Nim
Nim
type Point = object x, y: float   Circle = object center: Point radius: float   # Constructors proc createPoint(x, y = 0.0): Point = result.x = x result.y = y   proc createCircle(x, y = 0.0, radius = 1.0): Circle = result.center.x = x result.center.y = y result.radius = radius   var p1 = createPoint() echo "p1: ", p1 # We use the default $ operator for printing var p2 = createPoint(3, 4.2) var p3 = createPoint(x = 2) var p4 = createPoint(y = 2.5)   p2 = p4 p3 = createPoint()   var c1 = createCircle() echo "c1: ", c1 var c2 = createCircle(2, 0.5, 4.2) var c3 = createCircle(x = 2.1, y = 2) var c4 = createCircle(radius = 10)   c1.center.x = 12 c1.radius = 5.2
http://rosettacode.org/wiki/Polymorphism
Polymorphism
Task Create two classes   Point(x,y)   and   Circle(x,y,r)   with a polymorphic function print, accessors for (x,y,r), copy constructor, assignment and destructor and every possible default constructors
#Objeck
Objeck
  bundle Default { class Point { @x : Int; @y : Int;   New() { @x := 0; @y := 0; }   New(x : Int, y : Int) { @x := x; @y := y; }   New(p : Point) { @x := p->GetX(); @y := p->GetY(); }   method : public : GetX() ~ Int { return @x; }   method : public : GetY() ~ Int { return @y; }   method : public : SetX(x : Int) ~ Nil { @x := x; }   method : public : SetY(y : Int) ~ Nil { @y := y; }   method : public : Print() ~ Nil { "Point"->PrintLine(); } }     class Circle from Point { @r : Int;   New() { Parent(); @r := 0; }   New(p : Point) { Parent(p); @r := 0; }   New(c : Circle) { Parent(c->GetX(), c->GetY()); @r := c->GetR(); }   method : public : GetR() ~ Int { return @r; }   method : public : SetR(r : Int) ~ Nil { @r := r; }   method : public : Print() ~ Nil { "Circle"->PrintLine(); } }   class Poly { function : Main(args : String[]) ~ Nil { p := Point->New(); c := Circle->New(); p->Print(); c->Print(); } } }  
http://rosettacode.org/wiki/Poker_hand_analyser
Poker hand analyser
Task Create a program to parse a single five card poker hand and rank it according to this list of poker hands. A poker hand is specified as a space separated list of five playing cards. Each input card has two characters indicating face and suit. Example 2d       (two of diamonds). Faces are:    a, 2, 3, 4, 5, 6, 7, 8, 9, 10, j, q, k Suits are:    h (hearts),   d (diamonds),   c (clubs),   and   s (spades),   or alternatively,   the unicode card-suit characters:    ♥ ♦ ♣ ♠ Duplicate cards are illegal. The program should analyze a single hand and produce one of the following outputs: straight-flush four-of-a-kind full-house flush straight three-of-a-kind two-pair one-pair high-card invalid Examples 2♥ 2♦ 2♣ k♣ q♦: three-of-a-kind 2♥ 5♥ 7♦ 8♣ 9♠: high-card a♥ 2♦ 3♣ 4♣ 5♦: straight 2♥ 3♥ 2♦ 3♣ 3♦: full-house 2♥ 7♥ 2♦ 3♣ 3♦: two-pair 2♥ 7♥ 7♦ 7♣ 7♠: four-of-a-kind 10♥ j♥ q♥ k♥ a♥: straight-flush 4♥ 4♠ k♠ 5♦ 10♠: one-pair q♣ 10♣ 7♣ 6♣ q♣: invalid The programs output for the above examples should be displayed here on this page. Extra credit use the playing card characters introduced with Unicode 6.0 (U+1F0A1 - U+1F0DE). allow two jokers use the symbol   joker duplicates would be allowed (for jokers only) five-of-a-kind would then be the highest hand More extra credit examples joker 2♦ 2♠ k♠ q♦: three-of-a-kind joker 5♥ 7♦ 8♠ 9♦: straight joker 2♦ 3♠ 4♠ 5♠: straight joker 3♥ 2♦ 3♠ 3♦: four-of-a-kind joker 7♥ 2♦ 3♠ 3♦: three-of-a-kind joker 7♥ 7♦ 7♠ 7♣: five-of-a-kind joker j♥ q♥ k♥ A♥: straight-flush joker 4♣ k♣ 5♦ 10♠: one-pair joker k♣ 7♣ 6♣ 4♣: flush joker 2♦ joker 4♠ 5♠: straight joker Q♦ joker A♠ 10♠: straight joker Q♦ joker A♦ 10♦: straight-flush joker 2♦ 2♠ joker q♦: four-of-a-kind Related tasks Playing cards Card shuffles Deal cards_for_FreeCell War Card_Game Go Fish
#Scala
Scala
val faces = "23456789TJQKA" val suits = "CHSD" sealed trait Card object Joker extends Card case class RealCard(face: Int, suit: Char) extends Card val allRealCards = for { face <- 0 until faces.size suit <- suits } yield RealCard(face, suit)   def parseCard(str: String): Card = { if (str == "joker") { Joker } else { RealCard(faces.indexOf(str(0)), str(1)) } }   def parseHand(str: String): List[Card] = { str.split(" ").map(parseCard).toList }   trait HandType { def name: String def check(hand: List[RealCard]): Boolean }   case class And(x: HandType, y: HandType, name: String) extends HandType { def check(hand: List[RealCard]) = x.check(hand) && y.check(hand) }   object Straight extends HandType { val name = "straight" def check(hand: List[RealCard]): Boolean = { val faces = hand.map(_.face).toSet faces.size == 5 && (faces.min == faces.max - 4 || faces == Set(0, 1, 2, 3, 12)) } }   object Flush extends HandType { val name = "flush" def check(hand: List[RealCard]): Boolean = { hand.map(_.suit).toSet.size == 1 } }   case class NOfAKind(n: Int, name: String = "", nOccur: Int = 1) extends HandType { def check(hand: List[RealCard]): Boolean = { hand.groupBy(_.face).values.count(_.size == n) >= nOccur } }   val allHandTypes = List( NOfAKind(5, "five-of-a-kind"), And(Straight, Flush, "straight-flush"), NOfAKind(4, "four-of-a-kind"), And(NOfAKind(3), NOfAKind(2), "full-house"), Flush, Straight, NOfAKind(3, "three-of-a-kind"), NOfAKind(2, "two-pair", 2), NOfAKind(2, "one-pair") )   def possibleRealHands(hand: List[Card]): List[List[RealCard]] = { val realCards = hand.collect { case r: RealCard => r } val nJokers = hand.count(_ == Joker) allRealCards.toList.combinations(nJokers).map(_ ++ realCards).toList }   def analyzeHand(hand: List[Card]): String = { val possibleHands = possibleRealHands(hand) allHandTypes.find(t => possibleHands.exists(t.check)).map(_.name).getOrElse("high-card") }
http://rosettacode.org/wiki/Population_count
Population count
Population count You are encouraged to solve this task according to the task description, using any language you may know. The   population count   is the number of   1s   (ones)   in the binary representation of a non-negative integer. Population count   is also known as:   pop count   popcount   sideways sum   bit summation   Hamming weight For example,   5   (which is   101   in binary)   has a population count of   2. Evil numbers   are non-negative integers that have an   even   population count. Odious numbers     are  positive integers that have an    odd   population count. Task write a function (or routine) to return the population count of a non-negative integer. all computation of the lists below should start with   0   (zero indexed). display the   pop count   of the   1st   thirty powers of   3       (30,   31,   32,   33,   34,   ∙∙∙   329). display the   1st   thirty     evil     numbers. display the   1st   thirty   odious   numbers. display each list of integers on one line   (which may or may not include a title),   each set of integers being shown should be properly identified. See also The On-Line Encyclopedia of Integer Sequences:   A000120 population count. The On-Line Encyclopedia of Integer Sequences:   A000069 odious numbers. The On-Line Encyclopedia of Integer Sequences:   A001969 evil numbers.
#Gambas
Gambas
Public Sub Main() Dim sEvil, sOdious As String 'To store the output for printing Evil and Odious Dim iCount, iEvil, iOdious As Integer 'Counters   Print "First 30 numbers ^3\t"; 'Print title   For iCount = 0 To 29 'Count 30 times Print Len(Replace(Bin(3 ^ iCount), "0", ""));; 'Get the Bin of the number, take out the '0's and the remaining Next 'length is the Population count e.g. 3^2=9, Bin=1001, remove '0's='11', length=2   iCount = 0 'Reset iCount   Repeat 'Repeat/Until loop If Even(Len(Replace(Bin(iCount), "0", ""))) Then 'If (as above) the result is Even then sEvil &= Str(icount) & " " 'Add it to sEvil Inc iEvil 'Increase iEvil End If If Odd(Len(Replace(Bin(iCount), "0", ""))) Then 'If (as above) the result is Odd then sOdious &= Str(icount) & " " 'Add it to sOdious Inc iOdious 'Increase iOdious End If Inc iCount 'Increase iCount Until iEvil = 30 And iOdious = 30 'Until both iEvil and iOdious = 30 then exit the loop   Print "\n1st 30 Evil numbers =\t" & sEvil 'Print Evil Print "1st 30 Odious numbers =\t" & sOdious 'Print Odious   End
http://rosettacode.org/wiki/Polynomial_long_division
Polynomial long division
This page uses content from Wikipedia. The original article was at Polynomial long division. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance) In algebra, polynomial long division is an algorithm for dividing a polynomial by another polynomial of the same or lower degree. Let us suppose a polynomial is represented by a vector, x {\displaystyle x} (i.e., an ordered collection of coefficients) so that the i {\displaystyle i} th element keeps the coefficient of x i {\displaystyle x^{i}} , and the multiplication by a monomial is a shift of the vector's elements "towards right" (injecting ones from left) followed by a multiplication of each element by the coefficient of the monomial. Then a pseudocode for the polynomial long division using the conventions described above could be: degree(P): return the index of the last non-zero element of P; if all elements are 0, return -∞ polynomial_long_division(N, D) returns (q, r): // N, D, q, r are vectors if degree(D) < 0 then error q ← 0 while degree(N) ≥ degree(D) d ← D shifted right by (degree(N) - degree(D)) q(degree(N) - degree(D)) ← N(degree(N)) / d(degree(d)) // by construction, degree(d) = degree(N) of course d ← d * q(degree(N) - degree(D)) N ← N - d endwhile r ← N return (q, r) Note: vector * scalar multiplies each element of the vector by the scalar; vectorA - vectorB subtracts each element of the vectorB from the element of the vectorA with "the same index". The vectors in the pseudocode are zero-based. Error handling (for allocations or for wrong inputs) is not mandatory. Conventions can be different; in particular, note that if the first coefficient in the vector is the highest power of x for the polynomial represented by the vector, then the algorithm becomes simpler. Example for clarification This example is from Wikipedia, but changed to show how the given pseudocode works. 0 1 2 3 ---------------------- N: -42 0 -12 1 degree = 3 D: -3 1 0 0 degree = 1 d(N) - d(D) = 2, so let's shift D towards right by 2: N: -42 0 -12 1 d: 0 0 -3 1 N(3)/d(3) = 1, so d is unchanged. Now remember that "shifting by 2" is like multiplying by x2, and the final multiplication (here by 1) is the coefficient of this monomial. Let's store this into q: 0 1 2 --------------- q: 0 0 1 now compute N - d, and let it be the "new" N, and let's loop N: -42 0 -9 0 degree = 2 D: -3 1 0 0 degree = 1 d(N) - d(D) = 1, right shift D by 1 and let it be d N: -42 0 -9 0 d: 0 -3 1 0 * -9/1 = -9 q: 0 -9 1 d: 0 27 -9 0 N ← N - d N: -42 -27 0 0 degree = 1 D: -3 1 0 0 degree = 1 looping again... d(N)-d(D)=0, so no shift is needed; we multiply D by -27 (= -27/1) storing the result in d, then q: -27 -9 1 and N: -42 -27 0 0 - d: 81 -27 0 0 = N: -123 0 0 0 (last N) d(N) < d(D), so now r ← N, and the result is: 0 1 2 ------------- q: -27 -9 1 → x2 - 9x - 27 r: -123 0 0 → -123 Related task   Polynomial derivative
#Ruby
Ruby
def polynomial_long_division(numerator, denominator) dd = degree(denominator) raise ArgumentError, "denominator is zero" if dd < 0 if dd == 0 return [multiply(numerator, 1.0/denominator[0]), [0]*numerator.length] end   q = [0] * numerator.length   while (dn = degree(numerator)) >= dd d = shift_right(denominator, dn - dd) q[dn-dd] = numerator[dn] / d[degree(d)] d = multiply(d, q[dn-dd]) numerator = subtract(numerator, d) end   [q, numerator] end   def degree(ary) idx = ary.rindex(&:nonzero?) idx ? idx : -1 end   def shift_right(ary, n) [0]*n + ary[0, ary.length - n] end   def subtract(a1, a2) a1.zip(a2).collect {|v1,v2| v1 - v2} end   def multiply(ary, num) ary.collect {|x| x * num} end   f = [-42, 0, -12, 1] g = [-3, 1, 0, 0] q, r = polynomial_long_division(f, g) puts "#{f} / #{g} => #{q} remainder #{r}" # => [-42, 0, -12, 1] / [-3, 1, 0, 0] => [-27, -9, 1, 0] remainder [-123, 0, 0, 0]   g = [-3, 1, 1, 0] q, r = polynomial_long_division(f, g) puts "#{f} / #{g} => #{q} remainder #{r}" # => [-42, 0, -12, 1] / [-3, 1, 1, 0] => [-13, 1, 0, 0] remainder [-81, 16, 0, 0]
http://rosettacode.org/wiki/Polynomial_regression
Polynomial regression
Find an approximating polynomial of known degree for a given data. Example: For input data: x = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; y = {1, 6, 17, 34, 57, 86, 121, 162, 209, 262, 321}; The approximating polynomial is: 3 x2 + 2 x + 1 Here, the polynomial's coefficients are (3, 2, 1). This task is intended as a subtask for Measure relative performance of sorting algorithms implementations.
#Python
Python
>>> x = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>> y = [1, 6, 17, 34, 57, 86, 121, 162, 209, 262, 321] >>> coeffs = numpy.polyfit(x,y,deg=2) >>> coeffs array([ 3., 2., 1.])
http://rosettacode.org/wiki/Power_set
Power set
A   set   is a collection (container) of certain values, without any particular order, and no repeated values. It corresponds with a finite set in mathematics. A set can be implemented as an associative array (partial mapping) in which the value of each key-value pair is ignored. Given a set S, the power set (or powerset) of S, written P(S), or 2S, is the set of all subsets of S. Task By using a library or built-in set type, or by defining a set type with necessary operations, write a function with a set S as input that yields the power set 2S of S. For example, the power set of     {1,2,3,4}     is {{}, {1}, {2}, {1,2}, {3}, {1,3}, {2,3}, {1,2,3}, {4}, {1,4}, {2,4}, {1,2,4}, {3,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}. For a set which contains n elements, the corresponding power set has 2n elements, including the edge cases of empty set. The power set of the empty set is the set which contains itself (20 = 1): P {\displaystyle {\mathcal {P}}} ( ∅ {\displaystyle \varnothing } ) = { ∅ {\displaystyle \varnothing } } And the power set of the set which contains only the empty set, has two subsets, the empty set and the set which contains the empty set (21 = 2): P {\displaystyle {\mathcal {P}}} ({ ∅ {\displaystyle \varnothing } }) = { ∅ {\displaystyle \varnothing } , { ∅ {\displaystyle \varnothing } } } Extra credit: Demonstrate that your language supports these last two powersets.
#JavaScript
JavaScript
function powerset(ary) { var ps = [[]]; for (var i=0; i < ary.length; i++) { for (var j = 0, len = ps.length; j < len; j++) { ps.push(ps[j].concat(ary[i])); } } return ps; }   var res = powerset([1,2,3,4]);   load('json2.js'); print(JSON.stringify(res));
http://rosettacode.org/wiki/Primality_by_trial_division
Primality by trial division
Task Write a boolean function that tells whether a given integer is prime. Remember that   1   and all non-positive numbers are not prime. Use trial division. Even numbers greater than   2   may be eliminated right away. A loop from   3   to   √ n    will suffice,   but other loops are allowed. Related tasks   count in factors   prime decomposition   AKS test for primes   factors of an integer   Sieve of Eratosthenes   factors of a Mersenne number   trial factoring of a Mersenne number   partition an integer X into N primes   sequence of primes by Trial Division
#Elixir
Elixir
defmodule RC do def is_prime(2), do: true def is_prime(n) when n<2 or rem(n,2)==0, do: false def is_prime(n), do: is_prime(n,3)   def is_prime(n,k) when n<k*k, do: true def is_prime(n,k) when rem(n,k)==0, do: false def is_prime(n,k), do: is_prime(n,k+2) end   IO.inspect for n <- 1..50, RC.is_prime(n), do: n
http://rosettacode.org/wiki/Price_fraction
Price fraction
A friend of mine runs a pharmacy.   He has a specialized function in his Dispensary application which receives a decimal value of currency and replaces it to a standard value.   This value is regulated by a government department. Task Given a floating point value between   0.00   and   1.00,   rescale according to the following table: >= 0.00 < 0.06  := 0.10 >= 0.06 < 0.11  := 0.18 >= 0.11 < 0.16  := 0.26 >= 0.16 < 0.21  := 0.32 >= 0.21 < 0.26  := 0.38 >= 0.26 < 0.31  := 0.44 >= 0.31 < 0.36  := 0.50 >= 0.36 < 0.41  := 0.54 >= 0.41 < 0.46  := 0.58 >= 0.46 < 0.51  := 0.62 >= 0.51 < 0.56  := 0.66 >= 0.56 < 0.61  := 0.70 >= 0.61 < 0.66  := 0.74 >= 0.66 < 0.71  := 0.78 >= 0.71 < 0.76  := 0.82 >= 0.76 < 0.81  := 0.86 >= 0.81 < 0.86  := 0.90 >= 0.86 < 0.91  := 0.94 >= 0.91 < 0.96  := 0.98 >= 0.96 < 1.01  := 1.00
#MUMPS
MUMPS
PRICFRAC(X)  ;Outputs a specified value dependent upon the input value  ;The non-inclusive upper limits are encoded in the PFMAX string, and the values  ;to convert to are encoded in the PFRES string. NEW PFMAX,PFRES,I,RESULT SET PFMAX=".06^.11^.16^.21^.26^.31^.36^.41^.46^.51^.56^.61^.66^.71^.76^.81^.86^.91^.96^1.01" SET PFRES=".10^.18^.26^.32^.38^.44^.50^.54^.58^.62^.66^.70^.74^.78^.82^.86^.90^.94^.98^1.00" Q:(X<0)!(X>1.01) "" FOR I=1:1:$LENGTH(PFMAX,"^") Q:($DATA(RESULT)'=0) SET:X<$P(PFMAX,"^",I) RESULT=$P(PFRES,"^",I) KILL PFMAX,PFRES,I QUIT RESULT
http://rosettacode.org/wiki/Price_fraction
Price fraction
A friend of mine runs a pharmacy.   He has a specialized function in his Dispensary application which receives a decimal value of currency and replaces it to a standard value.   This value is regulated by a government department. Task Given a floating point value between   0.00   and   1.00,   rescale according to the following table: >= 0.00 < 0.06  := 0.10 >= 0.06 < 0.11  := 0.18 >= 0.11 < 0.16  := 0.26 >= 0.16 < 0.21  := 0.32 >= 0.21 < 0.26  := 0.38 >= 0.26 < 0.31  := 0.44 >= 0.31 < 0.36  := 0.50 >= 0.36 < 0.41  := 0.54 >= 0.41 < 0.46  := 0.58 >= 0.46 < 0.51  := 0.62 >= 0.51 < 0.56  := 0.66 >= 0.56 < 0.61  := 0.70 >= 0.61 < 0.66  := 0.74 >= 0.66 < 0.71  := 0.78 >= 0.71 < 0.76  := 0.82 >= 0.76 < 0.81  := 0.86 >= 0.81 < 0.86  := 0.90 >= 0.86 < 0.91  := 0.94 >= 0.91 < 0.96  := 0.98 >= 0.96 < 1.01  := 1.00
#NetRexx
NetRexx
/* NetRexx */ options replace format comments java crossref symbols nobinary   runSample(arg) return   -- ----------------------------------------------------------------------------- method runSample(arg) public static parse arg in_val . if in_val \= '' then test_vals = [in_val] else test_vals = getTestData()   say 'Input Adjustment' loop tv = 0 to test_vals.length - 1 in_val = test_vals[tv] adjust = priceFraction(in_val) say in_val.format(null, 2).right(5) adjust.format(null, 2).right(10) end tv   return   -- ----------------------------------------------------------------------------- method priceFraction(in_val) public static out_val = -1 limit_table = getLimitTable() limit_table_K = limit_table.length loop p1 = 0 to limit_table_K - 1 pair = limit_table[p1] hi_limit = pair[0] adjustmt = pair[1] if in_val < hi_limit then do out_val = adjustmt leave p1 end end p1 if out_val = -1 then signal IllegalArgumentException('Input' in_val 'is outside of acceptable range.')   return out_val   -- ----------------------------------------------------------------------------- method getLimitTable() public static returns Rexx[,] limit_table = [ - [0.06, 0.10], [0.11, 0.18], [0.16, 0.26], [0.21, 0.32], [0.26, 0.38], - [0.31, 0.44], [0.36, 0.50], [0.41, 0.54], [0.46, 0.58], [0.51, 0.62], - [0.56, 0.66], [0.61, 0.70], [0.66, 0.74], [0.71, 0.78], [0.76, 0.82], - [0.81, 0.86], [0.86, 0.90], [0.91, 0.94], [0.96, 0.98], [1.01, 1.00] - ] return limit_table   -- ----------------------------------------------------------------------------- method getTestData() private static returns Rexx[] test_vals = Rexx[5] rng = Random(1024) loop tv = 0 to test_vals.length - 1 test_vals[tv] = rng.nextFloat() end tv return test_vals  
http://rosettacode.org/wiki/Proper_divisors
Proper divisors
The   proper divisors   of a positive integer N are those numbers, other than N itself, that divide N without remainder. For N > 1 they will always include 1,   but for N == 1 there are no proper divisors. Examples The proper divisors of     6     are   1, 2, and 3. The proper divisors of   100   are   1, 2, 4, 5, 10, 20, 25, and 50. Task Create a routine to generate all the proper divisors of a number. use it to show the proper divisors of the numbers 1 to 10 inclusive. Find a number in the range 1 to 20,000 with the most proper divisors. Show the number and just the count of how many proper divisors it has. Show all output here. Related tasks   Amicable pairs   Abundant, deficient and perfect number classifications   Aliquot sequence classifications   Factors of an integer   Prime decomposition
#PL.2FI
PL/I
*process source xref; (subrg): cpd: Proc Options(main); p9a=time(); Dcl (p9a,p9b) Pic'(9)9'; Dcl cnt(3) Bin Fixed(31) Init((3)0); Dcl x Bin Fixed(31); Dcl pd(300) Bin Fixed(31); Dcl sumpd Bin Fixed(31); Dcl npd Bin Fixed(31); Dcl hi Bin Fixed(31) Init(0); Dcl (xl(10),xi) Bin Fixed(31); Dcl i Bin Fixed(31); Do x=1 To 10; Call proper_divisors(x,pd,npd); Put Edit(x,' -> ',(pd(i) Do i=1 To npd))(Skip,f(2),a,10(f(2))); End; xi=0; Do x=1 To 20000; Call proper_divisors(x,pd,npd); Select; When(npd>hi) Do; xi=1; xl(1)=x; hi=npd; End; When(npd=hi) Do; xi+=1; xl(xi)=x; End; Otherwise; End; End; Put Edit(hi,' -> ',(xl(i) Do i=1 To xi))(Skip,f(3),a,10(f(6)));   x= 166320; Call proper_divisors(x,pd,npd); Put Edit(x,' -> ',npd)(Skip,f(8),a,f(4)); x=1441440; Call proper_divisors(x,pd,npd); Put Edit(x,' -> ',npd)(Skip,f(8),a,f(4));     p9b=time(); Put Edit((p9b-p9a)/1000,' seconds elapsed')(Skip,f(6,3),a); Return;   proper_divisors: Proc(n,pd,npd); Dcl (n,pd(300),npd) Bin Fixed(31); Dcl (d,delta) Bin Fixed(31); npd=0; If n>1 Then Do; If mod(n,2)=1 Then /* odd number */ delta=2; Else /* even number */ delta=1; Do d=1 To n/2 By delta; If mod(n,d)=0 Then Do; npd+=1; pd(npd)=d; End; End; End; End;   End;
http://rosettacode.org/wiki/Priority_queue
Priority queue
A priority queue is somewhat similar to a queue, with an important distinction: each item is added to a priority queue with a priority level, and will be later removed from the queue with the highest priority element first. That is, the items are (conceptually) stored in the queue in priority order instead of in insertion order. Task Create a priority queue.   The queue must support at least two operations:   Insertion.   An element is added to the queue with a priority (a numeric value).   Top item removal.   Deletes the element or one of the elements with the current top priority and return it. Optionally, other operations may be defined, such as peeking (find what current top priority/top element is), merging (combining two priority queues into one), etc. To test your implementation, insert a number of elements into the queue, each with some random priority. Then dequeue them sequentially; now the elements should be sorted by priority. You can use the following task/priority items as input data: Priority Task ══════════ ════════════════ 3 Clear drains 4 Feed cat 5 Make tea 1 Solve RC tasks 2 Tax return The implementation should try to be efficient.   A typical implementation has   O(log n)   insertion and extraction time,   where   n   is the number of items in the queue. You may choose to impose certain limits such as small range of allowed priority levels, limited capacity, etc.   If so, discuss the reasons behind it.
#Racket
Racket
  #lang racket (require data/heap)   (define pq (make-heap (λ(x y) (<= (second x) (second y)))))   (define (insert! x pri) (heap-add! pq (list pri x)))   (define (remove-min!) (begin0 (first (heap-min pq)) (heap-remove-min! pq)))   (insert! 3 "Clear drains") (insert! 4 "Feed cat") (insert! 5 "Make tea") (insert! 1 "Solve RC tasks") (insert! 2 "Tax return")   (remove-min!) (remove-min!) (remove-min!) (remove-min!) (remove-min!)  
http://rosettacode.org/wiki/Prime_decomposition
Prime decomposition
The prime decomposition of a number is defined as a list of prime numbers which when all multiplied together, are equal to that number. Example 12 = 2 × 2 × 3, so its prime decomposition is {2, 2, 3} Task Write a function which returns an array or collection which contains the prime decomposition of a given number   n {\displaystyle n}   greater than   1. If your language does not have an isPrime-like function available, you may assume that you have a function which determines whether a number is prime (note its name before your code). If you would like to test code from this task, you may use code from trial division or the Sieve of Eratosthenes. Note: The program must not be limited by the word size of your computer or some other artificial limit; it should work for any number regardless of size (ignoring the physical limits of RAM etc). Related tasks   count in factors   factors of an integer   Sieve of Eratosthenes   primality by trial division   factors of a Mersenne number   trial factoring of a Mersenne number   partition an integer X into N primes   sequence of primes by Trial Division
#jq
jq
def factors: . as $in | [2, $in, false] | recurse( . as [$p, $q, $valid, $s] | if $q == 1 then empty elif $q % $p == 0 then [$p, $q/$p, true] elif $p == 2 then [3, $q, false, $s] else ($s // ($q | sqrt)) as $s | if $p + 2 <= $s then [$p + 2, $q, false, $s] else [$q, 1, true] end end ) | if .[2] then .[0] else empty end ;