task_url stringlengths 30 116 | task_name stringlengths 2 86 | task_description stringlengths 0 14.4k | language_url stringlengths 2 53 | language_name stringlengths 1 52 | code stringlengths 0 61.9k |
|---|---|---|---|---|---|
http://rosettacode.org/wiki/Parametric_polymorphism | Parametric polymorphism | Parametric Polymorphism
type variables
Task
Write a small example for a type declaration that is parametric over another type, together with a short bit of code (and its type signature) that uses it.
A good example is a container type, let's say a binary tree, together with some function that traverses the tree, say, a map-function that operates on every element of the tree.
This language feature only applies to statically-typed languages.
| #Julia | Julia | module BinaryTrees
mutable struct BinaryTree{V}
v::V
l::Union{BinaryTree{V}, Nothing}
r::Union{BinaryTree{V}, Nothing}
end
BinaryTree(v) = BinaryTree(v, nothing, nothing)
map(f, bt::BinaryTree) = BinaryTree(f(bt.v), map(f, bt.l), map(f, bt.r))
map(f, bt::Nothing) = nothing
let inttree = BinaryTree(
0,
BinaryTree(
1,
BinaryTree(3),
BinaryTree(5),
),
BinaryTree(
2,
BinaryTree(4),
nothing,
),
)
map(x -> 2x^2, inttree)
end
let strtree = BinaryTree(
"hello",
BinaryTree(
"world!",
BinaryTree("Julia"),
nothing,
),
BinaryTree(
"foo",
BinaryTree("bar"),
BinaryTree("baz"),
),
)
map(uppercase, strtree)
end
end |
http://rosettacode.org/wiki/Parsing/RPN_to_infix_conversion | Parsing/RPN to infix conversion | Parsing/RPN to infix conversion
You are encouraged to solve this task according to the task description, using any language you may know.
Task
Create a program that takes an RPN representation of an expression formatted as a space separated sequence of tokens and generates the equivalent expression in infix notation.
Assume an input of a correct, space separated, string of tokens
Generate a space separated output string representing the same expression in infix notation
Show how the major datastructure of your algorithm changes with each new token parsed.
Test with the following input RPN strings then print and display the output here.
RPN input
sample output
3 4 2 * 1 5 - 2 3 ^ ^ / +
3 + 4 * 2 / ( 1 - 5 ) ^ 2 ^ 3
1 2 + 3 4 + ^ 5 6 + ^
( ( 1 + 2 ) ^ ( 3 + 4 ) ) ^ ( 5 + 6 )
Operator precedence and operator associativity is given in this table:
operator
precedence
associativity
operation
^
4
right
exponentiation
*
3
left
multiplication
/
3
left
division
+
2
left
addition
-
2
left
subtraction
See also
Parsing/Shunting-yard algorithm for a method of generating an RPN from an infix expression.
Parsing/RPN calculator algorithm for a method of calculating a final value from this output RPN expression.
Postfix to infix from the RubyQuiz site.
| #EchoLisp | EchoLisp |
(require 'hash)
(string-delimiter "")
(define (^ a b ) (expt a b)) ;; add this not-native function
(define-syntax-rule (r-assoc? op) (= op "^"))
(define-syntax-rule (l-assoc? op) (not ( = op "^" )))
(define PRECEDENCES (list->hash
'(("+" . 2) ("-" . 2) ("*" . 3) ("/" . 3) ("//" . 3) ("^" . 4))
(make-hash)))
;; RPN vector or list -> infix tree -> (a op (b op c) d) ..
(define (rpn->infix rpn)
(define S (stack 'S))
(for ((token rpn))
(if (procedure? token)
(let [(op2 (pop S)) (op1 (pop S))]
(unless (and op1 op2) (error "cannot translate expression" rpn))
(push S (list op1 token op2))
)
(push S token ))
(writeln 'token (string token) 'stack (stack->list S)))
(begin0
(pop S) ;; return (top S)
(unless (stack-empty? S) (error "ill-formed rpn" rpn)))
)
;; a node tree is (left op right) or a number
(define-syntax-id _.left (first _)) ; mynode.left expands to (first mynode)
(define-syntax-id _.right (third _))
(define-syntax-id _.op (string (second _ )))
(define-syntax-rule (precedence node) (hash-ref PRECEDENCES (string (second node))))
(define (left-par? node) ; does lhs needs ( lhs ) ?
(cond
[(number? node.left) #f]
[(< (precedence node.left) (precedence node)) #t]
[(and
(r-assoc? node.op)
(= (precedence node.left) (precedence node))) #t]
[else #f]))
(define (right-par? node)
(cond
[(number? node.right) #f]
[(< (precedence node.right) (precedence node)) #t]
[(and
(l-assoc? node.op)
(= (precedence node.right) (precedence node))) #t]
[else #f]))
;; infix tree -> char string
(define (infix->string node)
(cond
[(number? node) (string node)]
[else (let
[(lhs (infix->string node.left))
(rhs (infix->string node.right))]
(when (left-par? node) (set! lhs (string-append "(" lhs ")")))
(when (right-par? node) (set! rhs (string-append "(" rhs ")")))
(string-append lhs " " node.op " " rhs))]))
|
http://rosettacode.org/wiki/Partial_function_application | Partial function application | Partial function application is the ability to take a function of many
parameters and apply arguments to some of the parameters to create a new
function that needs only the application of the remaining arguments to
produce the equivalent of applying all arguments to the original function.
E.g:
Given values v1, v2
Given f(param1, param2)
Then partial(f, param1=v1) returns f'(param2)
And f(param1=v1, param2=v2) == f'(param2=v2) (for any value v2)
Note that in the partial application of a parameter, (in the above case param1), other parameters are not explicitly mentioned. This is a recurring feature of partial function application.
Task
Create a function fs( f, s ) that takes a function, f( n ), of one value and a sequence of values s.
Function fs should return an ordered sequence of the result of applying function f to every value of s in turn.
Create function f1 that takes a value and returns it multiplied by 2.
Create function f2 that takes a value and returns it squared.
Partially apply f1 to fs to form function fsf1( s )
Partially apply f2 to fs to form function fsf2( s )
Test fsf1 and fsf2 by evaluating them with s being the sequence of integers from 0 to 3 inclusive and then the sequence of even integers from 2 to 8 inclusive.
Notes
In partially applying the functions f1 or f2 to fs, there should be no explicit mention of any other parameters to fs, although introspection of fs within the partial applicator to find its parameters is allowed.
This task is more about how results are generated rather than just getting results.
| #Groovy | Groovy | def fs = { fn, values -> values.collect { fn(it) } }
def f1 = { v -> v * 2 }
def f2 = { v -> v ** 2 }
def fsf1 = fs.curry(f1)
def fsf2 = fs.curry(f2) |
http://rosettacode.org/wiki/Partial_function_application | Partial function application | Partial function application is the ability to take a function of many
parameters and apply arguments to some of the parameters to create a new
function that needs only the application of the remaining arguments to
produce the equivalent of applying all arguments to the original function.
E.g:
Given values v1, v2
Given f(param1, param2)
Then partial(f, param1=v1) returns f'(param2)
And f(param1=v1, param2=v2) == f'(param2=v2) (for any value v2)
Note that in the partial application of a parameter, (in the above case param1), other parameters are not explicitly mentioned. This is a recurring feature of partial function application.
Task
Create a function fs( f, s ) that takes a function, f( n ), of one value and a sequence of values s.
Function fs should return an ordered sequence of the result of applying function f to every value of s in turn.
Create function f1 that takes a value and returns it multiplied by 2.
Create function f2 that takes a value and returns it squared.
Partially apply f1 to fs to form function fsf1( s )
Partially apply f2 to fs to form function fsf2( s )
Test fsf1 and fsf2 by evaluating them with s being the sequence of integers from 0 to 3 inclusive and then the sequence of even integers from 2 to 8 inclusive.
Notes
In partially applying the functions f1 or f2 to fs, there should be no explicit mention of any other parameters to fs, although introspection of fs within the partial applicator to find its parameters is allowed.
This task is more about how results are generated rather than just getting results.
| #Haskell | Haskell | fs = map
f1 = (* 2)
f2 = (^ 2)
fsf1 = fs f1
fsf2 = fs f2
main :: IO ()
main = do
print $ fsf1 [0, 1, 2, 3] -- prints [0, 2, 4, 6]
print $ fsf2 [0, 1, 2, 3] -- prints [0, 1, 4, 9]
print $ fsf1 [2, 4, 6, 8] -- prints [4, 8, 12, 16]
print $ fsf2 [2, 4, 6, 8] -- prints [4, 16, 36, 64] |
http://rosettacode.org/wiki/Partition_an_integer_x_into_n_primes | Partition an integer x into n primes | Task
Partition a positive integer X into N distinct primes.
Or, to put it in another way:
Find N unique primes such that they add up to X.
Show in the output section the sum X and the N primes in ascending order separated by plus (+) signs:
• partition 99809 with 1 prime.
• partition 18 with 2 primes.
• partition 19 with 3 primes.
• partition 20 with 4 primes.
• partition 2017 with 24 primes.
• partition 22699 with 1, 2, 3, and 4 primes.
• partition 40355 with 3 primes.
The output could/should be shown in a format such as:
Partitioned 19 with 3 primes: 3+5+11
Use any spacing that may be appropriate for the display.
You need not validate the input(s).
Use the lowest primes possible; use 18 = 5+13, not 18 = 7+11.
You only need to show one solution.
This task is similar to factoring an integer.
Related tasks
Count in factors
Prime decomposition
Factors of an integer
Sieve of Eratosthenes
Primality by trial division
Factors of a Mersenne number
Factors of a Mersenne number
Sequence of primes by trial division
| #Prolog | Prolog | prime_partition(N, 1, [N], Min):-
is_prime(N),
N > Min,
!.
prime_partition(N, K, [P|Rest], Min):-
K > 1,
is_prime(P),
P > Min,
P < N,
K1 is K - 1,
N1 is N - P,
prime_partition(N1, K1, Rest, P),
!.
prime_partition(N, K, Primes):-
prime_partition(N, K, Primes, 1).
print_primes([Prime]):-
!,
writef('%w\n', [Prime]).
print_primes([Prime|Primes]):-
writef('%w + ', [Prime]),
print_primes(Primes).
print_prime_partition(N, K):-
prime_partition(N, K, Primes),
!,
writef('%w = ', [N]),
print_primes(Primes).
print_prime_partition(N, K):-
writef('%w cannot be partitioned into %w primes.\n', [N, K]).
main:-
find_prime_numbers(100000),
print_prime_partition(99809, 1),
print_prime_partition(18, 2),
print_prime_partition(19, 3),
print_prime_partition(20, 4),
print_prime_partition(2017, 24),
print_prime_partition(22699, 1),
print_prime_partition(22699, 2),
print_prime_partition(22699, 3),
print_prime_partition(22699, 4),
print_prime_partition(40355, 3). |
http://rosettacode.org/wiki/Partition_function_P | Partition function P |
The Partition Function P, often notated P(n) is the number of solutions where n∈ℤ can be expressed as the sum of a set of positive integers.
Example
P(4) = 5 because 4 = Σ(4) = Σ(3,1) = Σ(2,2) = Σ(2,1,1) = Σ(1,1,1,1)
P(n) can be expressed as the recurrence relation:
P(n) = P(n-1) +P(n-2) -P(n-5) -P(n-7) +P(n-12) +P(n-15) -P(n-22) -P(n-26) +P(n-35) +P(n-40) ...
The successive numbers in the above equation have the differences: 1, 3, 2, 5, 3, 7, 4, 9, 5, 11, 6, 13, 7, 15, 8 ...
This task may be of popular interest because Mathologer made the video, The hardest "What comes next?" (Euler's pentagonal formula), where he asks the programmers among his viewers to calculate P(666). The video has been viewed more than 100,000 times in the first couple of weeks since its release.
In Wolfram Language, this function has been implemented as PartitionsP.
Task
Write a function which returns the value of PartitionsP(n). Solutions can be iterative or recursive.
Bonus task: show how long it takes to compute PartitionsP(6666).
References
The hardest "What comes next?" (Euler's pentagonal formula) The explanatory video by Mathologer that makes this task a popular interest.
Partition Function P Mathworld entry for the Partition function.
Partition function (number theory) Wikipedia entry for the Partition function.
Related tasks
9 billion names of God the integer
| #Wren | Wren | import "/big" for BigInt
var p = []
var pd = []
var partDiffDiff = Fn.new { |n| (n&1 == 1) ? (n + 1)/2 : n + 1 }
var partDiff = Fn.new { |n|
if (n < 2) return 1
pd[n] = pd[n-1] + partDiffDiff.call(n-1)
return pd[n]
}
var partitionsP = Fn.new { |n|
if (n < 2) return
var psum = BigInt.zero
for (i in 1..n) {
var pdi = partDiff.call(i)
if (pdi > n) break
var sign = (i-1)%4 < 2 ? 1 : -1
psum = psum + p[n-pdi] * sign
}
p[n] = psum
}
var start = System.clock
var N = 6666
p = List.filled(N+1, null)
pd = List.filled(N+1, 0)
p[0] = BigInt.one
p[1] = BigInt.one
pd[0] = 1
pd[1] = 1
for (n in 2..N) partitionsP.call(n)
System.print("p[%(N)] = %(p[N])")
System.print("Took %(System.clock - start) seconds") |
http://rosettacode.org/wiki/Pascal%27s_triangle/Puzzle | Pascal's triangle/Puzzle | This puzzle involves a Pascals Triangle, also known as a Pyramid of Numbers.
[ 151]
[ ][ ]
[40][ ][ ]
[ ][ ][ ][ ]
[ X][11][ Y][ 4][ Z]
Each brick of the pyramid is the sum of the two bricks situated below it.
Of the three missing numbers at the base of the pyramid,
the middle one is the sum of the other two (that is, Y = X + Z).
Task
Write a program to find a solution to this puzzle.
| #Picat | Picat | import cp.
go =>
puzzle(T, X, Y,Z),
foreach(TT in T)
println(TT)
end,
println([x=X,y=Y,z=Z]),
nl,
fail, % are there any more solutions?
nl.
% Port of the Prolog solution
puzzle(Ts, X, Y, Z) :-
Ts = [ [151],
[_, _],
[40, _, _],
[_, _, _, _],
[X, 11, Y, 4, Z]],
Y #= X + Z,
triangle(Ts),
Vs = vars(Ts),
Vs :: 0..10000,
solve(Vs).
triangle([T|Ts]) :-
( Ts = [N|_] -> triangle_(T, N), triangle(Ts) ; true ).
triangle_([], _).
triangle_([T|Ts],[A,B|Rest]) :-
T #= A + B, triangle_(Ts, [B|Rest]). |
http://rosettacode.org/wiki/Pascal%27s_triangle/Puzzle | Pascal's triangle/Puzzle | This puzzle involves a Pascals Triangle, also known as a Pyramid of Numbers.
[ 151]
[ ][ ]
[40][ ][ ]
[ ][ ][ ][ ]
[ X][11][ Y][ 4][ Z]
Each brick of the pyramid is the sum of the two bricks situated below it.
Of the three missing numbers at the base of the pyramid,
the middle one is the sum of the other two (that is, Y = X + Z).
Task
Write a program to find a solution to this puzzle.
| #PicoLisp | PicoLisp | (be number (@N @Max)
(^ @C (box 0))
(repeat)
(or
((^ @ (>= (val (-> @C)) (-> @Max))) T (fail))
((^ @N (inc (-> @C)))) ) )
(be + (@A @B @Sum)
(^ @ (-> @A))
(^ @ (-> @B))
(^ @Sum (+ (-> @A) (-> @B))) )
(be + (@A @B @Sum)
(^ @ (-> @A))
(^ @ (-> @Sum))
(^ @B (- (-> @Sum) (-> @A)))
T
(^ @ (ge0 (-> @B))) )
(be + (@A @B @Sum)
(number @A @Sum)
(^ @B (- (-> @Sum) (-> @A))) )
#{
151
A B
40 C D
E F G H
X 11 Y 4 Z
}#
(be puzzle (@X @Y @Z)
(+ @A @B 151)
(+ 40 @C @A)
(+ @C @D @B)
(+ @E @F 40)
(+ @F @G @C)
(+ @G @H @D)
(+ @X 11 @E)
(+ 11 @Y @F)
(+ @Y 4 @G)
(+ 4 @Z @H)
(+ @X @Z @Y)
T ) |
http://rosettacode.org/wiki/Password_generator | Password generator | Create a password generation program which will generate passwords containing random ASCII characters from the following groups:
lower-case letters: a ──► z
upper-case letters: A ──► Z
digits: 0 ──► 9
other printable characters: !"#$%&'()*+,-./:;<=>?@[]^_{|}~
(the above character list excludes white-space, backslash and grave)
The generated password(s) must include at least one (of each of the four groups):
lower-case letter,
upper-case letter,
digit (numeral), and
one "other" character.
The user must be able to specify the password length and the number of passwords to generate.
The passwords should be displayed or written to a file, one per line.
The randomness should be from a system source or library.
The program should implement a help option or button which should describe the program and options when invoked.
You may also allow the user to specify a seed value, and give the option of excluding visually similar characters.
For example: Il1 O0 5S 2Z where the characters are:
capital eye, lowercase ell, the digit one
capital oh, the digit zero
the digit five, capital ess
the digit two, capital zee
| #Lua | Lua | function randPW (length)
local index, pw, rnd = 0, ""
local chars = {
"ABCDEFGHIJKLMNOPQRSTUVWXYZ",
"abcdefghijklmnopqrstuvwxyz",
"0123456789",
"!\"#$%&'()*+,-./:;<=>?@[]^_{|}~"
}
repeat
index = index + 1
rnd = math.random(chars[index]:len())
if math.random(2) == 1 then
pw = pw .. chars[index]:sub(rnd, rnd)
else
pw = chars[index]:sub(rnd, rnd) .. pw
end
index = index % #chars
until pw:len() >= length
return pw
end
math.randomseed(os.time())
if #arg ~= 2 then
print("\npwgen.lua")
print("=========\n")
print("A Lua script to generate random passwords.\n")
print("Usage: lua pwgen.lua [password length] [number of passwords to generate]\n")
os.exit()
end
for i = 1, arg[2] do print(randPW(tonumber(arg[1]))) end |
http://rosettacode.org/wiki/Permutations | Permutations | Task
Write a program that generates all permutations of n different objects. (Practically numerals!)
Related tasks
Find the missing permutation
Permutations/Derangements
The number of samples of size k from n objects.
With combinations and permutations generation tasks.
Order Unimportant
Order Important
Without replacement
(
n
k
)
=
n
C
k
=
n
(
n
−
1
)
…
(
n
−
k
+
1
)
k
(
k
−
1
)
…
1
{\displaystyle {\binom {n}{k}}=^{n}\operatorname {C} _{k}={\frac {n(n-1)\ldots (n-k+1)}{k(k-1)\dots 1}}}
n
P
k
=
n
⋅
(
n
−
1
)
⋅
(
n
−
2
)
⋯
(
n
−
k
+
1
)
{\displaystyle ^{n}\operatorname {P} _{k}=n\cdot (n-1)\cdot (n-2)\cdots (n-k+1)}
Task: Combinations
Task: Permutations
With replacement
(
n
+
k
−
1
k
)
=
n
+
k
−
1
C
k
=
(
n
+
k
−
1
)
!
(
n
−
1
)
!
k
!
{\displaystyle {\binom {n+k-1}{k}}=^{n+k-1}\operatorname {C} _{k}={(n+k-1)! \over (n-1)!k!}}
n
k
{\displaystyle n^{k}}
Task: Combinations with repetitions
Task: Permutations with repetitions
| #True_BASIC | True BASIC | SUB SWAP(vb1, vb2)
LET temp = vb1
LET vb1 = vb2
LET vb2 = temp
END SUB
LET n = 4
DIM a(4)
DIM c(4)
FOR i = 1 TO n
LET a(i) = i
NEXT i
PRINT
DO
FOR i = 1 TO n
PRINT a(i);
NEXT i
PRINT
LET i = n
DO
LET i = i - 1
LOOP UNTIL (i = 0) OR (a(i) < a(i + 1))
LET j = i + 1
LET k = n
DO WHILE j < k
CALL SWAP (a(j), a(k))
LET j = j + 1
LET k = k - 1
LOOP
IF i > 0 THEN
LET j = i + 1
DO WHILE a(j) < a(i)
LET j = j + 1
LOOP
CALL SWAP (a(i), a(j))
END IF
LOOP UNTIL i = 0
END |
http://rosettacode.org/wiki/Penney%27s_game | Penney's game | Penney's game is a game where two players bet on being the first to see a particular sequence of heads or tails in consecutive tosses of a fair coin.
It is common to agree on a sequence length of three then one player will openly choose a sequence, for example:
Heads, Tails, Heads, or HTH for short.
The other player on seeing the first players choice will choose his sequence. The coin is tossed and the first player to see his sequence in the sequence of coin tosses wins.
Example
One player might choose the sequence HHT and the other THT.
Successive coin tosses of HTTHT gives the win to the second player as the last three coin tosses are his sequence.
Task
Create a program that tosses the coin, keeps score and plays Penney's game against a human opponent.
Who chooses and shows their sequence of three should be chosen randomly.
If going first, the computer should randomly choose its sequence of three.
If going second, the computer should automatically play the optimum sequence.
Successive coin tosses should be shown.
Show output of a game where the computer chooses first and a game where the user goes first here on this page.
See also
The Penney Ante Part 1 (Video).
The Penney Ante Part 2 (Video).
| #zkl | zkl | fcn coinToss{ (0).random(2) and "H" or "T" } // (0).random(2) --> 0<=r<2
reg myBet, yourBet;
if(coinToss()=="H"){ // The toss says I go first
myBet=(3).pump(String,coinToss);
println("I bet ",myBet);
yourBet=ask("Your bet of three (H/T): ")[0,3].toUpper();
}else{
yourBet=ask("Your bet of three (H/T): ")[0,3].toUpper();
myBet=((yourBet[1]=="H") and "T" or "H") + yourBet[0,2];
println("I bet ",myBet);
}
print("Flipping: "); coins:="";
while(1){
print(toss:=coinToss()); coins=coins + toss;
if(Void!=coins.find(yourBet)){ println(" You win!"); break; }
if(Void!=coins.find(myBet)) { println(" I win!"); break; }
// ignore we both won
} |
http://rosettacode.org/wiki/Pathological_floating_point_problems | Pathological floating point problems | Most programmers are familiar with the inexactness of floating point calculations in a binary processor.
The classic example being:
0.1 + 0.2 = 0.30000000000000004
In many situations the amount of error in such calculations is very small and can be overlooked or eliminated with rounding.
There are pathological problems however, where seemingly simple, straight-forward calculations are extremely sensitive to even tiny amounts of imprecision.
This task's purpose is to show how your language deals with such classes of problems.
A sequence that seems to converge to a wrong limit.
Consider the sequence:
v1 = 2
v2 = -4
vn = 111 - 1130 / vn-1 + 3000 / (vn-1 * vn-2)
As n grows larger, the series should converge to 6 but small amounts of error will cause it to approach 100.
Task 1
Display the values of the sequence where n = 3, 4, 5, 6, 7, 8, 20, 30, 50 & 100 to at least 16 decimal places.
n = 3 18.5
n = 4 9.378378
n = 5 7.801153
n = 6 7.154414
n = 7 6.806785
n = 8 6.5926328
n = 20 6.0435521101892689
n = 30 6.006786093031205758530554
n = 50 6.0001758466271871889456140207471954695237
n = 100 6.000000019319477929104086803403585715024350675436952458072592750856521767230266
Task 2
The Chaotic Bank Society is offering a new investment account to their customers.
You first deposit $e - 1 where e is 2.7182818... the base of natural logarithms.
After each year, your account balance will be multiplied by the number of years that have passed, and $1 in service charges will be removed.
So ...
after 1 year, your balance will be multiplied by 1 and $1 will be removed for service charges.
after 2 years your balance will be doubled and $1 removed.
after 3 years your balance will be tripled and $1 removed.
...
after 10 years, multiplied by 10 and $1 removed, and so on.
What will your balance be after 25 years?
Starting balance: $e-1
Balance = (Balance * year) - 1 for 25 years
Balance after 25 years: $0.0399387296732302
Task 3, extra credit
Siegfried Rump's example. Consider the following function, designed by Siegfried Rump in 1988.
f(a,b) = 333.75b6 + a2( 11a2b2 - b6 - 121b4 - 2 ) + 5.5b8 + a/(2b)
compute f(a,b) where a=77617.0 and b=33096.0
f(77617.0, 33096.0) = -0.827396059946821
Demonstrate how to solve at least one of the first two problems, or both, and the third if you're feeling particularly jaunty.
See also;
Floating-Point Arithmetic Section 1.3.2 Difficult problems.
| #Sidef | Sidef | func series (n) {
var (u, v) = (2, -4)
(n-2).times { (u, v) = (v, 111 - 1130/v + 3000/(v * u)) }
return v
}
[(3..8)..., 20, 30, 50, 100].each {|n|
printf("n = %3d -> %s\n", n, series(n))
} |
http://rosettacode.org/wiki/Parallel_brute_force | Parallel brute force | Task
Find, through brute force, the five-letter passwords corresponding with the following SHA-256 hashes:
1. 1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad
2. 3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b
3. 74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f
Your program should naively iterate through all possible passwords consisting only of five lower-case ASCII English letters. It should use concurrent or parallel processing, if your language supports that feature. You may calculate SHA-256 hashes by calling a library or through a custom implementation. Print each matching password, along with its SHA-256 hash.
Related task: SHA-256
| #Common_Lisp | Common Lisp | (defpackage #:parallel-brute-force
(:use #:cl
#:lparallel))
(in-package #:parallel-brute-force)
(defparameter *alphabet* "abcdefghijklmnopqrstuvwxyz")
(defparameter *hash0* "1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad")
(defparameter *hash1* "3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b")
(defparameter *hash2* "74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f")
(defparameter *kernel-size* 7)
(defun sha-256 (input)
(ironclad:byte-array-to-hex-string
(ironclad:digest-sequence :sha256 (ironclad:ascii-string-to-byte-array input))))
(defun call-with-5-char-string (fun first-char)
(loop with str = (make-array 5 :element-type 'character :initial-element first-char)
for c1 across *alphabet*
do (setf (char str 1) c1)
(loop for c2 across *alphabet*
do (setf (char str 2) c2)
(loop for c3 across *alphabet*
do (setf (char str 3) c3)
(loop for c4 across *alphabet*
do (setf (char str 4) c4)
(funcall fun (copy-seq str)))))))
(defmacro with-5-char-string ((str first-char) &body body)
`(call-with-5-char-string (lambda (,str) ,@body) ,first-char))
(defun find-passwords-with (first-char)
(let (results)
(with-5-char-string (str first-char)
(let ((hash (sha-256 str)))
(when (or (string= hash *hash0*) (string= hash *hash1*) (string= hash *hash2*))
(push (list str hash) results))))
(nreverse results)))
(defun find-passwords ()
(setf *kernel* (make-kernel *kernel-size*))
(let ((results (unwind-protect
(pmapcan #'find-passwords-with *alphabet*)
(end-kernel))))
(dolist (r results)
(format t "~A: ~A~%" (first r) (second r))))) |
http://rosettacode.org/wiki/Parallel_calculations | Parallel calculations | Many programming languages allow you to specify computations to be run in parallel.
While Concurrent computing is focused on concurrency,
the purpose of this task is to distribute time-consuming calculations
on as many CPUs as possible.
Assume we have a collection of numbers, and want to find the one
with the largest minimal prime factor
(that is, the one that contains relatively large factors).
To speed up the search, the factorization should be done
in parallel using separate threads or processes,
to take advantage of multi-core CPUs.
Show how this can be formulated in your language.
Parallelize the factorization of those numbers,
then search the returned list of numbers and factors
for the largest minimal factor,
and return that number and its prime factors.
For the prime number decomposition
you may use the solution of the Prime decomposition task.
| #Delphi | Delphi |
program Parallel_calculations;
{$APPTYPE CONSOLE}
uses
System.SysUtils,
System.Threading,
Velthuis.BigIntegers;
function IsPrime(n: BigInteger): Boolean;
var
i: BigInteger;
begin
if n <= 1 then
exit(False);
i := 2;
while i < BigInteger.Sqrt(n) do
begin
if n mod i = 0 then
exit(False);
inc(i);
end;
Result := True;
end;
function GetPrimes(n: BigInteger): TArray<BigInteger>;
var
divisor, next, rest: BigInteger;
begin
divisor := 2;
next := 3;
rest := n;
while (rest <> 1) do
begin
while (rest mod divisor = 0) do
begin
SetLength(Result, Length(Result) + 1);
Result[High(Result)] := divisor;
rest := rest div divisor;
end;
divisor := next;
next := next + 2;
end;
end;
function Min(l: TArray<BigInteger>): BigInteger;
begin
if Length(l) = 0 then
exit(0);
Result := l[0];
for var v in l do
if v < result then
Result := v;
end;
const
n: array of Uint64 = [12757923, 12878611, 12757923, 15808973, 15780709, 197622519];
var
m: BigInteger;
len, j, i: Uint64;
l: TArray<TArray<BigInteger>>;
begin
j := 0;
m := 0;
len := length(n);
SetLength(l, len);
TParallel.for (0, len - 1,
procedure(i: Integer)
begin
l[i] := getPrimes(n[i]);
end);
for i := 0 to len - 1 do
begin
var _min := Min(l[i]);
if _min > m then
begin
m := _min;
j := i;
end;
end;
writeln('Number ', n[j].ToString, ' has largest minimal factor:');
for var v in l[j] do
write(' ', v.ToString);
readln;
end. |
http://rosettacode.org/wiki/Parsing/RPN_calculator_algorithm | Parsing/RPN calculator algorithm | Task
Create a stack-based evaluator for an expression in reverse Polish notation (RPN) that also shows the changes in the stack as each individual token is processed as a table.
Assume an input of a correct, space separated, string of tokens of an RPN expression
Test with the RPN expression generated from the Parsing/Shunting-yard algorithm task:
3 4 2 * 1 5 - 2 3 ^ ^ / +
Print or display the output here
Notes
^ means exponentiation in the expression above.
/ means division.
See also
Parsing/Shunting-yard algorithm for a method of generating an RPN from an infix expression.
Several solutions to 24 game/Solve make use of RPN evaluators (although tracing how they work is not a part of that task).
Parsing/RPN to infix conversion.
Arithmetic evaluation.
| #C | C | #include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
void die(const char *msg)
{
fprintf(stderr, "%s", msg);
abort();
}
#define MAX_D 256
double stack[MAX_D];
int depth;
void push(double v)
{
if (depth >= MAX_D) die("stack overflow\n");
stack[depth++] = v;
}
double pop()
{
if (!depth) die("stack underflow\n");
return stack[--depth];
}
double rpn(char *s)
{
double a, b;
int i;
char *e, *w = " \t\n\r\f";
for (s = strtok(s, w); s; s = strtok(0, w)) {
a = strtod(s, &e);
if (e > s) printf(" :"), push(a);
#define binop(x) printf("%c:", *s), b = pop(), a = pop(), push(x)
else if (*s == '+') binop(a + b);
else if (*s == '-') binop(a - b);
else if (*s == '*') binop(a * b);
else if (*s == '/') binop(a / b);
else if (*s == '^') binop(pow(a, b));
#undef binop
else {
fprintf(stderr, "'%c': ", *s);
die("unknown oeprator\n");
}
for (i = depth; i-- || 0 * putchar('\n'); )
printf(" %g", stack[i]);
}
if (depth != 1) die("stack leftover\n");
return pop();
}
int main(void)
{
char s[] = " 3 4 2 * 1 5 - 2 3 ^ ^ / + ";
printf("%g\n", rpn(s));
return 0;
} |
http://rosettacode.org/wiki/Pancake_numbers | Pancake numbers | Adrian Monk has problems and an assistant, Sharona Fleming. Sharona can deal with most of Adrian's problems except his lack of punctuality paying her remuneration. 2 pay checks down and she prepares him pancakes for breakfast. Knowing that he will be unable to eat them unless they are stacked in ascending order of size she leaves him only a skillet which he can insert at any point in the pile and flip all the above pancakes, repeating until the pile is sorted. Sharona has left the pile of n pancakes such that the maximum number of flips is required. Adrian is determined to do this in as few flips as possible. This sequence n->p(n) is known as the Pancake numbers.
The task is to determine p(n) for n = 1 to 9, and for each show an example requiring p(n) flips.
Sorting_algorithms/Pancake_sort actually performs the sort some giving the number of flips used. How do these compare with p(n)?
Few people know p(20), generously I shall award an extra credit for anyone doing more than p(16).
References
Bill Gates and the pancake problem
A058986
| #F.23 | F# |
// Pancake numbers. Nigel Galloway: October 28th., 2020
let pKake z=let n=[for n in 1..z-1->Array.ofList([n.. -1..0]@[n+1..z-1])]
let e=let rec fG n g=match g with 0->n |_->fG (n*g) (g-1) in fG 1 z
let rec fN i g l=match (Set.count g)-e with 0->(i,List.last l)
|_->let l=l|>List.collect(fun g->[for n in n->List.permute(fun g->n.[g]) g])|>Set.ofList
fN (i+1) (Set.union g l) (Set.difference l g|>Set.toList)
fN 0 (set[[1..z]]) [[1..z]]
[1..9]|>List.iter(fun n->let i,g=pKake n in printfn "Maximum number of flips to sort %d elements is %d. e.g %A->%A" n i g [1..n])
|
http://rosettacode.org/wiki/Parsing/Shunting-yard_algorithm | Parsing/Shunting-yard algorithm | Task
Given the operator characteristics and input from the Shunting-yard algorithm page and tables, use the algorithm to show the changes in the operator stack and RPN output
as each individual token is processed.
Assume an input of a correct, space separated, string of tokens representing an infix expression
Generate a space separated output string representing the RPN
Test with the input string:
3 + 4 * 2 / ( 1 - 5 ) ^ 2 ^ 3
print and display the output here.
Operator precedence is given in this table:
operator
precedence
associativity
operation
^
4
right
exponentiation
*
3
left
multiplication
/
3
left
division
+
2
left
addition
-
2
left
subtraction
Extra credit
Add extra text explaining the actions and an optional comment for the action on receipt of each token.
Note
The handling of functions and arguments is not required.
See also
Parsing/RPN calculator algorithm for a method of calculating a final value from this output RPN expression.
Parsing/RPN to infix conversion.
| #EchoLisp | EchoLisp |
(require 'hash)
(require 'tree)
(define OPS (make-hash))
(hash-set OPS "^" '( 4 #f)) ;; right assoc
(hash-set OPS "*" '( 3 #t)) ;; left assoc
(hash-set OPS "/" '( 3 #t))
(hash-set OPS "+" '( 2 #t))
(hash-set OPS "-" '( 2 #t))
;; helpers
(define (is-right-par? token) (string=? token ")"))
(define (is-left-par? token) (string=? token "("))
(define (is-num? op) (not (hash-ref OPS op))) ;; crude
(define (is-op? op) (hash-ref OPS op))
(define (is-left? op) (second (hash-ref OPS op)))
(define (is-right? op) (not (is-left? op)))
(define (op-prec op) (first (hash-ref OPS op)))
;; Wikipedia algorithm, translated as it is
(define (shunt tokens S Q)
(for ((token tokens))
(writeln "S: " (stack->list S) "Q: " (queue->list Q) "token: "token)
(cond
[(is-left-par? token) (push S token) ]
[(is-right-par? token)
(while (and (stack-top S) (not (is-left-par? (stack-top S))))
(q-push Q ( pop S)))
(when (stack-empty? S) (error 'misplaced-parenthesis "()" ))
(pop S)] ; // left par
[(is-op? token)
(while (and
(is-op? (stack-top S))
(or
(and (is-left? token) (<= (op-prec token) (op-prec (stack-top S))))
(and (is-right? token) (< (op-prec token) (op-prec (stack-top S))))))
(q-push Q (pop S)))
(push S token)]
[(is-num? token) (q-push Q token)]
[else (error 'bad-token token)])) ; for
(while (stack-top S) (q-push Q (pop S))))
(string-delimiter "")
(define (task infix)
(define S (stack 'S))
(define Q (queue 'Q))
(shunt (text-parse infix) S Q)
(writeln 'infix infix)
(writeln 'RPN (queue->list Q)))
|
http://rosettacode.org/wiki/Paraffins | Paraffins |
This organic chemistry task is essentially to implement a tree enumeration algorithm.
Task
Enumerate, without repetitions and in order of increasing size, all possible paraffin molecules (also known as alkanes).
Paraffins are built up using only carbon atoms, which has four bonds, and hydrogen, which has one bond. All bonds for each atom must be used, so it is easiest to think of an alkane as linked carbon atoms forming the "backbone" structure, with adding hydrogen atoms linking the remaining unused bonds.
In a paraffin, one is allowed neither double bonds (two bonds between the same pair of atoms), nor cycles of linked carbons. So all paraffins with n carbon atoms share the empirical formula CnH2n+2
But for all n ≥ 4 there are several distinct molecules ("isomers") with the same formula but different structures.
The number of isomers rises rather rapidly when n increases.
In counting isomers it should be borne in mind that the four bond positions on a given carbon atom can be freely interchanged and bonds rotated (including 3-D "out of the paper" rotations when it's being observed on a flat diagram), so rotations or re-orientations of parts of the molecule (without breaking bonds) do not give different isomers. So what seem at first to be different molecules may in fact turn out to be different orientations of the same molecule.
Example
With n = 3 there is only one way of linking the carbons despite the different orientations the molecule can be drawn; and with n = 4 there are two configurations:
a straight chain: (CH3)(CH2)(CH2)(CH3)
a branched chain: (CH3)(CH(CH3))(CH3)
Due to bond rotations, it doesn't matter which direction the branch points in.
The phenomenon of "stereo-isomerism" (a molecule being different from its mirror image due to the actual 3-D arrangement of bonds) is ignored for the purpose of this task.
The input is the number n of carbon atoms of a molecule (for instance 17).
The output is how many different different paraffins there are with n carbon atoms (for instance 24,894 if n = 17).
The sequence of those results is visible in the OEIS entry:
oeis:A00602 number of n-node unrooted quartic trees; number of n-carbon alkanes C(n)H(2n+2) ignoring stereoisomers.
The sequence is (the index starts from zero, and represents the number of carbon atoms):
1, 1, 1, 1, 2, 3, 5, 9, 18, 35, 75, 159, 355, 802, 1858, 4347, 10359,
24894, 60523, 148284, 366319, 910726, 2278658, 5731580, 14490245,
36797588, 93839412, 240215803, 617105614, 1590507121, 4111846763,
10660307791, 27711253769, ...
Extra credit
Show the paraffins in some way.
A flat 1D representation, with arrays or lists is enough, for instance:
*Main> all_paraffins 1
[CCP H H H H]
*Main> all_paraffins 2
[BCP (C H H H) (C H H H)]
*Main> all_paraffins 3
[CCP H H (C H H H) (C H H H)]
*Main> all_paraffins 4
[BCP (C H H (C H H H)) (C H H (C H H H)),
CCP H (C H H H) (C H H H) (C H H H)]
*Main> all_paraffins 5
[CCP H H (C H H (C H H H)) (C H H (C H H H)),
CCP H (C H H H) (C H H H) (C H H (C H H H)),
CCP (C H H H) (C H H H) (C H H H) (C H H H)]
*Main> all_paraffins 6
[BCP (C H H (C H H (C H H H))) (C H H (C H H (C H H H))),
BCP (C H H (C H H (C H H H))) (C H (C H H H) (C H H H)),
BCP (C H (C H H H) (C H H H)) (C H (C H H H) (C H H H)),
CCP H (C H H H) (C H H (C H H H)) (C H H (C H H H)),
CCP (C H H H) (C H H H) (C H H H) (C H H (C H H H))]
Showing a basic 2D ASCII-art representation of the paraffins is better; for instance (molecule names aren't necessary):
methane ethane propane isobutane
H H H H H H H H H
│ │ │ │ │ │ │ │ │
H ─ C ─ H H ─ C ─ C ─ H H ─ C ─ C ─ C ─ H H ─ C ─ C ─ C ─ H
│ │ │ │ │ │ │ │ │
H H H H H H H │ H
│
H ─ C ─ H
│
H
Links
A paper that explains the problem and its solution in a functional language:
http://www.cs.wright.edu/~tkprasad/courses/cs776/paraffins-turner.pdf
A Haskell implementation:
https://github.com/ghc/nofib/blob/master/imaginary/paraffins/Main.hs
A Scheme implementation:
http://www.ccs.neu.edu/home/will/Twobit/src/paraffins.scm
A Fortress implementation: (this site has been closed)
http://java.net/projects/projectfortress/sources/sources/content/ProjectFortress/demos/turnersParaffins0.fss?rev=3005
| #D | D | import std.stdio, std.bigint;
enum uint nMax = 250;
enum uint nBranches = 4;
__gshared BigInt[nMax + 1] rooted = [1.BigInt, 1.BigInt /*...*/],
unrooted = [1.BigInt, 1.BigInt /*...*/];
void tree(in uint br, in uint n, in uint l, in uint inSum,
in BigInt cnt) nothrow {
__gshared static BigInt[nBranches] c;
uint sum = inSum;
foreach (immutable b; br + 1 .. nBranches + 1) {
sum += n;
if (sum > nMax || (l * 2 >= sum && b >= nBranches))
return;
if (b == br + 1) {
c[br] = rooted[n] * cnt;
} else {
c[br] *= rooted[n] + b - br - 1;
c[br] /= b - br;
}
if (l * 2 < sum)
unrooted[sum] += c[br];
if (b < nBranches)
rooted[sum] += c[br];
foreach_reverse (immutable m; 1 .. n)
tree(b, m, l, sum, c[br]);
}
}
void bicenter(in uint s) nothrow {
if ((s & 1) == 0)
unrooted[s] += rooted[s / 2] * (rooted[s / 2] + 1) / 2;
}
void main() {
foreach (immutable n; 1 .. nMax + 1) {
tree(0, n, n, 1, 1.BigInt);
n.bicenter;
writeln(n, ": ", unrooted[n]);
}
} |
http://rosettacode.org/wiki/Pangram_checker | Pangram checker | Pangram checker
You are encouraged to solve this task according to the task description, using any language you may know.
A pangram is a sentence that contains all the letters of the English alphabet at least once.
For example: The quick brown fox jumps over the lazy dog.
Task
Write a function or method to check a sentence to see if it is a pangram (or not) and show its use.
Related tasks
determine if a string has all the same characters
determine if a string has all unique characters
| #Action.21 | Action! | INCLUDE "D2:CHARTEST.ACT" ;from the Action! Tool Kit
DEFINE CHAR_COUNT="26"
BYTE FUNC IsPangram(CHAR ARRAY t)
BYTE ARRAY tab(CHAR_COUNT)
BYTE i,c
FOR i=0 TO CHAR_COUNT-1
DO tab(i)=0 OD
FOR i=1 TO t(0)
DO
c=ToLower(t(i))
IF c>='a AND c<='z THEN
tab(c-'a)=1
FI
OD
FOR i=0 TO CHAR_COUNT-1
DO
IF tab(i)=0 THEN
RETURN (0)
FI
OD
RETURN (1)
PROC Test(CHAR ARRAY t)
BYTE res
res=IsPangram(t)
PrintF("""%S"" is ",t)
IF res=0 THEN
Print("not ")
FI
PrintE("a pangram.")
PutE()
RETURN
PROC Main()
Put(125) PutE() ;clear screen
Test("The quick brown fox jumps over the lazy dog.")
Test("QwErTyUiOpAsDfGhJkLzXcVbNm")
Test("Not a pangram")
Test("")
RETURN |
http://rosettacode.org/wiki/Pangram_checker | Pangram checker | Pangram checker
You are encouraged to solve this task according to the task description, using any language you may know.
A pangram is a sentence that contains all the letters of the English alphabet at least once.
For example: The quick brown fox jumps over the lazy dog.
Task
Write a function or method to check a sentence to see if it is a pangram (or not) and show its use.
Related tasks
determine if a string has all the same characters
determine if a string has all unique characters
| #ActionScript | ActionScript | function pangram(k:string):Boolean {
var lowerK:String = k.toLowerCase();
var has:Object = {}
for (var i:Number=0; i<=k.length-1; i++) {
has[lowerK.charAt(i)] = true;
}
var result:Boolean = true;
for (var ch:String='a'; ch <= 'z'; ch=String.fromCharCode(ch.charCodeAt(0)+1)) {
result = result && has[ch]
}
return result || false;
} |
http://rosettacode.org/wiki/Pascal_matrix_generation | Pascal matrix generation | A pascal matrix is a two-dimensional square matrix holding numbers from Pascal's triangle, also known as binomial coefficients and which can be shown as nCr.
Shown below are truncated 5-by-5 matrices M[i, j] for i,j in range 0..4.
A Pascal upper-triangular matrix that is populated with jCi:
[[1, 1, 1, 1, 1],
[0, 1, 2, 3, 4],
[0, 0, 1, 3, 6],
[0, 0, 0, 1, 4],
[0, 0, 0, 0, 1]]
A Pascal lower-triangular matrix that is populated with iCj (the transpose of the upper-triangular matrix):
[[1, 0, 0, 0, 0],
[1, 1, 0, 0, 0],
[1, 2, 1, 0, 0],
[1, 3, 3, 1, 0],
[1, 4, 6, 4, 1]]
A Pascal symmetric matrix that is populated with i+jCi:
[[1, 1, 1, 1, 1],
[1, 2, 3, 4, 5],
[1, 3, 6, 10, 15],
[1, 4, 10, 20, 35],
[1, 5, 15, 35, 70]]
Task
Write functions capable of generating each of the three forms of n-by-n matrices.
Use those functions to display upper, lower, and symmetric Pascal 5-by-5 matrices on this page.
The output should distinguish between different matrices and the rows of each matrix (no showing a list of 25 numbers assuming the reader should split it into rows).
Note
The Cholesky decomposition of a Pascal symmetric matrix is the Pascal lower-triangle matrix of the same size.
| #Common_Lisp | Common Lisp | (defun pascal-lower (n &aux (a (make-array (list n n) :initial-element 0)))
(dotimes (i n)
(setf (aref a i 0) 1))
(dotimes (i (1- n) a)
(dotimes (j (1- n))
(setf (aref a (1+ i) (1+ j))
(+ (aref a i j)
(aref a i (1+ j)))))))
(defun pascal-upper (n &aux (a (make-array (list n n) :initial-element 0)))
(dotimes (i n)
(setf (aref a 0 i) 1))
(dotimes (i (1- n) a)
(dotimes (j (1- n))
(setf (aref a (1+ j) (1+ i))
(+ (aref a j i)
(aref a (1+ j) i))))))
(defun pascal-symmetric (n &aux (a (make-array (list n n) :initial-element 0)))
(dotimes (i n)
(setf (aref a i 0) 1 (aref a 0 i) 1))
(dotimes (i (1- n) a)
(dotimes (j (1- n))
(setf (aref a (1+ i) (1+ j))
(+ (aref a (1+ i) j)
(aref a i (1+ j)))))))
? (pascal-lower 4)
#2A((1 0 0 0) (1 1 0 0) (1 2 1 0) (1 3 3 1))
? (pascal-upper 4)
#2A((1 1 1 1) (0 1 2 3) (0 0 1 3) (0 0 0 1))
? (pascal-symmetric 4)
#2A((1 1 1 1) (1 2 3 4) (1 3 6 10) (1 4 10 20))
;In case one really insists in printing the array row by row:
(defun print-matrix (a)
(let ((p (array-dimension a 0))
(q (array-dimension a 1)))
(dotimes (i p)
(dotimes (j q)
(princ (aref a i j))
(princ #\Space))
(terpri))))
? (print-matrix (pascal-lower 5))
1 0 0 0 0
1 1 0 0 0
1 2 1 0 0
1 3 3 1 0
1 4 6 4 1
? (print-matrix (pascal-upper 5))
1 1 1 1 1
0 1 2 3 4
0 0 1 3 6
0 0 0 1 4
0 0 0 0 1
? (print-matrix (pascal-symmetric 5))
1 1 1 1 1
1 2 3 4 5
1 3 6 10 15
1 4 10 20 35
1 5 15 35 70 |
http://rosettacode.org/wiki/Parameterized_SQL_statement | Parameterized SQL statement | SQL injection
Using a SQL update statement like this one (spacing is optional):
UPDATE players
SET name = 'Smith, Steve', score = 42, active = TRUE
WHERE jerseyNum = 99
Non-parameterized SQL is the GoTo statement of database programming. Don't do it, and make sure your coworkers don't either. | #Huginn | Huginn | import Database as db;
import Algorithms as algo;
import FileSystem as fs;
main() {
dbPath = "/tmp/parametrized-sql.sqlite";
fs.remove( dbPath );
fs.open( dbPath, fs.OPEN_MODE.WRITE );
conn = db.connect( "sqlite3:///" + dbPath );
// Setup...
conn.query(
"CREATE TABLE Players (\n"
"\tname VARCHAR(64),\n"
"\tscore FLOAT,\n"
"\tactive INTEGER,\n"
"\tno VARCHAR(8)\n"
");"
).execute();
conn.query(
"INSERT INTO Players VALUES ( 'name', 0, 'false', 99 );"
).execute();
conn.query(
"INSERT INTO Players VALUES ( 'name', 0, 'false', 100 );"
).execute();
// Demonstrate parameterized SQL...
parametrizedQuery = conn.query(
"UPDATE Players SET name=?, score=?, active=? WHERE no=?"
);
for ( i, v : algo.enumerate( ( "Smith, Steve", 42, true, 99 ) ) ) {
parametrizedQuery.bind( i + 1, string( v ) );
}
parametrizedQuery.execute();
// and show the results...
for ( record : conn.query( "SELECT * FROM Players;" ).execute() ) {
print( "{}\n".format( record ) );
}
return ( 0 );
} |
http://rosettacode.org/wiki/Parameterized_SQL_statement | Parameterized SQL statement | SQL injection
Using a SQL update statement like this one (spacing is optional):
UPDATE players
SET name = 'Smith, Steve', score = 42, active = TRUE
WHERE jerseyNum = 99
Non-parameterized SQL is the GoTo statement of database programming. Don't do it, and make sure your coworkers don't either. | #Java | Java |
import java.sql.DriverManager;
import java.sql.Connection;
import java.sql.PreparedStatement;
public class DBDemo{
private String protocol; //set this to some connection protocol like "jdbc:sqlserver://"
private String dbName; //set this to the name of your database
private String username;
private String password;
PreparedStatement query;
public int setUpAndExecPS(){
try {
Connection conn = DriverManager.getConnection(protocol + dbName, username, password);
query = conn.prepareStatement(
"UPDATE players SET name = ?, score = ?, active = ? WHERE jerseyNum = ?");
query.setString(1, "Smith, Steve");//automatically sanitizes and adds quotes
query.setInt(2, 42);
query.setBoolean(3, true);
query.setInt(4, 99);
//there are similar methods for other SQL types in PerparedStatement
return query.executeUpdate();//returns the number of rows changed
//PreparedStatement.executeQuery() will return a java.sql.ResultSet,
//execute() will simply return a boolean saying whether it succeeded or not
} catch (Exception e) {
e.printStackTrace();
}
return 0;
}
}
|
http://rosettacode.org/wiki/Pascal%27s_triangle | Pascal's triangle | Pascal's triangle is an arithmetic and geometric figure often associated with the name of Blaise Pascal, but also studied centuries earlier in India, Persia, China and elsewhere.
Its first few rows look like this:
1
1 1
1 2 1
1 3 3 1
where each element of each row is either 1 or the sum of the two elements right above it.
For example, the next row of the triangle would be:
1 (since the first element of each row doesn't have two elements above it)
4 (1 + 3)
6 (3 + 3)
4 (3 + 1)
1 (since the last element of each row doesn't have two elements above it)
So the triangle now looks like this:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
Each row n (starting with row 0 at the top) shows the coefficients of the binomial expansion of (x + y)n.
Task
Write a function that prints out the first n rows of the triangle (with f(1) yielding the row consisting of only the element 1).
This can be done either by summing elements from the previous rows or using a binary coefficient or combination function.
Behavior for n ≤ 0 does not need to be uniform, but should be noted.
See also
Evaluate binomial coefficients
| #BCPL | BCPL | get "libhdr"
let pascal(n) be
for i=0 to n-1
$( let c = 1
for j=1 to 2*(n-1-i) do wrch(' ')
for k=0 to i
$( writef("%I3 ",c)
c := c*(i-k)/(k+1)
$)
wrch('*N')
$)
let start() be pascal(8) |
http://rosettacode.org/wiki/Parse_an_IP_Address | Parse an IP Address | The purpose of this task is to demonstrate parsing of text-format IP addresses, using IPv4 and IPv6.
Taking the following as inputs:
127.0.0.1
The "localhost" IPv4 address
127.0.0.1:80
The "localhost" IPv4 address, with a specified port (80)
::1
The "localhost" IPv6 address
[::1]:80
The "localhost" IPv6 address, with a specified port (80)
2605:2700:0:3::4713:93e3
Rosetta Code's primary server's public IPv6 address
[2605:2700:0:3::4713:93e3]:80
Rosetta Code's primary server's public IPv6 address, with a specified port (80)
Task
Emit each described IP address as a hexadecimal integer representing the address, the address space, and the port number specified, if any.
In languages where variant result types are clumsy, the result should be ipv4 or ipv6 address number, something which says which address space was represented, port number and something that says if the port was specified.
Example
127.0.0.1 has the address number 7F000001 (2130706433 decimal)
in the ipv4 address space.
::ffff:127.0.0.1 represents the same address in the ipv6 address space where it has the
address number FFFF7F000001 (281472812449793 decimal).
::1 has address number 1 and serves the same purpose in the ipv6 address
space that 127.0.0.1 serves in the ipv4 address space.
| #Julia | Julia |
const testdata = ["127.0.0.1", "127.0.0.1:80", "::1", "[::1]:80",
"2605:2700:0:3::4713:93e3", "[2605:2700:0:3::4713:93e3]:80",
"::ffff:192.168.173.22", "[::ffff:192.168.173.22]:80",
"1::", "[1::]:80", "::", "[::]:80"]
maybev4(ip) = search(ip, '.') > 0 && length(matchall(r":", ip)) < 2
maybev6(ip) = length(matchall(r":", ip)) > 1
function parseip(ip)
if (mat = match(r"^\[([:.\da-fA-F]+)\]:(\d+)$", ip))!= nothing ||
(mat = match(r"^([\d.]+)[:/](\d+)$", ip)) != nothing
port = mat.captures[2]
ip = mat.captures[1]
else
port = "none"
end
if maybev4(ip)
println("Processing ip v4 $ip")
iphex = hex(Int(Base.IPv4(ip)))
addresspace = "IPv4"
elseif maybev6(ip)
println("Processing ip v6 $ip")
iphex = hex(UInt128(Base.IPv6(ip)))
addresspace = "IPv6"
else
throw("Bad IP address argument $ip")
end
iphex, addresspace, port
end
for ip in testdata
hx, add, por = parseip(ip)
println("For input $ip, IP in hex is $hx, address space $add, port $por.")
end
|
http://rosettacode.org/wiki/Parse_an_IP_Address | Parse an IP Address | The purpose of this task is to demonstrate parsing of text-format IP addresses, using IPv4 and IPv6.
Taking the following as inputs:
127.0.0.1
The "localhost" IPv4 address
127.0.0.1:80
The "localhost" IPv4 address, with a specified port (80)
::1
The "localhost" IPv6 address
[::1]:80
The "localhost" IPv6 address, with a specified port (80)
2605:2700:0:3::4713:93e3
Rosetta Code's primary server's public IPv6 address
[2605:2700:0:3::4713:93e3]:80
Rosetta Code's primary server's public IPv6 address, with a specified port (80)
Task
Emit each described IP address as a hexadecimal integer representing the address, the address space, and the port number specified, if any.
In languages where variant result types are clumsy, the result should be ipv4 or ipv6 address number, something which says which address space was represented, port number and something that says if the port was specified.
Example
127.0.0.1 has the address number 7F000001 (2130706433 decimal)
in the ipv4 address space.
::ffff:127.0.0.1 represents the same address in the ipv6 address space where it has the
address number FFFF7F000001 (281472812449793 decimal).
::1 has address number 1 and serves the same purpose in the ipv6 address
space that 127.0.0.1 serves in the ipv4 address space.
| #Kotlin | Kotlin | // version 1.1.3
import java.math.BigInteger
enum class AddressSpace { IPv4, IPv6, Invalid }
data class IPAddressComponents(
val address: BigInteger,
val addressSpace: AddressSpace,
val port: Int // -1 denotes 'not specified'
)
val INVALID = IPAddressComponents(BigInteger.ZERO, AddressSpace.Invalid, 0)
fun ipAddressParse(ipAddress: String): IPAddressComponents {
var addressSpace = AddressSpace.IPv4
var ipa = ipAddress.toLowerCase()
var port = -1
var trans = false
if (ipa.startsWith("::ffff:") && '.' in ipa) {
addressSpace = AddressSpace.IPv6
trans = true
ipa = ipa.drop(7)
}
else if (ipa.startsWith("[::ffff:") && '.' in ipa) {
addressSpace = AddressSpace.IPv6
trans = true
ipa = ipa.drop(8).replace("]", "")
}
val octets = ipa.split('.').reversed().toTypedArray()
var address = BigInteger.ZERO
if (octets.size == 4) {
val split = octets[0].split(':')
if (split.size == 2) {
val temp = split[1].toIntOrNull()
if (temp == null || temp !in 0..65535) return INVALID
port = temp
octets[0] = split[0]
}
for (i in 0..3) {
val num = octets[i].toLongOrNull()
if (num == null || num !in 0..255) return INVALID
val bigNum = BigInteger.valueOf(num)
address = address.or(bigNum.shiftLeft(i * 8))
}
if (trans) address += BigInteger("ffff00000000", 16)
}
else if (octets.size == 1) {
addressSpace = AddressSpace.IPv6
if (ipa[0] == '[') {
ipa = ipa.drop(1)
val split = ipa.split("]:")
if (split.size != 2) return INVALID
val temp = split[1].toIntOrNull()
if (temp == null || temp !in 0..65535) return INVALID
port = temp
ipa = ipa.dropLast(2 + split[1].length)
}
val hextets = ipa.split(':').reversed().toMutableList()
val len = hextets.size
if (ipa.startsWith("::"))
hextets[len - 1] = "0"
else if (ipa.endsWith("::"))
hextets[0] = "0"
if (ipa == "::") hextets[1] = "0"
if (len > 8 || (len == 8 && hextets.any { it == "" }) || hextets.count { it == "" } > 1)
return INVALID
if (len < 8) {
var insertions = 8 - len
for (i in 0..7) {
if (hextets[i] == "") {
hextets[i] = "0"
while (insertions-- > 0) hextets.add(i, "0")
break
}
}
}
for (j in 0..7) {
val num = hextets[j].toLongOrNull(16)
if (num == null || num !in 0x0..0xFFFF) return INVALID
val bigNum = BigInteger.valueOf(num)
address = address.or(bigNum.shiftLeft(j * 16))
}
}
else return INVALID
return IPAddressComponents(address, addressSpace, port)
}
fun main(args: Array<String>) {
val ipas = listOf(
"127.0.0.1",
"127.0.0.1:80",
"::1",
"[::1]:80",
"2605:2700:0:3::4713:93e3",
"[2605:2700:0:3::4713:93e3]:80",
"::ffff:192.168.173.22",
"[::ffff:192.168.173.22]:80",
"1::",
"::",
"256.0.0.0",
"::ffff:127.0.0.0.1"
)
for (ipa in ipas) {
val (address, addressSpace, port) = ipAddressParse(ipa)
println("IP address : $ipa")
println("Address : ${"%X".format(address)}")
println("Address Space : $addressSpace")
println("Port : ${if (port == -1) "not specified" else port.toString()}")
println()
}
} |
http://rosettacode.org/wiki/Parametric_polymorphism | Parametric polymorphism | Parametric Polymorphism
type variables
Task
Write a small example for a type declaration that is parametric over another type, together with a short bit of code (and its type signature) that uses it.
A good example is a container type, let's say a binary tree, together with some function that traverses the tree, say, a map-function that operates on every element of the tree.
This language feature only applies to statically-typed languages.
| #Kotlin | Kotlin | // version 1.0.6
class BinaryTree<T>(var value: T) {
var left : BinaryTree<T>? = null
var right: BinaryTree<T>? = null
fun <U> map(f: (T) -> U): BinaryTree<U> {
val tree = BinaryTree<U>(f(value))
if (left != null) tree.left = left?.map(f)
if (right != null) tree.right = right?.map(f)
return tree
}
fun showTopThree() = "(${left?.value}, $value, ${right?.value})"
}
fun main(args: Array<String>) {
val b = BinaryTree(6)
b.left = BinaryTree(5)
b.right = BinaryTree(7)
println(b.showTopThree())
val b2 = b.map { it * 10.0 }
println(b2.showTopThree())
} |
http://rosettacode.org/wiki/Parsing/RPN_to_infix_conversion | Parsing/RPN to infix conversion | Parsing/RPN to infix conversion
You are encouraged to solve this task according to the task description, using any language you may know.
Task
Create a program that takes an RPN representation of an expression formatted as a space separated sequence of tokens and generates the equivalent expression in infix notation.
Assume an input of a correct, space separated, string of tokens
Generate a space separated output string representing the same expression in infix notation
Show how the major datastructure of your algorithm changes with each new token parsed.
Test with the following input RPN strings then print and display the output here.
RPN input
sample output
3 4 2 * 1 5 - 2 3 ^ ^ / +
3 + 4 * 2 / ( 1 - 5 ) ^ 2 ^ 3
1 2 + 3 4 + ^ 5 6 + ^
( ( 1 + 2 ) ^ ( 3 + 4 ) ) ^ ( 5 + 6 )
Operator precedence and operator associativity is given in this table:
operator
precedence
associativity
operation
^
4
right
exponentiation
*
3
left
multiplication
/
3
left
division
+
2
left
addition
-
2
left
subtraction
See also
Parsing/Shunting-yard algorithm for a method of generating an RPN from an infix expression.
Parsing/RPN calculator algorithm for a method of calculating a final value from this output RPN expression.
Postfix to infix from the RubyQuiz site.
| #F.23 | F# | type ast =
| Num of int
| Add of ast * ast | Sub of ast * ast
| Mul of ast * ast | Div of ast * ast
| Exp of ast * ast
let (|Int|_|) = System.Int32.TryParse >> function | (true, v) -> Some(v) | (false, _) -> None
let rec parse =
function
| [] -> failwith "Not enough tokens"
| (Int head)::tail -> (Num(head), tail)
| op::tail ->
let (left, rest1) = parse tail
let (right, rest2) = parse rest1
match op with
| "+" -> (Add (right, left)), rest2
| "-" -> (Sub (right, left)), rest2
| "*" -> (Mul (right, left)), rest2
| "/" -> (Div (right, left)), rest2
| "^" -> (Exp (right, left)), rest2
| _ -> failwith ("unknown op: " + op)
let rec infix p x =
let brackets (x : ast) = seq { yield "("; yield! infix 0 x; yield ")" }
let left op context l r = seq { yield! infix context l; yield op; yield! infix context r }
let right op context l r = seq { yield! brackets l; yield op; yield! infix context r }
seq {
match x with
| Num (n) -> yield n.ToString()
| Add (l, r) when p <= 2 -> yield! left "+" 2 l r
| Sub (l, r) when p <= 2 -> yield! left "-" 2 l r
| Mul (l, r) when p <= 3 -> yield! left "*" 3 l r
| Div (l, r) when p <= 3 -> yield! left "/" 3 l r
| Exp (Exp(ll, lr), r) -> yield! right "^" 4 (Exp(ll,lr)) r
| Exp (l, r) -> yield! left "^" 4 l r
| _ -> yield! brackets x
}
[<EntryPoint>]
let main argv =
let (tree, rest) =
argv |> Array.rev |> Seq.toList |> parse
match rest with
| [] -> printfn "%A" tree
| _ -> failwith ("not a valid RPN expression (excess tokens): " + System.String.Join(" ", argv))
Seq.iter (printf " %s") (infix 0 tree); printfn ""
0
|
http://rosettacode.org/wiki/Partial_function_application | Partial function application | Partial function application is the ability to take a function of many
parameters and apply arguments to some of the parameters to create a new
function that needs only the application of the remaining arguments to
produce the equivalent of applying all arguments to the original function.
E.g:
Given values v1, v2
Given f(param1, param2)
Then partial(f, param1=v1) returns f'(param2)
And f(param1=v1, param2=v2) == f'(param2=v2) (for any value v2)
Note that in the partial application of a parameter, (in the above case param1), other parameters are not explicitly mentioned. This is a recurring feature of partial function application.
Task
Create a function fs( f, s ) that takes a function, f( n ), of one value and a sequence of values s.
Function fs should return an ordered sequence of the result of applying function f to every value of s in turn.
Create function f1 that takes a value and returns it multiplied by 2.
Create function f2 that takes a value and returns it squared.
Partially apply f1 to fs to form function fsf1( s )
Partially apply f2 to fs to form function fsf2( s )
Test fsf1 and fsf2 by evaluating them with s being the sequence of integers from 0 to 3 inclusive and then the sequence of even integers from 2 to 8 inclusive.
Notes
In partially applying the functions f1 or f2 to fs, there should be no explicit mention of any other parameters to fs, although introspection of fs within the partial applicator to find its parameters is allowed.
This task is more about how results are generated rather than just getting results.
| #Icon_and_Unicon | Icon and Unicon | link printf
procedure main()
fsf1 := partial(fs,f1)
fsf2 := partial(fs,f2)
every s := [ 0, 1, 2, 3 ] |
[ 2, 4, 6, 8 ] do {
printf("\ns := %s\n",list2string(s))
printf("fsf1(s) := %s\n",list2string(fsf1(s)))
printf("fsf2(s) := %s\n",list2string(fsf2(s)))
}
end
procedure partial(f,g) #: partial application of f & g
@( p := create repeat {
s := (r@&source)[1] # return r / get argument s
r := f(g,s) # apply f(g,...)
}
) # create and activate procedure p
return p
end
procedure fs(f,s) #: return list where f is applied to each element of s
every put(r := [], f(!s))
return r
end
procedure f1(n) # double
return n * 2
end
procedure f2(n) #: square
return n ^ 2
end
procedure list2string(L) #: format list as a string
every (s := "[ ") ||:= !L || " "
return s || "]"
end |
http://rosettacode.org/wiki/Partition_an_integer_x_into_n_primes | Partition an integer x into n primes | Task
Partition a positive integer X into N distinct primes.
Or, to put it in another way:
Find N unique primes such that they add up to X.
Show in the output section the sum X and the N primes in ascending order separated by plus (+) signs:
• partition 99809 with 1 prime.
• partition 18 with 2 primes.
• partition 19 with 3 primes.
• partition 20 with 4 primes.
• partition 2017 with 24 primes.
• partition 22699 with 1, 2, 3, and 4 primes.
• partition 40355 with 3 primes.
The output could/should be shown in a format such as:
Partitioned 19 with 3 primes: 3+5+11
Use any spacing that may be appropriate for the display.
You need not validate the input(s).
Use the lowest primes possible; use 18 = 5+13, not 18 = 7+11.
You only need to show one solution.
This task is similar to factoring an integer.
Related tasks
Count in factors
Prime decomposition
Factors of an integer
Sieve of Eratosthenes
Primality by trial division
Factors of a Mersenne number
Factors of a Mersenne number
Sequence of primes by trial division
| #Python | Python | from itertools import combinations as cmb
def isP(n):
if n == 2:
return True
if n % 2 == 0:
return False
return all(n % x > 0 for x in range(3, int(n ** 0.5) + 1, 2))
def genP(n):
p = [2]
p.extend([x for x in range(3, n + 1, 2) if isP(x)])
return p
data = [
(99809, 1), (18, 2), (19, 3), (20, 4), (2017, 24),
(22699, 1), (22699, 2), (22699, 3), (22699, 4), (40355, 3)]
for n, cnt in data:
ci = iter(cmb(genP(n), cnt))
while True:
try:
c = next(ci)
if sum(c) == n:
print(' '.join(
[repr((n, cnt)), "->", '+'.join(str(s) for s in c)]
))
break
except StopIteration:
print(repr((n, cnt)) + " -> Not possible")
break |
http://rosettacode.org/wiki/Pascal%27s_triangle/Puzzle | Pascal's triangle/Puzzle | This puzzle involves a Pascals Triangle, also known as a Pyramid of Numbers.
[ 151]
[ ][ ]
[40][ ][ ]
[ ][ ][ ][ ]
[ X][11][ Y][ 4][ Z]
Each brick of the pyramid is the sum of the two bricks situated below it.
Of the three missing numbers at the base of the pyramid,
the middle one is the sum of the other two (that is, Y = X + Z).
Task
Write a program to find a solution to this puzzle.
| #Prolog | Prolog | :- use_module(library(clpfd)).
puzzle(Ts, X, Y, Z) :-
Ts = [ [151],
[_, _],
[40, _, _],
[_, _, _, _],
[X, 11, Y, 4, Z]],
Y #= X + Z, triangle(Ts), append(Ts, Vs), Vs ins 0..sup, label(Vs).
triangle([T|Ts]) :- ( Ts = [N|_] -> triangle_(T, N), triangle(Ts) ; true ).
triangle_([], _).
triangle_([T|Ts], [A,B|Rest]) :- T #= A + B, triangle_(Ts, [B|Rest]).
% ?- puzzle(_,X,Y,Z).
% X = 5,
% Y = 13,
% Z = 8 ; |
http://rosettacode.org/wiki/Password_generator | Password generator | Create a password generation program which will generate passwords containing random ASCII characters from the following groups:
lower-case letters: a ──► z
upper-case letters: A ──► Z
digits: 0 ──► 9
other printable characters: !"#$%&'()*+,-./:;<=>?@[]^_{|}~
(the above character list excludes white-space, backslash and grave)
The generated password(s) must include at least one (of each of the four groups):
lower-case letter,
upper-case letter,
digit (numeral), and
one "other" character.
The user must be able to specify the password length and the number of passwords to generate.
The passwords should be displayed or written to a file, one per line.
The randomness should be from a system source or library.
The program should implement a help option or button which should describe the program and options when invoked.
You may also allow the user to specify a seed value, and give the option of excluding visually similar characters.
For example: Il1 O0 5S 2Z where the characters are:
capital eye, lowercase ell, the digit one
capital oh, the digit zero
the digit five, capital ess
the digit two, capital zee
| #Mathematica.2FWolfram_Language | Mathematica/Wolfram Language | (* Length is the Length of the password, num is the number you want, \
and similar=1 if you want similar characters, 0 if not. True and \
False, should work in place of 1/0 *)
pwgen[length_Integer: 5, num_Integer: 1, similars_Boolean: 1] :=
pwgenerator[length, num, similars] =
Module[{list, valid, validchars, similar, k, j, password, pwlist,
lcase, ucase, digits, spec, s}, lcase = Alphabet[];
ucase = Capitalize[lcase];
digits = Range[0, 9];
spec = StringPartition["!:#$%\'()*+,-./:;>=<?@{}[]^_|~", 1];
validchars = Flatten[Union[lcase, ucase, digits, spec]];
similar = StringPartition["Il10O5S2Z", 1];
list = {};
Table[valid = 0;
While[valid == 0,
For[j = 0; k = {};, j < length, j++,
AppendTo[k, RandomInteger[{1, Length[validchars]}]]];
k = Flatten[k];
password = validchars[[k]];
Which[(Intersection[password, similar] >= 1 && similars == 0 ),
valid = 0;, (Intersection[password, similar] == 0 &&
similars == 1),
valid = 0;, (Intersection[password, similar] == 0 &&
similars == 0 ), valid = 1;
Return[password], (Intersection[password, similar] >= 1 &&
similars == 1), valid = 1; Return[password]];
], {num}];
] |
http://rosettacode.org/wiki/Permutations | Permutations | Task
Write a program that generates all permutations of n different objects. (Practically numerals!)
Related tasks
Find the missing permutation
Permutations/Derangements
The number of samples of size k from n objects.
With combinations and permutations generation tasks.
Order Unimportant
Order Important
Without replacement
(
n
k
)
=
n
C
k
=
n
(
n
−
1
)
…
(
n
−
k
+
1
)
k
(
k
−
1
)
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{\displaystyle {\binom {n}{k}}=^{n}\operatorname {C} _{k}={\frac {n(n-1)\ldots (n-k+1)}{k(k-1)\dots 1}}}
n
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{\displaystyle ^{n}\operatorname {P} _{k}=n\cdot (n-1)\cdot (n-2)\cdots (n-k+1)}
Task: Combinations
Task: Permutations
With replacement
(
n
+
k
−
1
k
)
=
n
+
k
−
1
C
k
=
(
n
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{\displaystyle {\binom {n+k-1}{k}}=^{n+k-1}\operatorname {C} _{k}={(n+k-1)! \over (n-1)!k!}}
n
k
{\displaystyle n^{k}}
Task: Combinations with repetitions
Task: Permutations with repetitions
| #Ursala | Ursala | #import std
permutations =
~&itB^?a( # are both the input argument list and its tail non-empty?
@ahPfatPRD *= refer ^C( # yes, recursively generate all permutations of the tail, and for each one
~&a, # insert the head at the first position
~&ar&& ~&arh2falrtPXPRD), # if the rest is non-empty, recursively insert at all subsequent positions
~&aNC) # no, return the singleton list of the argument |
http://rosettacode.org/wiki/Pathological_floating_point_problems | Pathological floating point problems | Most programmers are familiar with the inexactness of floating point calculations in a binary processor.
The classic example being:
0.1 + 0.2 = 0.30000000000000004
In many situations the amount of error in such calculations is very small and can be overlooked or eliminated with rounding.
There are pathological problems however, where seemingly simple, straight-forward calculations are extremely sensitive to even tiny amounts of imprecision.
This task's purpose is to show how your language deals with such classes of problems.
A sequence that seems to converge to a wrong limit.
Consider the sequence:
v1 = 2
v2 = -4
vn = 111 - 1130 / vn-1 + 3000 / (vn-1 * vn-2)
As n grows larger, the series should converge to 6 but small amounts of error will cause it to approach 100.
Task 1
Display the values of the sequence where n = 3, 4, 5, 6, 7, 8, 20, 30, 50 & 100 to at least 16 decimal places.
n = 3 18.5
n = 4 9.378378
n = 5 7.801153
n = 6 7.154414
n = 7 6.806785
n = 8 6.5926328
n = 20 6.0435521101892689
n = 30 6.006786093031205758530554
n = 50 6.0001758466271871889456140207471954695237
n = 100 6.000000019319477929104086803403585715024350675436952458072592750856521767230266
Task 2
The Chaotic Bank Society is offering a new investment account to their customers.
You first deposit $e - 1 where e is 2.7182818... the base of natural logarithms.
After each year, your account balance will be multiplied by the number of years that have passed, and $1 in service charges will be removed.
So ...
after 1 year, your balance will be multiplied by 1 and $1 will be removed for service charges.
after 2 years your balance will be doubled and $1 removed.
after 3 years your balance will be tripled and $1 removed.
...
after 10 years, multiplied by 10 and $1 removed, and so on.
What will your balance be after 25 years?
Starting balance: $e-1
Balance = (Balance * year) - 1 for 25 years
Balance after 25 years: $0.0399387296732302
Task 3, extra credit
Siegfried Rump's example. Consider the following function, designed by Siegfried Rump in 1988.
f(a,b) = 333.75b6 + a2( 11a2b2 - b6 - 121b4 - 2 ) + 5.5b8 + a/(2b)
compute f(a,b) where a=77617.0 and b=33096.0
f(77617.0, 33096.0) = -0.827396059946821
Demonstrate how to solve at least one of the first two problems, or both, and the third if you're feeling particularly jaunty.
See also;
Floating-Point Arithmetic Section 1.3.2 Difficult problems.
| #Stata | Stata | clear
set obs 100
gen n=_n
gen v=.
replace v=2 in 1
replace v=-4 in 2
replace v=111-1130/v[_n-1]+3000/(v[_n-1]*v[_n-2]) in 3/l
format %20.16f v
list if inlist(n,3,4,5,6,7,8,20,30,50,100), noobs |
http://rosettacode.org/wiki/Pathological_floating_point_problems | Pathological floating point problems | Most programmers are familiar with the inexactness of floating point calculations in a binary processor.
The classic example being:
0.1 + 0.2 = 0.30000000000000004
In many situations the amount of error in such calculations is very small and can be overlooked or eliminated with rounding.
There are pathological problems however, where seemingly simple, straight-forward calculations are extremely sensitive to even tiny amounts of imprecision.
This task's purpose is to show how your language deals with such classes of problems.
A sequence that seems to converge to a wrong limit.
Consider the sequence:
v1 = 2
v2 = -4
vn = 111 - 1130 / vn-1 + 3000 / (vn-1 * vn-2)
As n grows larger, the series should converge to 6 but small amounts of error will cause it to approach 100.
Task 1
Display the values of the sequence where n = 3, 4, 5, 6, 7, 8, 20, 30, 50 & 100 to at least 16 decimal places.
n = 3 18.5
n = 4 9.378378
n = 5 7.801153
n = 6 7.154414
n = 7 6.806785
n = 8 6.5926328
n = 20 6.0435521101892689
n = 30 6.006786093031205758530554
n = 50 6.0001758466271871889456140207471954695237
n = 100 6.000000019319477929104086803403585715024350675436952458072592750856521767230266
Task 2
The Chaotic Bank Society is offering a new investment account to their customers.
You first deposit $e - 1 where e is 2.7182818... the base of natural logarithms.
After each year, your account balance will be multiplied by the number of years that have passed, and $1 in service charges will be removed.
So ...
after 1 year, your balance will be multiplied by 1 and $1 will be removed for service charges.
after 2 years your balance will be doubled and $1 removed.
after 3 years your balance will be tripled and $1 removed.
...
after 10 years, multiplied by 10 and $1 removed, and so on.
What will your balance be after 25 years?
Starting balance: $e-1
Balance = (Balance * year) - 1 for 25 years
Balance after 25 years: $0.0399387296732302
Task 3, extra credit
Siegfried Rump's example. Consider the following function, designed by Siegfried Rump in 1988.
f(a,b) = 333.75b6 + a2( 11a2b2 - b6 - 121b4 - 2 ) + 5.5b8 + a/(2b)
compute f(a,b) where a=77617.0 and b=33096.0
f(77617.0, 33096.0) = -0.827396059946821
Demonstrate how to solve at least one of the first two problems, or both, and the third if you're feeling particularly jaunty.
See also;
Floating-Point Arithmetic Section 1.3.2 Difficult problems.
| #Swift | Swift | extension Numeric where Self: Strideable {
@inlinable
public func power(_ n: Self) -> Self {
return stride(from: 0, to: n, by: 1).lazy.map({_ in self }).reduce(1, *)
}
}
protocol PathologicalFloat: SignedNumeric, Strideable, ExpressibleByFloatLiteral {
static var e: Self { get }
static func /(_ lhs: Self, _ rhs: Self) -> Self
}
extension Double: PathologicalFloat {
static var e: Double { Double("2.71828182845904523536028747135266249")! }
}
extension Float: PathologicalFloat {
static var e: Float { Float("2.7182818284590")! }
}
extension Decimal: PathologicalFloat {
static var e: Decimal { Decimal(string: "2.71828182845904523536028747135266249")! }
}
extension BDouble: PathologicalFloat {
static var e: BDouble { BDouble("2.71828182845904523536028747135266249")! }
public func advanced(by n: BDouble) -> BDouble { self + n }
public func distance(to other: BDouble) -> BDouble { abs(self - other) }
}
func badSequence<T: PathologicalFloat>(n: Int) -> T {
guard n != 1 else { return 2 }
guard n != 2 else { return -4 }
var a: T = 2, b: T = -4
for _ in stride(from: 2, to: n, by: 1) {
(a, b) = (b, 111 - 1130 / b + 3000 / (a * b))
}
return b
}
func chaoticBank<T: PathologicalFloat>(years: T) -> T {
var balance = T.e - 1
for year: T in stride(from: 1, through: 25, by: 1) {
balance = (balance * year) - 1
}
return balance
}
func rumpFunction<T: PathologicalFloat>(_ a: T, _ b: T) -> T {
let aSquared = a.power(2)
let bSix = b.power(6)
let f1 = 333.75 * bSix
let f2 = aSquared * (11 * aSquared * b.power(2) - bSix - 121 * b.power(4) - 2)
let f3 = 5.5 * b.power(8) + a / (2 * b)
return f1 + f2 + f3
}
func fmt<T: CVarArg>(_ n: T) -> String { String(format: "%16.16f", n) }
print("Bad sequence")
for i in [3, 4, 5, 6, 7, 8, 20, 30, 50, 100] {
let vFloat: Float = badSequence(n: i)
let vDouble: Double = badSequence(n: i)
let vDecimal: Decimal = badSequence(n: i)
let vBigDouble: BDouble = badSequence(n: i)
print("v(\(i)) as Float \(fmt(vFloat)); as Double = \(fmt(vDouble)); as Decimal = \(vDecimal); as BDouble = \(vBigDouble.decimalExpansion(precisionAfterDecimalPoint: 16, rounded: false))")
}
let bankFloat: Float = chaoticBank(years: 25)
let bankDouble: Double = chaoticBank(years: 25)
let bankDecimal: Decimal = chaoticBank(years: 25)
let bankBigDouble: BDouble = chaoticBank(years: 25)
print("\nChaotic bank")
print("After 25 years your bank will be \(bankFloat) if stored as a Float")
print("After 25 years your bank will be \(bankDouble) if stored as a Double")
print("After 25 years your bank will be \(bankDecimal) if stored as a Decimal")
print("After 25 years your bank will be \(bankBigDouble.decimalExpansion(precisionAfterDecimalPoint: 16, rounded: false)) if stored as a BigDouble")
let rumpFloat: Float = rumpFunction(77617.0, 33096.0)
let rumpDouble: Double = rumpFunction(77617.0, 33096.0)
let rumpDecimal: Decimal = rumpFunction(77617.0, 33096.0)
let rumpBigDouble: BDouble = rumpFunction(77617.0, 33096.0)
print("\nRump's function")
print("rump(77617.0, 33096.0) as Float \(rumpFloat); as Double = \(rumpDouble); as Decimal = \(rumpDecimal); as BDouble = \(rumpBigDouble.decimalExpansion(precisionAfterDecimalPoint: 16, rounded: false))") |
http://rosettacode.org/wiki/Parallel_brute_force | Parallel brute force | Task
Find, through brute force, the five-letter passwords corresponding with the following SHA-256 hashes:
1. 1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad
2. 3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b
3. 74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f
Your program should naively iterate through all possible passwords consisting only of five lower-case ASCII English letters. It should use concurrent or parallel processing, if your language supports that feature. You may calculate SHA-256 hashes by calling a library or through a custom implementation. Print each matching password, along with its SHA-256 hash.
Related task: SHA-256
| #D | D | import std.digest.sha;
import std.parallelism;
import std.range;
import std.stdio;
// Find the five lower-case letter strings representing the following sha256 hashes
immutable p1 = cast(ubyte[32]) x"1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad";
immutable p2 = cast(ubyte[32]) x"3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b";
immutable p3 = cast(ubyte[32]) x"74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f";
void main() {
import std.datetime.stopwatch;
auto sw = StopWatch(AutoStart.yes);
// Switch these top loops to toggle between non-parallel and parrallel solutions.
// foreach(char a; 'a'..'z'+1) {
foreach(i, a; taskPool.parallel(iota('a', 'z'+1))) {
char[5] psw;
psw[0] = cast(char) a;
foreach(char b; 'a'..'z'+1) {
psw[1] = b;
foreach(char c; 'a'..'z'+1) {
psw[2] = c;
foreach(char d; 'a'..'z'+1) {
psw[3] = d;
foreach(char e; 'a'..'z'+1) {
psw[4] = e;
auto hash = psw.sha256Of;
if (equal(hash, p1) || equal(hash, p2) || equal(hash, p3)) {
writefln("%s <=> %(%x%)", psw, hash);
}
}
}
}
}
}
sw.stop;
writeln(sw.peek);
}
//Specialization that supports static arrays too
bool equal(T)(const T[] p, const T[] q) {
if (p.length != q.length) {
return false;
}
for(int i=0; i<p.length; i++) {
if (p[i] != q[i]) {
return false;
}
}
return true;
} |
http://rosettacode.org/wiki/Parallel_calculations | Parallel calculations | Many programming languages allow you to specify computations to be run in parallel.
While Concurrent computing is focused on concurrency,
the purpose of this task is to distribute time-consuming calculations
on as many CPUs as possible.
Assume we have a collection of numbers, and want to find the one
with the largest minimal prime factor
(that is, the one that contains relatively large factors).
To speed up the search, the factorization should be done
in parallel using separate threads or processes,
to take advantage of multi-core CPUs.
Show how this can be formulated in your language.
Parallelize the factorization of those numbers,
then search the returned list of numbers and factors
for the largest minimal factor,
and return that number and its prime factors.
For the prime number decomposition
you may use the solution of the Prime decomposition task.
| #Erlang | Erlang |
-module( parallel_calculations ).
-export( [fun_results/2, task/0] ).
fun_results( Fun, Datas ) ->
My_pid = erlang:self(),
Pids = [fun_spawn( Fun, X, My_pid ) || X <- Datas],
[fun_receive(X) || X <- Pids].
task() ->
Numbers = [12757923, 12878611, 12757923, 15808973, 15780709, 197622519],
Results = fun_results( fun factors/1, Numbers ),
Min_results = [lists:min(X) || X <- Results],
{_Max_min_factor, Number} = lists:max( lists:zip(Min_results, Numbers) ),
{Number, Factors} = lists:keyfind( Number, 1, lists:zip(Numbers, Results) ),
io:fwrite( "~p has largest minimal factor among its prime factors ~p~n", [Number, Factors] ).
factors(N) -> factors(N,2,[]).
factors(1,_,Acc) -> Acc;
factors(N,K,Acc) when N rem K == 0 -> factors(N div K,K, [K|Acc]);
factors(N,K,Acc) -> factors(N,K+1,Acc).
fun_receive( Pid ) ->
receive
{ok, Result, Pid} -> Result;
{Type, Error, Pid} -> erlang:Type( Error )
end.
fun_spawn( Fun, Data, My_pid ) ->
erlang:spawn( fun() ->
Result = try
{ok, Fun(Data), erlang:self()}
catch
Type:Error -> {Type, Error, erlang:self()}
end,
My_pid ! Result
end ).
|
http://rosettacode.org/wiki/Parsing/RPN_calculator_algorithm | Parsing/RPN calculator algorithm | Task
Create a stack-based evaluator for an expression in reverse Polish notation (RPN) that also shows the changes in the stack as each individual token is processed as a table.
Assume an input of a correct, space separated, string of tokens of an RPN expression
Test with the RPN expression generated from the Parsing/Shunting-yard algorithm task:
3 4 2 * 1 5 - 2 3 ^ ^ / +
Print or display the output here
Notes
^ means exponentiation in the expression above.
/ means division.
See also
Parsing/Shunting-yard algorithm for a method of generating an RPN from an infix expression.
Several solutions to 24 game/Solve make use of RPN evaluators (although tracing how they work is not a part of that task).
Parsing/RPN to infix conversion.
Arithmetic evaluation.
| #C.23 | C# | using System;
using System.Collections.Generic;
using System.Linq;
using System.Globalization;
using System.Threading;
namespace RPNEvaluator
{
class RPNEvaluator
{
static void Main(string[] args)
{
Thread.CurrentThread.CurrentCulture = CultureInfo.InvariantCulture;
string rpn = "3 4 2 * 1 5 - 2 3 ^ ^ / +";
Console.WriteLine("{0}\n", rpn);
decimal result = CalculateRPN(rpn);
Console.WriteLine("\nResult is {0}", result);
}
static decimal CalculateRPN(string rpn)
{
string[] rpnTokens = rpn.Split(' ');
Stack<decimal> stack = new Stack<decimal>();
decimal number = decimal.Zero;
foreach (string token in rpnTokens)
{
if (decimal.TryParse(token, out number))
{
stack.Push(number);
}
else
{
switch (token)
{
case "^":
case "pow":
{
number = stack.Pop();
stack.Push((decimal)Math.Pow((double)stack.Pop(), (double)number));
break;
}
case "ln":
{
stack.Push((decimal)Math.Log((double)stack.Pop(), Math.E));
break;
}
case "sqrt":
{
stack.Push((decimal)Math.Sqrt((double)stack.Pop()));
break;
}
case "*":
{
stack.Push(stack.Pop() * stack.Pop());
break;
}
case "/":
{
number = stack.Pop();
stack.Push(stack.Pop() / number);
break;
}
case "+":
{
stack.Push(stack.Pop() + stack.Pop());
break;
}
case "-":
{
number = stack.Pop();
stack.Push(stack.Pop() - number);
break;
}
default:
Console.WriteLine("Error in CalculateRPN(string) Method!");
break;
}
}
PrintState(stack);
}
return stack.Pop();
}
static void PrintState(Stack<decimal> stack)
{
decimal[] arr = stack.ToArray();
for (int i = arr.Length - 1; i >= 0; i--)
{
Console.Write("{0,-8:F3}", arr[i]);
}
Console.WriteLine();
}
}
} |
http://rosettacode.org/wiki/Pancake_numbers | Pancake numbers | Adrian Monk has problems and an assistant, Sharona Fleming. Sharona can deal with most of Adrian's problems except his lack of punctuality paying her remuneration. 2 pay checks down and she prepares him pancakes for breakfast. Knowing that he will be unable to eat them unless they are stacked in ascending order of size she leaves him only a skillet which he can insert at any point in the pile and flip all the above pancakes, repeating until the pile is sorted. Sharona has left the pile of n pancakes such that the maximum number of flips is required. Adrian is determined to do this in as few flips as possible. This sequence n->p(n) is known as the Pancake numbers.
The task is to determine p(n) for n = 1 to 9, and for each show an example requiring p(n) flips.
Sorting_algorithms/Pancake_sort actually performs the sort some giving the number of flips used. How do these compare with p(n)?
Few people know p(20), generously I shall award an extra credit for anyone doing more than p(16).
References
Bill Gates and the pancake problem
A058986
| #FreeBASIC | FreeBASIC |
Function pancake(n As Integer) As Integer
Dim As Integer gap = 2, sum = 2, adj = -1
While (sum < n)
adj += 1
gap = gap * 2 - 1
sum += gap
Wend
Return n + adj
End Function
For n As Integer = 1 To 20
Print Using "p(##) = ## "; n; pancake(n);
If n Mod 5 = 0 Then ?
Next n
Sleep
|
http://rosettacode.org/wiki/Pancake_numbers | Pancake numbers | Adrian Monk has problems and an assistant, Sharona Fleming. Sharona can deal with most of Adrian's problems except his lack of punctuality paying her remuneration. 2 pay checks down and she prepares him pancakes for breakfast. Knowing that he will be unable to eat them unless they are stacked in ascending order of size she leaves him only a skillet which he can insert at any point in the pile and flip all the above pancakes, repeating until the pile is sorted. Sharona has left the pile of n pancakes such that the maximum number of flips is required. Adrian is determined to do this in as few flips as possible. This sequence n->p(n) is known as the Pancake numbers.
The task is to determine p(n) for n = 1 to 9, and for each show an example requiring p(n) flips.
Sorting_algorithms/Pancake_sort actually performs the sort some giving the number of flips used. How do these compare with p(n)?
Few people know p(20), generously I shall award an extra credit for anyone doing more than p(16).
References
Bill Gates and the pancake problem
A058986
| #Go | Go | package main
import "fmt"
func pancake(n int) int {
gap, sum, adj := 2, 2, -1
for sum < n {
adj++
gap = gap*2 - 1
sum += gap
}
return n + adj
}
func main() {
for i := 0; i < 4; i++ {
for j := 1; j < 6; j++ {
n := i*5 + j
fmt.Printf("p(%2d) = %2d ", n, pancake(n))
}
fmt.Println()
}
} |
http://rosettacode.org/wiki/Pancake_numbers | Pancake numbers | Adrian Monk has problems and an assistant, Sharona Fleming. Sharona can deal with most of Adrian's problems except his lack of punctuality paying her remuneration. 2 pay checks down and she prepares him pancakes for breakfast. Knowing that he will be unable to eat them unless they are stacked in ascending order of size she leaves him only a skillet which he can insert at any point in the pile and flip all the above pancakes, repeating until the pile is sorted. Sharona has left the pile of n pancakes such that the maximum number of flips is required. Adrian is determined to do this in as few flips as possible. This sequence n->p(n) is known as the Pancake numbers.
The task is to determine p(n) for n = 1 to 9, and for each show an example requiring p(n) flips.
Sorting_algorithms/Pancake_sort actually performs the sort some giving the number of flips used. How do these compare with p(n)?
Few people know p(20), generously I shall award an extra credit for anyone doing more than p(16).
References
Bill Gates and the pancake problem
A058986
| #Java | Java | public class Pancake {
private static int pancake(int n) {
int gap = 2;
int sum = 2;
int adj = -1;
while (sum < n) {
adj++;
gap = 2 * gap - 1;
sum += gap;
}
return n + adj;
}
public static void main(String[] args) {
for (int i = 0; i < 4; i++) {
for (int j = 1; j < 6; j++) {
int n = 5 * i + j;
System.out.printf("p(%2d) = %2d ", n, pancake(n));
}
System.out.println();
}
}
} |
http://rosettacode.org/wiki/Parsing/Shunting-yard_algorithm | Parsing/Shunting-yard algorithm | Task
Given the operator characteristics and input from the Shunting-yard algorithm page and tables, use the algorithm to show the changes in the operator stack and RPN output
as each individual token is processed.
Assume an input of a correct, space separated, string of tokens representing an infix expression
Generate a space separated output string representing the RPN
Test with the input string:
3 + 4 * 2 / ( 1 - 5 ) ^ 2 ^ 3
print and display the output here.
Operator precedence is given in this table:
operator
precedence
associativity
operation
^
4
right
exponentiation
*
3
left
multiplication
/
3
left
division
+
2
left
addition
-
2
left
subtraction
Extra credit
Add extra text explaining the actions and an optional comment for the action on receipt of each token.
Note
The handling of functions and arguments is not required.
See also
Parsing/RPN calculator algorithm for a method of calculating a final value from this output RPN expression.
Parsing/RPN to infix conversion.
| #F.23 | F# | open System
type action = Shift | ReduceStack | ReduceInput
type State (input : string list, stack : string list, output : string list) =
member x.Input with get() = input
member x.Stack with get() = stack
member x.Output with get() = output
member x.report act =
let markTop = function | [] -> [] | x::xs -> ("["+x+"]")::xs
let (s, text) =
if x.Stack = [] && x.Input = [] then x, ""
else
match act with
| Shift -> State((markTop x.Input), x.Stack, x.Output), "shift"
| ReduceStack -> State(x.Input, (markTop x.Stack), x.Output), "reduce"
| ReduceInput -> State((markTop x.Input), x.Stack, x.Output), "reduce"
let lstr (x :string list) = String.Join(" ", (List.toArray x))
printfn "%25s %-9s %6s %s" (lstr (List.rev s.Output)) (lstr (List.rev s.Stack)) text (lstr s.Input)
member x.shift =
x.report Shift
State(x.Input.Tail, (x.Input.Head)::x.Stack, x.Output)
member x.reduce =
x.report ReduceStack
State (x.Input, (x.Stack.Tail), (x.Stack.Head)::x.Output)
member x.reduceNumber =
x.report ReduceInput
State(x.Input.Tail, x.Stack, (x.Input.Head)::x.Output)
let prec = function
| "^" -> 4 | "*" | "/" -> 3 | "+" | "-" -> 2 | "(" -> 1
| x -> failwith ("No precedence! Not an operator: " + x)
type assocKind = | Left | Right
let assoc = function | "^" -> Right | _ -> Left
let (|Number|Open|Close|Operator|) x =
if (Double.TryParse >> fst) x then Number
elif x = "(" then Open
elif x = ")" then Close
else Operator
let rec shunting_yard (s : State) =
let rec reduce_to_Open (s : State) =
match s.Stack with
| [] -> failwith "mismatched parentheses!"
| "("::xs -> State(s.Input.Tail, xs, s.Output)
| _ ->
reduce_to_Open s.reduce
let reduce_by_prec_and_shift x s =
let (xPrec, xAssoc) = (prec x, assoc x)
let rec loop (s : State) =
match s.Stack with
| [] -> s
| x::xs ->
let topPrec = prec x
if xAssoc = Left && xPrec <= topPrec || xAssoc = Right && xPrec < topPrec then
loop s.reduce
else
s
(loop s).shift
let rec reduce_rest (s : State) =
match s.Stack with
| [] -> s
| "("::_ -> failwith "mismatched parentheses!"
| x::_ ->
reduce_rest s.reduce
match s.Input with
| x::inputRest ->
match x with
| Number ->
shunting_yard s.reduceNumber
| Open ->
shunting_yard s.shift
| Close ->
shunting_yard (reduce_to_Open s)
| Operator ->
shunting_yard (reduce_by_prec_and_shift x s)
| [] -> reduce_rest s
[<EntryPoint>]
let main argv =
let input = if argv.Length = 0 then "3 + 4 * 2 / ( 1 - 5 ) ^ 2 ^ 3".Split() else argv
shunting_yard (State((Array.toList input), [], []))
|> (fun s -> s.report ReduceStack)
0 |
http://rosettacode.org/wiki/Paraffins | Paraffins |
This organic chemistry task is essentially to implement a tree enumeration algorithm.
Task
Enumerate, without repetitions and in order of increasing size, all possible paraffin molecules (also known as alkanes).
Paraffins are built up using only carbon atoms, which has four bonds, and hydrogen, which has one bond. All bonds for each atom must be used, so it is easiest to think of an alkane as linked carbon atoms forming the "backbone" structure, with adding hydrogen atoms linking the remaining unused bonds.
In a paraffin, one is allowed neither double bonds (two bonds between the same pair of atoms), nor cycles of linked carbons. So all paraffins with n carbon atoms share the empirical formula CnH2n+2
But for all n ≥ 4 there are several distinct molecules ("isomers") with the same formula but different structures.
The number of isomers rises rather rapidly when n increases.
In counting isomers it should be borne in mind that the four bond positions on a given carbon atom can be freely interchanged and bonds rotated (including 3-D "out of the paper" rotations when it's being observed on a flat diagram), so rotations or re-orientations of parts of the molecule (without breaking bonds) do not give different isomers. So what seem at first to be different molecules may in fact turn out to be different orientations of the same molecule.
Example
With n = 3 there is only one way of linking the carbons despite the different orientations the molecule can be drawn; and with n = 4 there are two configurations:
a straight chain: (CH3)(CH2)(CH2)(CH3)
a branched chain: (CH3)(CH(CH3))(CH3)
Due to bond rotations, it doesn't matter which direction the branch points in.
The phenomenon of "stereo-isomerism" (a molecule being different from its mirror image due to the actual 3-D arrangement of bonds) is ignored for the purpose of this task.
The input is the number n of carbon atoms of a molecule (for instance 17).
The output is how many different different paraffins there are with n carbon atoms (for instance 24,894 if n = 17).
The sequence of those results is visible in the OEIS entry:
oeis:A00602 number of n-node unrooted quartic trees; number of n-carbon alkanes C(n)H(2n+2) ignoring stereoisomers.
The sequence is (the index starts from zero, and represents the number of carbon atoms):
1, 1, 1, 1, 2, 3, 5, 9, 18, 35, 75, 159, 355, 802, 1858, 4347, 10359,
24894, 60523, 148284, 366319, 910726, 2278658, 5731580, 14490245,
36797588, 93839412, 240215803, 617105614, 1590507121, 4111846763,
10660307791, 27711253769, ...
Extra credit
Show the paraffins in some way.
A flat 1D representation, with arrays or lists is enough, for instance:
*Main> all_paraffins 1
[CCP H H H H]
*Main> all_paraffins 2
[BCP (C H H H) (C H H H)]
*Main> all_paraffins 3
[CCP H H (C H H H) (C H H H)]
*Main> all_paraffins 4
[BCP (C H H (C H H H)) (C H H (C H H H)),
CCP H (C H H H) (C H H H) (C H H H)]
*Main> all_paraffins 5
[CCP H H (C H H (C H H H)) (C H H (C H H H)),
CCP H (C H H H) (C H H H) (C H H (C H H H)),
CCP (C H H H) (C H H H) (C H H H) (C H H H)]
*Main> all_paraffins 6
[BCP (C H H (C H H (C H H H))) (C H H (C H H (C H H H))),
BCP (C H H (C H H (C H H H))) (C H (C H H H) (C H H H)),
BCP (C H (C H H H) (C H H H)) (C H (C H H H) (C H H H)),
CCP H (C H H H) (C H H (C H H H)) (C H H (C H H H)),
CCP (C H H H) (C H H H) (C H H H) (C H H (C H H H))]
Showing a basic 2D ASCII-art representation of the paraffins is better; for instance (molecule names aren't necessary):
methane ethane propane isobutane
H H H H H H H H H
│ │ │ │ │ │ │ │ │
H ─ C ─ H H ─ C ─ C ─ H H ─ C ─ C ─ C ─ H H ─ C ─ C ─ C ─ H
│ │ │ │ │ │ │ │ │
H H H H H H H │ H
│
H ─ C ─ H
│
H
Links
A paper that explains the problem and its solution in a functional language:
http://www.cs.wright.edu/~tkprasad/courses/cs776/paraffins-turner.pdf
A Haskell implementation:
https://github.com/ghc/nofib/blob/master/imaginary/paraffins/Main.hs
A Scheme implementation:
http://www.ccs.neu.edu/home/will/Twobit/src/paraffins.scm
A Fortress implementation: (this site has been closed)
http://java.net/projects/projectfortress/sources/sources/content/ProjectFortress/demos/turnersParaffins0.fss?rev=3005
| #FreeBASIC | FreeBASIC | ' version 31-12-2016
' compile with: fbc -s console
' uses gmp, translation from pascal
#Include Once "gmp.bi"
Const As Integer max_n = 500, branch = 4
Dim Shared As mpz_ptr rooted(), unrooted(), c()
Dim Shared As mpz_ptr cnt, tmp
Sub tree(br As UInteger, n As UInteger, l As UInteger, sum As UInteger, cnt As mpz_ptr)
Dim As UInteger b, m
For b = br +1 To branch
sum = sum + n
If sum > max_n Then Return
' prevent unneeded long math
If (l * 2 >= sum) And (b >= branch) Then Return
If b = (br +1) Then
mpz_mul(c(br), rooted(n), cnt)
Else
mpz_add_ui(tmp, rooted(n), b - br -1)
mpz_mul(c(br), c(br), tmp)
mpz_divexact_ui(c(br), c(br), b - br)
End If
If l * 2 < sum Then
mpz_add(unrooted(sum), unrooted(sum), c(br))
End If
If b < branch Then
mpz_add(rooted(sum), rooted(sum), c(br))
For m = n -1 To 1 Step -1
tree(b, m, l, sum, c(br))
Next
End If
Next
End Sub
Sub bicenter(s As UInteger)
If (s And 1) = 1 Then Return
mpz_add_ui(tmp, rooted(s \ 2), 1)
mpz_mul(tmp, rooted(s \ 2), tmp)
mpz_tdiv_q_2exp(tmp, tmp, 1)
mpz_add(unrooted(s), unrooted(s), tmp)
End Sub
' ------=< MAIN >=------
Dim As UInteger n, sum
Dim As ZString Ptr ans
ReDim rooted(max_n), unrooted(max_n)
For n = 0 To max_n
rooted(n) = Allocate(Len(__mpz_struct)) : Mpz_init( rooted(n))
unrooted(n) = Allocate(Len(__mpz_struct)) : Mpz_init(unrooted(n))
Next
For n = 0 To 1
mpz_set_ui( rooted(n), 1)
mpz_set_ui(unrooted(n), 1)
Next
ReDim c(branch -1)
For n = 0 To branch -1
c(n) = Allocate(Len(__mpz_struct)) : Mpz_init(c(n))
Next
cnt = Allocate(Len(__mpz_struct)) : Mpz_init_set_ui(cnt, 1)
tmp = Allocate(Len(__mpz_struct)) : Mpz_init(tmp)
sum = 1
For n = 1 To max_n
tree(0, n, n, sum, cnt)
bicenter(n)
'gmp_printf("%d: %Zd"+Chr(13)+Chr(10), n, unrooted(n))
ans = Mpz_get_str (0, 10, unrooted(n))
Print Using "###: "; n; : Print *ans
Next
For n = 0 To max_n
mpz_Clear( rooted(n))
mpz_Clear(unrooted(n))
Next
For n = 0 To branch -1
mpz_clear(c(n))
Next
mpz_clear(cnt)
mpz_clear(tmp)
' empty keyboard buffer
While Inkey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End |
http://rosettacode.org/wiki/Pangram_checker | Pangram checker | Pangram checker
You are encouraged to solve this task according to the task description, using any language you may know.
A pangram is a sentence that contains all the letters of the English alphabet at least once.
For example: The quick brown fox jumps over the lazy dog.
Task
Write a function or method to check a sentence to see if it is a pangram (or not) and show its use.
Related tasks
determine if a string has all the same characters
determine if a string has all unique characters
| #Ada | Ada | with Ada.Text_IO; use Ada.Text_IO;
with Ada.Strings.Maps; use Ada.Strings.Maps;
with Ada.Characters.Handling; use Ada.Characters.Handling;
procedure pangram is
function ispangram(txt: String) return Boolean is
(Is_Subset(To_Set(Span => ('a','z')), To_Set(To_Lower(txt))));
begin
put_line(Boolean'Image(ispangram("This is a test")));
put_line(Boolean'Image(ispangram("The quick brown fox jumps over the lazy dog")));
put_line(Boolean'Image(ispangram("NOPQRSTUVWXYZ abcdefghijklm")));
put_line(Boolean'Image(ispangram("abcdefghijklopqrstuvwxyz"))); --Missing m, n
end pangram;
|
http://rosettacode.org/wiki/Pascal_matrix_generation | Pascal matrix generation | A pascal matrix is a two-dimensional square matrix holding numbers from Pascal's triangle, also known as binomial coefficients and which can be shown as nCr.
Shown below are truncated 5-by-5 matrices M[i, j] for i,j in range 0..4.
A Pascal upper-triangular matrix that is populated with jCi:
[[1, 1, 1, 1, 1],
[0, 1, 2, 3, 4],
[0, 0, 1, 3, 6],
[0, 0, 0, 1, 4],
[0, 0, 0, 0, 1]]
A Pascal lower-triangular matrix that is populated with iCj (the transpose of the upper-triangular matrix):
[[1, 0, 0, 0, 0],
[1, 1, 0, 0, 0],
[1, 2, 1, 0, 0],
[1, 3, 3, 1, 0],
[1, 4, 6, 4, 1]]
A Pascal symmetric matrix that is populated with i+jCi:
[[1, 1, 1, 1, 1],
[1, 2, 3, 4, 5],
[1, 3, 6, 10, 15],
[1, 4, 10, 20, 35],
[1, 5, 15, 35, 70]]
Task
Write functions capable of generating each of the three forms of n-by-n matrices.
Use those functions to display upper, lower, and symmetric Pascal 5-by-5 matrices on this page.
The output should distinguish between different matrices and the rows of each matrix (no showing a list of 25 numbers assuming the reader should split it into rows).
Note
The Cholesky decomposition of a Pascal symmetric matrix is the Pascal lower-triangle matrix of the same size.
| #D | D | import std.stdio, std.bigint, std.range, std.algorithm;
auto binomialCoeff(in uint n, in uint k) pure nothrow {
BigInt result = 1;
foreach (immutable i; 1 .. k + 1)
result = result * (n - i + 1) / i;
return result;
}
auto pascalUpp(in uint n) pure nothrow {
return n.iota.map!(i => n.iota.map!(j => binomialCoeff(j, i)));
}
auto pascalLow(in uint n) pure nothrow {
return n.iota.map!(i => n.iota.map!(j => binomialCoeff(i, j)));
}
auto pascalSym(in uint n) pure nothrow {
return n.iota.map!(i => n.iota.map!(j => binomialCoeff(i + j, i)));
}
void main() {
enum n = 5;
writefln("Upper:\n%(%(%2d %)\n%)", pascalUpp(n));
writefln("\nLower:\n%(%(%2d %)\n%)", pascalLow(n));
writefln("\nSymmetric:\n%(%(%2d %)\n%)", pascalSym(n));
} |
http://rosettacode.org/wiki/Parameterized_SQL_statement | Parameterized SQL statement | SQL injection
Using a SQL update statement like this one (spacing is optional):
UPDATE players
SET name = 'Smith, Steve', score = 42, active = TRUE
WHERE jerseyNum = 99
Non-parameterized SQL is the GoTo statement of database programming. Don't do it, and make sure your coworkers don't either. | #Julia | Julia | using SQLite
name = "Smith, Steve"
jerseys = Dict("Smith, Steve" => 99)
sqlbool(tf::Bool) = if(tf) "TRUE" else "FALSE" end
db = SQLite.DB() # no filename given, so create an in-memory temporary
SQLite.execute!(db, "create table players (id integer primary key,
name text,
score number,
active bool,
jerseynum integer)")
SQLite.query(db, "INSERT INTO players (name, score, active, jerseynum) values ('Jones, James', 9, 'FALSE', 99)")
SQLite.query(db, "UPDATE players SET name = ?, score = ?, active = ? WHERE jerseynum = ?";
values = ["Smith, Steve", 42, sqlbool(true), jerseys[name]])
tbl = SQLite.query(db, "SELECT * from players")
println(tbl) |
http://rosettacode.org/wiki/Parameterized_SQL_statement | Parameterized SQL statement | SQL injection
Using a SQL update statement like this one (spacing is optional):
UPDATE players
SET name = 'Smith, Steve', score = 42, active = TRUE
WHERE jerseyNum = 99
Non-parameterized SQL is the GoTo statement of database programming. Don't do it, and make sure your coworkers don't either. | #Kotlin | Kotlin | // Version 1.2.41
import java.sql.DriverManager
import java.sql.Connection
fun main(args: Array<String>) {
val url = "jdbc:mysql://localhost:3306/test"
val username = "example"
val password = "password123"
val conn = DriverManager.getConnection(url, username, password)
val query = conn.prepareStatement(
"UPDATE players SET name = ?, score = ?, active = ? WHERE jerseyNum = ?"
)
with (query) {
setString(1, "Smith, Steve")
setInt(2, 42)
setBoolean(3, true)
setInt(4, 99)
val rowCount = executeUpdate()
if (rowCount == 0) println("Update failed")
close()
}
conn.close()
} |
http://rosettacode.org/wiki/Pascal%27s_triangle | Pascal's triangle | Pascal's triangle is an arithmetic and geometric figure often associated with the name of Blaise Pascal, but also studied centuries earlier in India, Persia, China and elsewhere.
Its first few rows look like this:
1
1 1
1 2 1
1 3 3 1
where each element of each row is either 1 or the sum of the two elements right above it.
For example, the next row of the triangle would be:
1 (since the first element of each row doesn't have two elements above it)
4 (1 + 3)
6 (3 + 3)
4 (3 + 1)
1 (since the last element of each row doesn't have two elements above it)
So the triangle now looks like this:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
Each row n (starting with row 0 at the top) shows the coefficients of the binomial expansion of (x + y)n.
Task
Write a function that prints out the first n rows of the triangle (with f(1) yielding the row consisting of only the element 1).
This can be done either by summing elements from the previous rows or using a binary coefficient or combination function.
Behavior for n ≤ 0 does not need to be uniform, but should be noted.
See also
Evaluate binomial coefficients
| #Befunge | Befunge | 0" :swor fo rebmuN">:#,_&> 55+, v
v01*p00-1:g00.:<1p011p00:\-1_v#:<
>g:1+10p/48*,:#^_$ 55+,1+\: ^>$$@ |
http://rosettacode.org/wiki/Parse_an_IP_Address | Parse an IP Address | The purpose of this task is to demonstrate parsing of text-format IP addresses, using IPv4 and IPv6.
Taking the following as inputs:
127.0.0.1
The "localhost" IPv4 address
127.0.0.1:80
The "localhost" IPv4 address, with a specified port (80)
::1
The "localhost" IPv6 address
[::1]:80
The "localhost" IPv6 address, with a specified port (80)
2605:2700:0:3::4713:93e3
Rosetta Code's primary server's public IPv6 address
[2605:2700:0:3::4713:93e3]:80
Rosetta Code's primary server's public IPv6 address, with a specified port (80)
Task
Emit each described IP address as a hexadecimal integer representing the address, the address space, and the port number specified, if any.
In languages where variant result types are clumsy, the result should be ipv4 or ipv6 address number, something which says which address space was represented, port number and something that says if the port was specified.
Example
127.0.0.1 has the address number 7F000001 (2130706433 decimal)
in the ipv4 address space.
::ffff:127.0.0.1 represents the same address in the ipv6 address space where it has the
address number FFFF7F000001 (281472812449793 decimal).
::1 has address number 1 and serves the same purpose in the ipv6 address
space that 127.0.0.1 serves in the ipv4 address space.
| #Nim | Nim | import net, sequtils, strscans, strutils
const NoPort = -1
func parseIpAddressAndPort(str: string): tuple[ipAddr: IpAddress; port: int] =
var ipString: string
var port: int
if str.scanf("[$+]:$i", ipString, port):
# IP v6 address with port.
return (ipString.parseIpAddress(), port)
if '.' in str and str.scanf("$+:$i", ipString, port):
# IP v4 address with port.
return (ipString.parseIpAddress(), port)
# IP address without port.
result = (str.parseIpAddress(), NoPort)
const Inputs = ["127.0.0.1", "127.0.0.1:80",
"::1", "[::1]:80",
"2605:2700:0:3::4713:93e3",
"[2605:2700:0:3::4713:93e3]:80"]
# Room to reserve to display the input.
const InputSize = Inputs.mapIt(it.len).foldl(max(a, b)) + 2
echo "Input".alignLeft(InputSize), "Address".align(32), " Space", " Port"
for input in Inputs:
try:
let (ipAddress, port) = input.parseIpAddressAndPort()
let portStr = if port == NoPort: "" else: $port
case ipAddress.family
of IPv6:
echo input.alignLeft(InputSize),
ipAddress.address_v6.map(toHex).join().align(32),
"IP v6".align(7),
portStr.align(6)
of IPv4:
echo input.alignLeft(InputSize),
ipAddress.address_v4.map(toHex).join().align(32),
"IP v4".align(7),
portStr.align(6)
except ValueError:
echo "Invalid IP address: ", input |
http://rosettacode.org/wiki/Parse_an_IP_Address | Parse an IP Address | The purpose of this task is to demonstrate parsing of text-format IP addresses, using IPv4 and IPv6.
Taking the following as inputs:
127.0.0.1
The "localhost" IPv4 address
127.0.0.1:80
The "localhost" IPv4 address, with a specified port (80)
::1
The "localhost" IPv6 address
[::1]:80
The "localhost" IPv6 address, with a specified port (80)
2605:2700:0:3::4713:93e3
Rosetta Code's primary server's public IPv6 address
[2605:2700:0:3::4713:93e3]:80
Rosetta Code's primary server's public IPv6 address, with a specified port (80)
Task
Emit each described IP address as a hexadecimal integer representing the address, the address space, and the port number specified, if any.
In languages where variant result types are clumsy, the result should be ipv4 or ipv6 address number, something which says which address space was represented, port number and something that says if the port was specified.
Example
127.0.0.1 has the address number 7F000001 (2130706433 decimal)
in the ipv4 address space.
::ffff:127.0.0.1 represents the same address in the ipv6 address space where it has the
address number FFFF7F000001 (281472812449793 decimal).
::1 has address number 1 and serves the same purpose in the ipv6 address
space that 127.0.0.1 serves in the ipv4 address space.
| #Perl | Perl | sub parse_v4 {
my ($ip, $port) = @_;
my @quad = split(/\./, $ip);
return unless @quad == 4;
for (@quad) { return if ($_ > 255) }
if (!length $port) { $port = -1 }
elsif ($port =~ /^(\d+)$/) { $port = $1 }
else { return }
my $h = join '' => map(sprintf("%02x", $_), @quad);
return $h, $port
}
sub parse_v6 {
my $ip = shift;
my $omits;
return unless $ip =~ /^[\da-f:.]+$/i; # invalid char
$ip =~ s/^:/0:/;
$omits = 1 if $ip =~ s/::/:z:/g;
return if $ip =~ /z.*z/; # multiple omits illegal
my $v4 = '';
my $len = 8;
if ($ip =~ s/:((?:\d+\.){3}\d+)$//) {
# hybrid 4/6 ip
($v4) = parse_v4($1) or return;
$len -= 2;
}
# what's left should be v6 only
return unless $ip =~ /^[:a-fz\d]+$/i;
my @h = split(/:/, $ip);
return if @h + $omits > $len; # too many segments
@h = map( $_ eq 'z' ? (0) x ($len - @h + 1) : ($_), @h);
return join('' => map(sprintf("%04x", hex($_)), @h)).$v4;
}
sub parse_ip {
my $str = shift;
$str =~ s/^\s*//;
$str =~ s/\s*$//;
if ($str =~ s/^((?:\d+\.)+\d+)(?::(\d+))?$//) {
return 'v4', parse_v4($1, $2);
}
my ($ip, $port);
if ($str =~ /^\[(.*?)\]:(\d+)$/) {
$port = $2;
$ip = parse_v6($1);
} else {
$port = -1;
$ip = parse_v6($str);
}
return unless $ip;
return 'v6', $ip, $port;
}
for (qw/127.0.0.1 127.0.0.1:80
::1
[::1]:80
2605:2700:0:3::4713:93e3
[2605:2700:0:3::4713:93e3]:80
::ffff:192.168.0.1
[::ffff:192.168.0.1]:22
::ffff:127.0.0.0.1
a::b::1/)
{
print "$_\n\t";
my ($ver, $ip, $port) = parse_ip($_)
or print "parse error\n" and next;
print "$ver $ip\tport $port\n\n";
} |
http://rosettacode.org/wiki/Parametric_polymorphism | Parametric polymorphism | Parametric Polymorphism
type variables
Task
Write a small example for a type declaration that is parametric over another type, together with a short bit of code (and its type signature) that uses it.
A good example is a container type, let's say a binary tree, together with some function that traverses the tree, say, a map-function that operates on every element of the tree.
This language feature only applies to statically-typed languages.
| #Mathematica.2FWolfram_Language | Mathematica/Wolfram Language | f[a_] := Join[a, a]
f[{1, 2, 3}]
f[{"1", "2", "3"}]
f[{1.1, 2.1, 3.1}]
f[G[1, "a", Pi]]
g[x_] := x^2
g[2]
g[3.5]
g[Pi]
g["a"] |
http://rosettacode.org/wiki/Parametric_polymorphism | Parametric polymorphism | Parametric Polymorphism
type variables
Task
Write a small example for a type declaration that is parametric over another type, together with a short bit of code (and its type signature) that uses it.
A good example is a container type, let's say a binary tree, together with some function that traverses the tree, say, a map-function that operates on every element of the tree.
This language feature only applies to statically-typed languages.
| #Mercury | Mercury | :- type tree(A) ---> empty ; node(A, tree(A), tree(A)).
:- func map(func(A) = B, tree(A)) = tree(B).
map(_, empty) = empty.
map(F, node(A, Left, Right)) = node(F(A), map(F, Left), map(F, Right)). |
http://rosettacode.org/wiki/Parametric_polymorphism | Parametric polymorphism | Parametric Polymorphism
type variables
Task
Write a small example for a type declaration that is parametric over another type, together with a short bit of code (and its type signature) that uses it.
A good example is a container type, let's say a binary tree, together with some function that traverses the tree, say, a map-function that operates on every element of the tree.
This language feature only applies to statically-typed languages.
| #Nim | Nim | import strutils, sugar
type Tree[T] = ref object
value: T
left, right: Tree[T]
proc newTree[T](value = default(T)): Tree[T] =
## Create a tree with a single node with the given value.
Tree[T](value: value)
proc map[T, U](tree: Tree[T]; f: (T) -> U): Tree[U] =
## Apply function "f" to each element of a tree, building
## another tree.
result = newTree[U](f(tree.value))
if not tree.left.isNil:
result.left = tree.left.map(f)
if not tree.right.isNil:
result.right = tree.right.map(f)
proc print(tree: Tree; indent = 0) =
## Print a tree.
let start = repeat(' ', indent)
echo start, "value: ", tree.value
if tree.left.isNil:
echo start, " nil"
else:
print(tree.left, indent + 2)
if tree.right.isNil:
echo start, " nil"
else:
print(tree.right, indent + 2)
when isMainModule:
echo "Initial tree:"
var tree = newTree[int](5)
tree.left = newTree[int](2)
tree.right = newTree[int](7)
print(tree)
echo ""
echo "Tree created by applying a function to each node:"
let tree1 = tree.map((x) => 1 / x)
print(tree1) |
http://rosettacode.org/wiki/Parametric_polymorphism | Parametric polymorphism | Parametric Polymorphism
type variables
Task
Write a small example for a type declaration that is parametric over another type, together with a short bit of code (and its type signature) that uses it.
A good example is a container type, let's say a binary tree, together with some function that traverses the tree, say, a map-function that operates on every element of the tree.
This language feature only applies to statically-typed languages.
| #Objective-C | Objective-C | @interface Tree<T> : NSObject {
T value;
Tree<T> *left;
Tree<T> *right;
}
- (void)replaceAll:(T)v;
@end
@implementation Tree
- (void)replaceAll:(id)v {
value = v;
[left replaceAll:v];
[right replaceAll:v];
}
@end |
http://rosettacode.org/wiki/Parsing/RPN_to_infix_conversion | Parsing/RPN to infix conversion | Parsing/RPN to infix conversion
You are encouraged to solve this task according to the task description, using any language you may know.
Task
Create a program that takes an RPN representation of an expression formatted as a space separated sequence of tokens and generates the equivalent expression in infix notation.
Assume an input of a correct, space separated, string of tokens
Generate a space separated output string representing the same expression in infix notation
Show how the major datastructure of your algorithm changes with each new token parsed.
Test with the following input RPN strings then print and display the output here.
RPN input
sample output
3 4 2 * 1 5 - 2 3 ^ ^ / +
3 + 4 * 2 / ( 1 - 5 ) ^ 2 ^ 3
1 2 + 3 4 + ^ 5 6 + ^
( ( 1 + 2 ) ^ ( 3 + 4 ) ) ^ ( 5 + 6 )
Operator precedence and operator associativity is given in this table:
operator
precedence
associativity
operation
^
4
right
exponentiation
*
3
left
multiplication
/
3
left
division
+
2
left
addition
-
2
left
subtraction
See also
Parsing/Shunting-yard algorithm for a method of generating an RPN from an infix expression.
Parsing/RPN calculator algorithm for a method of calculating a final value from this output RPN expression.
Postfix to infix from the RubyQuiz site.
| #Go | Go | package main
import (
"fmt"
"strings"
)
var tests = []string{
"3 4 2 * 1 5 - 2 3 ^ ^ / +",
"1 2 + 3 4 + ^ 5 6 + ^",
}
var opa = map[string]struct {
prec int
rAssoc bool
}{
"^": {4, true},
"*": {3, false},
"/": {3, false},
"+": {2, false},
"-": {2, false},
}
const nPrec = 9
func main() {
for _, t := range tests {
parseRPN(t)
}
}
func parseRPN(e string) {
fmt.Println("\npostfix:", e)
type sf struct {
prec int
expr string
}
var stack []sf
for _, tok := range strings.Fields(e) {
fmt.Println("token:", tok)
if op, isOp := opa[tok]; isOp {
rhs := &stack[len(stack)-1]
stack = stack[:len(stack)-1]
lhs := &stack[len(stack)-1]
if lhs.prec < op.prec || (lhs.prec == op.prec && op.rAssoc) {
lhs.expr = "(" + lhs.expr + ")"
}
lhs.expr += " " + tok + " "
if rhs.prec < op.prec || (rhs.prec == op.prec && !op.rAssoc) {
lhs.expr += "(" + rhs.expr + ")"
} else {
lhs.expr += rhs.expr
}
lhs.prec = op.prec
} else {
stack = append(stack, sf{nPrec, tok})
}
for _, f := range stack {
fmt.Printf(" %d %q\n", f.prec, f.expr)
}
}
fmt.Println("infix:", stack[0].expr)
} |
http://rosettacode.org/wiki/Partial_function_application | Partial function application | Partial function application is the ability to take a function of many
parameters and apply arguments to some of the parameters to create a new
function that needs only the application of the remaining arguments to
produce the equivalent of applying all arguments to the original function.
E.g:
Given values v1, v2
Given f(param1, param2)
Then partial(f, param1=v1) returns f'(param2)
And f(param1=v1, param2=v2) == f'(param2=v2) (for any value v2)
Note that in the partial application of a parameter, (in the above case param1), other parameters are not explicitly mentioned. This is a recurring feature of partial function application.
Task
Create a function fs( f, s ) that takes a function, f( n ), of one value and a sequence of values s.
Function fs should return an ordered sequence of the result of applying function f to every value of s in turn.
Create function f1 that takes a value and returns it multiplied by 2.
Create function f2 that takes a value and returns it squared.
Partially apply f1 to fs to form function fsf1( s )
Partially apply f2 to fs to form function fsf2( s )
Test fsf1 and fsf2 by evaluating them with s being the sequence of integers from 0 to 3 inclusive and then the sequence of even integers from 2 to 8 inclusive.
Notes
In partially applying the functions f1 or f2 to fs, there should be no explicit mention of any other parameters to fs, although introspection of fs within the partial applicator to find its parameters is allowed.
This task is more about how results are generated rather than just getting results.
| #J | J | fs=:1 :'u"0 y'
f1=:*&2
f2=:^&2
fsf1=:f1 fs
fsf2=:f2 fs |
http://rosettacode.org/wiki/Partial_function_application | Partial function application | Partial function application is the ability to take a function of many
parameters and apply arguments to some of the parameters to create a new
function that needs only the application of the remaining arguments to
produce the equivalent of applying all arguments to the original function.
E.g:
Given values v1, v2
Given f(param1, param2)
Then partial(f, param1=v1) returns f'(param2)
And f(param1=v1, param2=v2) == f'(param2=v2) (for any value v2)
Note that in the partial application of a parameter, (in the above case param1), other parameters are not explicitly mentioned. This is a recurring feature of partial function application.
Task
Create a function fs( f, s ) that takes a function, f( n ), of one value and a sequence of values s.
Function fs should return an ordered sequence of the result of applying function f to every value of s in turn.
Create function f1 that takes a value and returns it multiplied by 2.
Create function f2 that takes a value and returns it squared.
Partially apply f1 to fs to form function fsf1( s )
Partially apply f2 to fs to form function fsf2( s )
Test fsf1 and fsf2 by evaluating them with s being the sequence of integers from 0 to 3 inclusive and then the sequence of even integers from 2 to 8 inclusive.
Notes
In partially applying the functions f1 or f2 to fs, there should be no explicit mention of any other parameters to fs, although introspection of fs within the partial applicator to find its parameters is allowed.
This task is more about how results are generated rather than just getting results.
| #Java | Java | import java.util.Arrays;
public class PartialApplication {
interface IntegerFunction {
int call(int arg);
}
// Original method fs(f, s).
static int[] fs(IntegerFunction f, int[] s) {
int[] r = new int[s.length];
for (int i = 0; i < s.length; i++)
r[i] = f.call(s[i]);
return r;
}
interface SequenceFunction {
int[] call(int[] arg);
}
// Curried method fs(f).call(s),
// necessary for partial application.
static SequenceFunction fs(final IntegerFunction f) {
return new SequenceFunction() {
public int[] call(int[] s) {
// Call original method.
return fs(f, s);
}
};
}
static IntegerFunction f1 = new IntegerFunction() {
public int call(int i) {
return i * 2;
}
};
static IntegerFunction f2 = new IntegerFunction() {
public int call(int i) {
return i * i;
}
};
static SequenceFunction fsf1 = fs(f1); // Partial application.
static SequenceFunction fsf2 = fs(f2);
public static void main(String[] args) {
int[][] sequences = {
{ 0, 1, 2, 3 },
{ 2, 4, 6, 8 },
};
for (int[] array : sequences) {
System.out.printf(
"array: %s\n" +
" fsf1(array): %s\n" +
" fsf2(array): %s\n",
Arrays.toString(array),
Arrays.toString(fsf1.call(array)),
Arrays.toString(fsf2.call(array)));
}
}
} |
http://rosettacode.org/wiki/Partition_an_integer_x_into_n_primes | Partition an integer x into n primes | Task
Partition a positive integer X into N distinct primes.
Or, to put it in another way:
Find N unique primes such that they add up to X.
Show in the output section the sum X and the N primes in ascending order separated by plus (+) signs:
• partition 99809 with 1 prime.
• partition 18 with 2 primes.
• partition 19 with 3 primes.
• partition 20 with 4 primes.
• partition 2017 with 24 primes.
• partition 22699 with 1, 2, 3, and 4 primes.
• partition 40355 with 3 primes.
The output could/should be shown in a format such as:
Partitioned 19 with 3 primes: 3+5+11
Use any spacing that may be appropriate for the display.
You need not validate the input(s).
Use the lowest primes possible; use 18 = 5+13, not 18 = 7+11.
You only need to show one solution.
This task is similar to factoring an integer.
Related tasks
Count in factors
Prime decomposition
Factors of an integer
Sieve of Eratosthenes
Primality by trial division
Factors of a Mersenne number
Factors of a Mersenne number
Sequence of primes by trial division
| #Racket | Racket | #lang racket
(require math/number-theory)
(define memoised-next-prime
(let ((m# (make-hash)))
(λ (P) (hash-ref! m# P (λ () (next-prime P))))))
(define (partition-X-into-N-primes X N)
(define (partition-x-into-n-primes-starting-at-P x n P result)
(cond ((= n x 0) result)
((or (= n 0) (= x 0) (> P x)) #f)
(else
(let ((P′ (memoised-next-prime P)))
(or (partition-x-into-n-primes-starting-at-P (- x P) (- n 1) P′ (cons P result))
(partition-x-into-n-primes-starting-at-P x n P′ result))))))
(reverse (or (partition-x-into-n-primes-starting-at-P X N 2 null) (list 'no-solution))))
(define (report-partition X N)
(let ((result (partition-X-into-N-primes X N)))
(printf "partition ~a\twith ~a\tprimes: ~a~%" X N (string-join (map ~a result) " + "))))
(module+ test
(report-partition 99809 1)
(report-partition 18 2)
(report-partition 19 3)
(report-partition 20 4)
(report-partition 2017 24)
(report-partition 22699 1)
(report-partition 22699 2)
(report-partition 22699 3)
(report-partition 22699 4)
(report-partition 40355 3)) |
http://rosettacode.org/wiki/Partition_an_integer_x_into_n_primes | Partition an integer x into n primes | Task
Partition a positive integer X into N distinct primes.
Or, to put it in another way:
Find N unique primes such that they add up to X.
Show in the output section the sum X and the N primes in ascending order separated by plus (+) signs:
• partition 99809 with 1 prime.
• partition 18 with 2 primes.
• partition 19 with 3 primes.
• partition 20 with 4 primes.
• partition 2017 with 24 primes.
• partition 22699 with 1, 2, 3, and 4 primes.
• partition 40355 with 3 primes.
The output could/should be shown in a format such as:
Partitioned 19 with 3 primes: 3+5+11
Use any spacing that may be appropriate for the display.
You need not validate the input(s).
Use the lowest primes possible; use 18 = 5+13, not 18 = 7+11.
You only need to show one solution.
This task is similar to factoring an integer.
Related tasks
Count in factors
Prime decomposition
Factors of an integer
Sieve of Eratosthenes
Primality by trial division
Factors of a Mersenne number
Factors of a Mersenne number
Sequence of primes by trial division
| #Raku | Raku | use Math::Primesieve;
my $sieve = Math::Primesieve.new;
# short circuit for '1' partition
multi partition ( Int $number, 1 ) { $number.is-prime ?? $number !! () }
multi partition ( Int $number, Int $parts where * > 0 = 2 ) {
my @these = $sieve.primes($number);
for @these.combinations($parts) { .return if .sum == $number };
()
}
# TESTING
(18,2, 19,3, 20,4, 99807,1, 99809,1, 2017,24, 22699,1, 22699,2, 22699,3, 22699,4, 40355,3)\
.race(:1batch).map: -> $number, $parts {
say (sprintf "Partition %5d into %2d prime piece", $number, $parts),
$parts == 1 ?? ': ' !! 's: ', join '+', partition($number, $parts) || 'not possible'
} |
http://rosettacode.org/wiki/Pascal%27s_triangle/Puzzle | Pascal's triangle/Puzzle | This puzzle involves a Pascals Triangle, also known as a Pyramid of Numbers.
[ 151]
[ ][ ]
[40][ ][ ]
[ ][ ][ ][ ]
[ X][11][ Y][ 4][ Z]
Each brick of the pyramid is the sum of the two bricks situated below it.
Of the three missing numbers at the base of the pyramid,
the middle one is the sum of the other two (that is, Y = X + Z).
Task
Write a program to find a solution to this puzzle.
| #PureBasic | PureBasic | ; Known;
; A.
; [ 151]
; [a ][b ]
; [40][c ][d ]
; [e ][f ][g ][h ]
; [ X][11][ Y][ 4][ Z]
;
; B.
; Y = X + Z
Procedure.i SolveForZ(x)
Protected a,b,c,d,e,f,g,h,z
For z=0 To 20
e=x+11: f=11+(x+z): g=(x+z)+4: h=4+z
If e+f=40
c=f+g : d=g+h: a=40+c: b=c+d
If a+b=151
ProcedureReturn z
EndIf
EndIf
Next z
ProcedureReturn -1
EndProcedure
Define x=-1, z=0, title$="Pascal's triangle/Puzzle in PureBasic"
Repeat
x+1
z=SolveForZ(x)
Until z>=0
MessageRequester(title$,"X="+Str(x)+#CRLF$+"Y="+Str(x+z)+#CRLF$+"Z="+Str(z)) |
http://rosettacode.org/wiki/Password_generator | Password generator | Create a password generation program which will generate passwords containing random ASCII characters from the following groups:
lower-case letters: a ──► z
upper-case letters: A ──► Z
digits: 0 ──► 9
other printable characters: !"#$%&'()*+,-./:;<=>?@[]^_{|}~
(the above character list excludes white-space, backslash and grave)
The generated password(s) must include at least one (of each of the four groups):
lower-case letter,
upper-case letter,
digit (numeral), and
one "other" character.
The user must be able to specify the password length and the number of passwords to generate.
The passwords should be displayed or written to a file, one per line.
The randomness should be from a system source or library.
The program should implement a help option or button which should describe the program and options when invoked.
You may also allow the user to specify a seed value, and give the option of excluding visually similar characters.
For example: Il1 O0 5S 2Z where the characters are:
capital eye, lowercase ell, the digit one
capital oh, the digit zero
the digit five, capital ess
the digit two, capital zee
| #Nim | Nim | import os, parseopt, random, sequtils, strformat, strutils
const Symbols = toSeq("!\"#$%&'()*+,-./:;<=>?@[]^_{|}~")
proc passGen(passLength = 10,
count = 1,
seed = 0,
excludeSimilars = false): seq[string] =
## Generate a sequence of passwords.
# Initialize the random number generator.
if seed == 0: randomize()
else: randomize(seed)
# Prepare list of chars per category.
var
lowerLetters = toSeq('a'..'z')
upperLetters = toSeq('A'..'Z')
digits = toSeq('0'..'9')
if excludeSimilars:
lowerLetters.delete(lowerLetters.find('l'))
for c in "IOSZ": upperLetters.delete(upperLetters.find(c))
for c in "012": digits.delete(digits.find(c))
let all = lowerLetters & upperLetters & digits & Symbols
# Generate the passwords.
for _ in 1..count:
var password = newString(passLength)
password[0] = lowerLetters.sample
password[1] = upperLetters.sample
password[2] = digits.sample
password[3] = Symbols.sample
for i in 4..<passLength:
password[i] = all.sample
password.shuffle()
result.add password
proc printHelp() =
## Print the help message.
echo &"Usage: {getAppFileName().lastPathPart} " &
"[-h] [-l:length] [-c:count] [-s:seed] [-x:(true|false)]"
echo " -h: display this help"
echo " -l: length of generated passwords"
echo " -c: number of passwords to generate"
echo " -s: seed for the random number generator"
echo " -x: exclude similar characters"
proc getIntValue(key, val: string): int =
## Get a positive integer value from a string.
try:
result = val.parseInt()
if result <= 0:
raise newException(ValueError, "")
except ValueError:
quit &"Wrong value for option -{key}: {val}", QuitFailure
var
passLength = 10
count = 1
seed = 0
excludeSimilars = false
# Process options.
var parser = initOptParser()
for kind, key, val in parser.getopt():
if kind != cmdShortOption:
printHelp()
quit(QuitFailure)
case key
of "h":
printHelp()
quit(QuitSuccess)
of "l":
passLength = getIntValue(key, val)
of "c":
count = getIntValue(key, val)
of "s":
seed = getIntValue(key, val)
of "x":
if val.toLowerAscii == "true":
excludeSimilars = true
elif val.toLowerAscii == "false":
excludeSimilars = false
else:
quit &"Wrong value for option -x: {val}", QuitFailure
else:
quit &"Wrong option: -{key}"
# Display the passwords.
for pw in passGen(passLength, count, seed, excludeSimilars):
echo pw |
http://rosettacode.org/wiki/Permutations | Permutations | Task
Write a program that generates all permutations of n different objects. (Practically numerals!)
Related tasks
Find the missing permutation
Permutations/Derangements
The number of samples of size k from n objects.
With combinations and permutations generation tasks.
Order Unimportant
Order Important
Without replacement
(
n
k
)
=
n
C
k
=
n
(
n
−
1
)
…
(
n
−
k
+
1
)
k
(
k
−
1
)
…
1
{\displaystyle {\binom {n}{k}}=^{n}\operatorname {C} _{k}={\frac {n(n-1)\ldots (n-k+1)}{k(k-1)\dots 1}}}
n
P
k
=
n
⋅
(
n
−
1
)
⋅
(
n
−
2
)
⋯
(
n
−
k
+
1
)
{\displaystyle ^{n}\operatorname {P} _{k}=n\cdot (n-1)\cdot (n-2)\cdots (n-k+1)}
Task: Combinations
Task: Permutations
With replacement
(
n
+
k
−
1
k
)
=
n
+
k
−
1
C
k
=
(
n
+
k
−
1
)
!
(
n
−
1
)
!
k
!
{\displaystyle {\binom {n+k-1}{k}}=^{n+k-1}\operatorname {C} _{k}={(n+k-1)! \over (n-1)!k!}}
n
k
{\displaystyle n^{k}}
Task: Combinations with repetitions
Task: Permutations with repetitions
| #VBA | VBA | Public Sub Permute(n As Integer, Optional printem As Boolean = True)
'Generate, count and print (if printem is not false) all permutations of first n integers
Dim P() As Integer
Dim t As Integer, i As Integer, j As Integer, k As Integer
Dim count As Long
Dim Last As Boolean
If n <= 1 Then
Debug.Print "Please give a number greater than 1"
Exit Sub
End If
'Initialize
ReDim P(n)
For i = 1 To n
P(i) = i
Next
count = 0
Last = False
Do While Not Last
'print?
If printem Then
For t = 1 To n
Debug.Print P(t);
Next
Debug.Print
End If
count = count + 1
Last = True
i = n - 1
Do While i > 0
If P(i) < P(i + 1) Then
Last = False
Exit Do
End If
i = i - 1
Loop
j = i + 1
k = n
While j < k
' Swap p(j) and p(k)
t = P(j)
P(j) = P(k)
P(k) = t
j = j + 1
k = k - 1
Wend
j = n
While P(j) > P(i)
j = j - 1
Wend
j = j + 1
'Swap p(i) and p(j)
t = P(i)
P(i) = P(j)
P(j) = t
Loop 'While not last
Debug.Print "Number of permutations: "; count
End Sub |
http://rosettacode.org/wiki/Pathological_floating_point_problems | Pathological floating point problems | Most programmers are familiar with the inexactness of floating point calculations in a binary processor.
The classic example being:
0.1 + 0.2 = 0.30000000000000004
In many situations the amount of error in such calculations is very small and can be overlooked or eliminated with rounding.
There are pathological problems however, where seemingly simple, straight-forward calculations are extremely sensitive to even tiny amounts of imprecision.
This task's purpose is to show how your language deals with such classes of problems.
A sequence that seems to converge to a wrong limit.
Consider the sequence:
v1 = 2
v2 = -4
vn = 111 - 1130 / vn-1 + 3000 / (vn-1 * vn-2)
As n grows larger, the series should converge to 6 but small amounts of error will cause it to approach 100.
Task 1
Display the values of the sequence where n = 3, 4, 5, 6, 7, 8, 20, 30, 50 & 100 to at least 16 decimal places.
n = 3 18.5
n = 4 9.378378
n = 5 7.801153
n = 6 7.154414
n = 7 6.806785
n = 8 6.5926328
n = 20 6.0435521101892689
n = 30 6.006786093031205758530554
n = 50 6.0001758466271871889456140207471954695237
n = 100 6.000000019319477929104086803403585715024350675436952458072592750856521767230266
Task 2
The Chaotic Bank Society is offering a new investment account to their customers.
You first deposit $e - 1 where e is 2.7182818... the base of natural logarithms.
After each year, your account balance will be multiplied by the number of years that have passed, and $1 in service charges will be removed.
So ...
after 1 year, your balance will be multiplied by 1 and $1 will be removed for service charges.
after 2 years your balance will be doubled and $1 removed.
after 3 years your balance will be tripled and $1 removed.
...
after 10 years, multiplied by 10 and $1 removed, and so on.
What will your balance be after 25 years?
Starting balance: $e-1
Balance = (Balance * year) - 1 for 25 years
Balance after 25 years: $0.0399387296732302
Task 3, extra credit
Siegfried Rump's example. Consider the following function, designed by Siegfried Rump in 1988.
f(a,b) = 333.75b6 + a2( 11a2b2 - b6 - 121b4 - 2 ) + 5.5b8 + a/(2b)
compute f(a,b) where a=77617.0 and b=33096.0
f(77617.0, 33096.0) = -0.827396059946821
Demonstrate how to solve at least one of the first two problems, or both, and the third if you're feeling particularly jaunty.
See also;
Floating-Point Arithmetic Section 1.3.2 Difficult problems.
| #TI-83_BASIC | TI-83 BASIC | u(1)=2
u(2)=-4
|
http://rosettacode.org/wiki/Pathological_floating_point_problems | Pathological floating point problems | Most programmers are familiar with the inexactness of floating point calculations in a binary processor.
The classic example being:
0.1 + 0.2 = 0.30000000000000004
In many situations the amount of error in such calculations is very small and can be overlooked or eliminated with rounding.
There are pathological problems however, where seemingly simple, straight-forward calculations are extremely sensitive to even tiny amounts of imprecision.
This task's purpose is to show how your language deals with such classes of problems.
A sequence that seems to converge to a wrong limit.
Consider the sequence:
v1 = 2
v2 = -4
vn = 111 - 1130 / vn-1 + 3000 / (vn-1 * vn-2)
As n grows larger, the series should converge to 6 but small amounts of error will cause it to approach 100.
Task 1
Display the values of the sequence where n = 3, 4, 5, 6, 7, 8, 20, 30, 50 & 100 to at least 16 decimal places.
n = 3 18.5
n = 4 9.378378
n = 5 7.801153
n = 6 7.154414
n = 7 6.806785
n = 8 6.5926328
n = 20 6.0435521101892689
n = 30 6.006786093031205758530554
n = 50 6.0001758466271871889456140207471954695237
n = 100 6.000000019319477929104086803403585715024350675436952458072592750856521767230266
Task 2
The Chaotic Bank Society is offering a new investment account to their customers.
You first deposit $e - 1 where e is 2.7182818... the base of natural logarithms.
After each year, your account balance will be multiplied by the number of years that have passed, and $1 in service charges will be removed.
So ...
after 1 year, your balance will be multiplied by 1 and $1 will be removed for service charges.
after 2 years your balance will be doubled and $1 removed.
after 3 years your balance will be tripled and $1 removed.
...
after 10 years, multiplied by 10 and $1 removed, and so on.
What will your balance be after 25 years?
Starting balance: $e-1
Balance = (Balance * year) - 1 for 25 years
Balance after 25 years: $0.0399387296732302
Task 3, extra credit
Siegfried Rump's example. Consider the following function, designed by Siegfried Rump in 1988.
f(a,b) = 333.75b6 + a2( 11a2b2 - b6 - 121b4 - 2 ) + 5.5b8 + a/(2b)
compute f(a,b) where a=77617.0 and b=33096.0
f(77617.0, 33096.0) = -0.827396059946821
Demonstrate how to solve at least one of the first two problems, or both, and the third if you're feeling particularly jaunty.
See also;
Floating-Point Arithmetic Section 1.3.2 Difficult problems.
| #Visual_Basic_.NET | Visual Basic .NET | Imports System.Globalization
Imports System.Numerics
Imports Numerics
#Const USE_BIGRATIONAL = True
#Const BANDED_ROWS = True
#Const INCREASED_LIMITS = True
#If Not USE_BIGRATIONAL Then
' Mock structure to make test code work.
Structure BigRational
Overrides Function ToString() As String
Return "NOT USING BIGRATIONAL"
End Function
Shared Narrowing Operator CType(value As BigRational) As Decimal
Return -1
End Operator
End Structure
#End If
Module Common
Public Const FMT_STR = "{0,4} {1,-15:G9} {2,-24:G17} {3,-32} {4,-32}"
Public ReadOnly Property Headings As String =
String.Format(CultureInfo.InvariantCulture,
FMT_STR,
{"N", "Single", "Double", "Decimal", "BigRational (rounded as decimal)"})
<Conditional("BANDED_ROWS")>
Sub SetConsoleFormat(n As Integer)
If n Mod 2 = 0 Then
Console.BackgroundColor = ConsoleColor.Black
Console.ForegroundColor = ConsoleColor.White
Else
Console.BackgroundColor = ConsoleColor.White
Console.ForegroundColor = ConsoleColor.Black
End If
End Sub
Function FormatOutput(n As Integer, x As (sn As Single, db As Double, dm As Decimal, br As BigRational)) As String
SetConsoleFormat(n)
Return String.Format(CultureInfo.CurrentCulture, FMT_STR, n, x.sn, x.db, x.dm, CDec(x.br))
End Function
Sub Main()
WrongConvergence()
Console.WriteLine()
ChaoticBankSociety()
Console.WriteLine()
SiegfriedRump()
SetConsoleFormat(0)
End Sub
End Module |
http://rosettacode.org/wiki/Parallel_brute_force | Parallel brute force | Task
Find, through brute force, the five-letter passwords corresponding with the following SHA-256 hashes:
1. 1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad
2. 3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b
3. 74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f
Your program should naively iterate through all possible passwords consisting only of five lower-case ASCII English letters. It should use concurrent or parallel processing, if your language supports that feature. You may calculate SHA-256 hashes by calling a library or through a custom implementation. Print each matching password, along with its SHA-256 hash.
Related task: SHA-256
| #Delphi | Delphi |
program Parallel_Brute_Force;
{$APPTYPE CONSOLE}
uses
System.SysUtils,
System.Threading,
DCPsha256;
function Sha256(W: string): string;
var
HashDigest: array[0..31] of byte;
d, i: Byte;
begin
Result := '';
with TDCP_sha256.Create(nil) do
begin
Init;
UpdateStr(W);
final(HashDigest[0]);
for i := 0 to High(HashDigest) do
Result := Result + lowercase(HashDigest[i].ToHexString(2));
end;
end;
procedure Force(a: int64);
var
password: string;
hash: string;
i, j, k, l: integer;
w: string;
const
Words: TArray<string> = ['1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad',
'3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b',
'74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f'];
begin
password := ' ';
password[1] := chr(97 + a);
for i := 97 to 122 do
begin
password[2] := chr(i);
for j := 97 to 122 do
begin
password[3] := chr(j);
for k := 97 to 122 do
begin
password[4] := chr(k);
for l := 97 to 122 do
begin
password[5] := chr(l);
hash := Sha256(password);
for w in Words do
begin
if SameText(hash, w) then
begin
Writeln('>>', password, ' => ', hash);
end;
end;
end;
end;
end;
end;
end;
var
s: string;
begin
TParallel.&For(0, 25, Force);
Writeln('Enter to exit');
readln;
end. |
http://rosettacode.org/wiki/Parallel_calculations | Parallel calculations | Many programming languages allow you to specify computations to be run in parallel.
While Concurrent computing is focused on concurrency,
the purpose of this task is to distribute time-consuming calculations
on as many CPUs as possible.
Assume we have a collection of numbers, and want to find the one
with the largest minimal prime factor
(that is, the one that contains relatively large factors).
To speed up the search, the factorization should be done
in parallel using separate threads or processes,
to take advantage of multi-core CPUs.
Show how this can be formulated in your language.
Parallelize the factorization of those numbers,
then search the returned list of numbers and factors
for the largest minimal factor,
and return that number and its prime factors.
For the prime number decomposition
you may use the solution of the Prime decomposition task.
| #F.23 | F# | open System
open PrimeDecomp // Has the decompose function from the Prime decomposition task
let data = [112272537195293L; 112582718962171L; 112272537095293L; 115280098190773L; 115797840077099L; 1099726829285419L]
let decomp num = decompose num 2L
let largestMinPrimeFactor (numbers: int64 list) =
let decompDetails = Async.Parallel [ for n in numbers -> async { return n, decomp n } ] // Compute the number and its prime decomposition list
|> Async.RunSynchronously // Start and wait for all parallel computations to complete.
|> Array.sortBy (snd >> List.min >> (~-)) // Sort in descending order, based on the min prime decomp number.
decompDetails.[0]
let showLargestMinPrimeFactor numbers =
let number, primeList = largestMinPrimeFactor numbers
printf "Number %d has largest minimal factor:\n " number
List.iter (printf "%d ") primeList
showLargestMinPrimeFactor data |
http://rosettacode.org/wiki/Parallel_calculations | Parallel calculations | Many programming languages allow you to specify computations to be run in parallel.
While Concurrent computing is focused on concurrency,
the purpose of this task is to distribute time-consuming calculations
on as many CPUs as possible.
Assume we have a collection of numbers, and want to find the one
with the largest minimal prime factor
(that is, the one that contains relatively large factors).
To speed up the search, the factorization should be done
in parallel using separate threads or processes,
to take advantage of multi-core CPUs.
Show how this can be formulated in your language.
Parallelize the factorization of those numbers,
then search the returned list of numbers and factors
for the largest minimal factor,
and return that number and its prime factors.
For the prime number decomposition
you may use the solution of the Prime decomposition task.
| #Factor | Factor |
USING: io kernel fry locals sequences arrays math.primes.factors math.parser channels threads prettyprint ;
IN: <filename>
:: map-parallel ( seq quot -- newseq )
<channel> :> ch
seq [ '[ _ quot call ch to ] "factors" spawn ] { } map-as
dup length [ ch from ] replicate nip ;
{ 576460752303423487 576460752303423487
576460752303423487 112272537195293
115284584522153 115280098190773
115797840077099 112582718962171
112272537095293 1099726829285419 }
dup [ factors ] map-parallel
dup [ infimum ] map dup supremum
swap index swap dupd nth -rot swap nth
"Number with largest min. factor is " swap number>string append
", with factors: " append write .
|
http://rosettacode.org/wiki/Parsing/RPN_calculator_algorithm | Parsing/RPN calculator algorithm | Task
Create a stack-based evaluator for an expression in reverse Polish notation (RPN) that also shows the changes in the stack as each individual token is processed as a table.
Assume an input of a correct, space separated, string of tokens of an RPN expression
Test with the RPN expression generated from the Parsing/Shunting-yard algorithm task:
3 4 2 * 1 5 - 2 3 ^ ^ / +
Print or display the output here
Notes
^ means exponentiation in the expression above.
/ means division.
See also
Parsing/Shunting-yard algorithm for a method of generating an RPN from an infix expression.
Several solutions to 24 game/Solve make use of RPN evaluators (although tracing how they work is not a part of that task).
Parsing/RPN to infix conversion.
Arithmetic evaluation.
| #C.2B.2B | C++ | #include <vector>
#include <string>
#include <sstream>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <iterator>
#include <cstdlib>
double rpn(const std::string &expr){
std::istringstream iss(expr);
std::vector<double> stack;
std::cout << "Input\tOperation\tStack after" << std::endl;
std::string token;
while (iss >> token) {
std::cout << token << "\t";
double tokenNum;
if (std::istringstream(token) >> tokenNum) {
std::cout << "Push\t\t";
stack.push_back(tokenNum);
} else {
std::cout << "Operate\t\t";
double secondOperand = stack.back();
stack.pop_back();
double firstOperand = stack.back();
stack.pop_back();
if (token == "*")
stack.push_back(firstOperand * secondOperand);
else if (token == "/")
stack.push_back(firstOperand / secondOperand);
else if (token == "-")
stack.push_back(firstOperand - secondOperand);
else if (token == "+")
stack.push_back(firstOperand + secondOperand);
else if (token == "^")
stack.push_back(std::pow(firstOperand, secondOperand));
else { //just in case
std::cerr << "Error" << std::endl;
std::exit(1);
}
}
std::copy(stack.begin(), stack.end(), std::ostream_iterator<double>(std::cout, " "));
std::cout << std::endl;
}
return stack.back();
}
int main() {
std::string s = " 3 4 2 * 1 5 - 2 3 ^ ^ / + ";
std::cout << "Final answer: " << rpn(s) << std::endl;
return 0;
} |
http://rosettacode.org/wiki/Pancake_numbers | Pancake numbers | Adrian Monk has problems and an assistant, Sharona Fleming. Sharona can deal with most of Adrian's problems except his lack of punctuality paying her remuneration. 2 pay checks down and she prepares him pancakes for breakfast. Knowing that he will be unable to eat them unless they are stacked in ascending order of size she leaves him only a skillet which he can insert at any point in the pile and flip all the above pancakes, repeating until the pile is sorted. Sharona has left the pile of n pancakes such that the maximum number of flips is required. Adrian is determined to do this in as few flips as possible. This sequence n->p(n) is known as the Pancake numbers.
The task is to determine p(n) for n = 1 to 9, and for each show an example requiring p(n) flips.
Sorting_algorithms/Pancake_sort actually performs the sort some giving the number of flips used. How do these compare with p(n)?
Few people know p(20), generously I shall award an extra credit for anyone doing more than p(16).
References
Bill Gates and the pancake problem
A058986
| #Julia | Julia | function pancake(len)
gap, gapsum, adj = 2, 2, -1
while gapsum < len
adj += 1
gap = gap * 2 - 1
gapsum += gap
end
return len + adj
end
for i in 1:25
print("pancake(", lpad(i, 2), ") = ", rpad(pancake(i), 5))
i % 5 == 0 && println()
end
|
http://rosettacode.org/wiki/Pancake_numbers | Pancake numbers | Adrian Monk has problems and an assistant, Sharona Fleming. Sharona can deal with most of Adrian's problems except his lack of punctuality paying her remuneration. 2 pay checks down and she prepares him pancakes for breakfast. Knowing that he will be unable to eat them unless they are stacked in ascending order of size she leaves him only a skillet which he can insert at any point in the pile and flip all the above pancakes, repeating until the pile is sorted. Sharona has left the pile of n pancakes such that the maximum number of flips is required. Adrian is determined to do this in as few flips as possible. This sequence n->p(n) is known as the Pancake numbers.
The task is to determine p(n) for n = 1 to 9, and for each show an example requiring p(n) flips.
Sorting_algorithms/Pancake_sort actually performs the sort some giving the number of flips used. How do these compare with p(n)?
Few people know p(20), generously I shall award an extra credit for anyone doing more than p(16).
References
Bill Gates and the pancake problem
A058986
| #Kotlin | Kotlin | fun pancake(n: Int): Int {
var gap = 2
var sum = 2
var adj = -1
while (sum < n) {
adj++
gap = gap * 2 - 1
sum += gap
}
return n + adj
}
fun main() {
(1 .. 20).map {"p(%2d) = %2d".format(it, pancake(it))}
val lines = results.chunked(5).map { it.joinToString(" ") }
lines.forEach { println(it) }
} |
http://rosettacode.org/wiki/Pancake_numbers | Pancake numbers | Adrian Monk has problems and an assistant, Sharona Fleming. Sharona can deal with most of Adrian's problems except his lack of punctuality paying her remuneration. 2 pay checks down and she prepares him pancakes for breakfast. Knowing that he will be unable to eat them unless they are stacked in ascending order of size she leaves him only a skillet which he can insert at any point in the pile and flip all the above pancakes, repeating until the pile is sorted. Sharona has left the pile of n pancakes such that the maximum number of flips is required. Adrian is determined to do this in as few flips as possible. This sequence n->p(n) is known as the Pancake numbers.
The task is to determine p(n) for n = 1 to 9, and for each show an example requiring p(n) flips.
Sorting_algorithms/Pancake_sort actually performs the sort some giving the number of flips used. How do these compare with p(n)?
Few people know p(20), generously I shall award an extra credit for anyone doing more than p(16).
References
Bill Gates and the pancake problem
A058986
| #MAD | MAD | NORMAL MODE IS INTEGER
VECTOR VALUES ROW = $5(2HP[,I2,4H] = ,I2,S2)*$
INTERNAL FUNCTION(N)
ENTRY TO P.
GAP = 2
ADJ = -1
THROUGH LOOP, FOR SUM=2, GAP, SUM.GE.N
ADJ = ADJ + 1
LOOP GAP = GAP * 2 - 1
FUNCTION RETURN N + ADJ
END OF FUNCTION
THROUGH OUTP, FOR R=1, 5, R.G.20
OUTP PRINT FORMAT ROW, R,P.(R), R+1,P.(R+1), R+2,P.(R+2),
0 R+3,P.(R+3), R+4,P.(R+4), R+5,P.(R+5)
END OF PROGRAM |
http://rosettacode.org/wiki/Pancake_numbers | Pancake numbers | Adrian Monk has problems and an assistant, Sharona Fleming. Sharona can deal with most of Adrian's problems except his lack of punctuality paying her remuneration. 2 pay checks down and she prepares him pancakes for breakfast. Knowing that he will be unable to eat them unless they are stacked in ascending order of size she leaves him only a skillet which he can insert at any point in the pile and flip all the above pancakes, repeating until the pile is sorted. Sharona has left the pile of n pancakes such that the maximum number of flips is required. Adrian is determined to do this in as few flips as possible. This sequence n->p(n) is known as the Pancake numbers.
The task is to determine p(n) for n = 1 to 9, and for each show an example requiring p(n) flips.
Sorting_algorithms/Pancake_sort actually performs the sort some giving the number of flips used. How do these compare with p(n)?
Few people know p(20), generously I shall award an extra credit for anyone doing more than p(16).
References
Bill Gates and the pancake problem
A058986
| #Nim | Nim | import strformat, strutils
func pancake(n: int): int =
var
gap, sumGaps = 2
pg = 1
adj = -1
while sumGaps < n:
inc adj
inc pg, gap
swap pg, gap
inc sumGaps, gap
result = n + adj
var line = ""
for n in 1..20:
line.addSep(" ")
line.add &"p({n:>2}) = {pancake(n):>2}"
if n mod 5 == 0: (echo line; line.setLen(0)) |
http://rosettacode.org/wiki/Pancake_numbers | Pancake numbers | Adrian Monk has problems and an assistant, Sharona Fleming. Sharona can deal with most of Adrian's problems except his lack of punctuality paying her remuneration. 2 pay checks down and she prepares him pancakes for breakfast. Knowing that he will be unable to eat them unless they are stacked in ascending order of size she leaves him only a skillet which he can insert at any point in the pile and flip all the above pancakes, repeating until the pile is sorted. Sharona has left the pile of n pancakes such that the maximum number of flips is required. Adrian is determined to do this in as few flips as possible. This sequence n->p(n) is known as the Pancake numbers.
The task is to determine p(n) for n = 1 to 9, and for each show an example requiring p(n) flips.
Sorting_algorithms/Pancake_sort actually performs the sort some giving the number of flips used. How do these compare with p(n)?
Few people know p(20), generously I shall award an extra credit for anyone doing more than p(16).
References
Bill Gates and the pancake problem
A058986
| #Perl | Perl | use strict;
use warnings;
use feature 'say';
sub pancake {
my($n) = @_;
my ($gap, $sum, $adj) = (2, 2, -1);
while ($sum < $n) { $sum += $gap = $gap * 2 - 1 and $adj++ }
$n + $adj;
}
my $out;
$out .= sprintf "p(%2d) = %2d ", $_, pancake $_ for 1..20;
say $out =~ s/.{1,55}\K /\n/gr;
# Maximum number of flips plus examples using exhaustive search
sub pancake2 {
my ($n) = @_;
my $numStacks = 1;
my @goalStack = 1 .. $n;
my %newStacks = my %stacks = (join(' ',@goalStack), 0);
for my $k (1..1000) {
my %nextStacks;
for my $pos (2..$n) {
for my $key (keys %newStacks) {
my @arr = split ' ', $key;
my $cakes = join ' ', (reverse @arr[0..$pos-1]), @arr[$pos..$#arr];
$nextStacks{$cakes} = $k unless $stacks{$cakes};
}
}
%stacks = (%stacks, (%newStacks = %nextStacks));
my $perms = scalar %stacks;
my %inverted = reverse %stacks;
return $k-1, $inverted{(sort keys %inverted)[-1]} if $perms == $numStacks;
$numStacks = $perms;
}
}
say "\nThe maximum number of flips to sort a given number of elements is:";
for my $n (1..9) {
my ($a,$b) = pancake2($n);
say "pancake($n) = $a example: $b";
} |
http://rosettacode.org/wiki/Palindromic_gapful_numbers | Palindromic gapful numbers | Palindromic gapful numbers
You are encouraged to solve this task according to the task description, using any language you may know.
Numbers (positive integers expressed in base ten) that are (evenly) divisible by the number formed by the
first and last digit are known as gapful numbers.
Evenly divisible means divisible with no remainder.
All one─ and two─digit numbers have this property and are trivially excluded. Only
numbers ≥ 100 will be considered for this Rosetta Code task.
Example
1037 is a gapful number because it is evenly divisible by the
number 17 which is formed by the first and last decimal digits
of 1037.
A palindromic number is (for this task, a positive integer expressed in base ten), when the number is
reversed, is the same as the original number.
Task
Show (nine sets) the first 20 palindromic gapful numbers that end with:
the digit 1
the digit 2
the digit 3
the digit 4
the digit 5
the digit 6
the digit 7
the digit 8
the digit 9
Show (nine sets, like above) of palindromic gapful numbers:
the last 15 palindromic gapful numbers (out of 100)
the last 10 palindromic gapful numbers (out of 1,000) {optional}
For other ways of expressing the (above) requirements, see the discussion page.
Note
All palindromic gapful numbers are divisible by eleven.
Related tasks
palindrome detection.
gapful numbers.
Also see
The OEIS entry: A108343 gapful numbers.
| #AppleScript | AppleScript | on doTask()
set part1 to {"First 20 palindromic gapful numbers > 100 ending with each digit from 1 to 9:"}
set part2 to {"86th to 100th such:"}
set part3 to {"991st to 1000th:"}
set astid to AppleScript's text item delimiters
set AppleScript's text item delimiters to " "
repeat with endDigit from 1 to 9
set {collector1, collector2, collector3} to {{}, {}, {}}
set outerNumber to endDigit * 11 -- Number formed from the palindromes' first and last digits.
set oddDigitCount to true -- Starting with palindromes in the hundreds.
set baseHi to endDigit * 10 -- Number formed from just the "high end" digits, initially endDigit and a middle 0.
set hi to baseHi
set carryCheck to hi + 10 -- Number reached when incrementing the "high end" number changes its first digit.
set inc to 10 -- Incrementor for the middle digit(s) of the palindromes themselves.
set counter to 0
set maxNeeded to 1000
set done to false
repeat until (done)
-- Work out every 10th palindrome (middle digit = 0) from the current "high end" number.
set pal to hi
if (oddDigitCount) then
set temp to hi div 10
else
set temp to hi
end if
repeat until (temp is 0)
set pal to pal * 10 + temp mod 10
set temp to temp div 10
end repeat
-- Check the result and the following 9 palindromes (derived by incrementing the middle digit(s))
-- and store as text any which are both gapful and the ones required.
repeat 10 times
if (pal mod outerNumber is 0) then
set counter to counter + 1
if (counter ≤ 20) then
set end of collector1 to intText(pal)
else if (counter < 86) then
else if (counter ≤ 100) then
set end of collector2 to intText(pal)
else if (counter < 991) then
else --if (counter ≤ 1000) then
set end of collector3 to intText(pal)
set done to (counter = maxNeeded)
if (done) then exit repeat
end if
end if
set pal to pal + inc
end repeat
-- Increment the high end number's penultimate digit after every 10th palindrome.
-- If a carry changes its first digit, reset for longer palindromes.
set hi to hi + 10
if (hi = carryCheck) then
set oddDigitCount to (not oddDigitCount)
if (oddDigitCount) then
set baseHi to baseHi * 10
set carryCheck to carryCheck * 10
set inc to inc div 11 * 10
else
set inc to inc * 11
end if
set hi to baseHi
end if
end repeat
set {end of part1, end of part2, end of part3} to {collector1 as text, collector2 as text, collector3 as text}
end repeat
set AppleScript's text item delimiters to linefeed
set output to {part1, "", part2, "", part3} as text
set AppleScript's text item delimiters to astid
return output
end doTask
on intText(n)
if (n < 100000000) then return n as text
set txt to text 2 thru end of ((100000000 + (n mod 100000000 as integer)) as text)
set n to n div 100000000
repeat
set lo to n mod 100000000 as integer
set n to n div 100000000
if (n is 0) then return (lo as text) & txt
set txt to (text 2 thru end of ((100000000 + lo) as text)) & txt
end repeat
end intText
return doTask() |
http://rosettacode.org/wiki/Parsing/Shunting-yard_algorithm | Parsing/Shunting-yard algorithm | Task
Given the operator characteristics and input from the Shunting-yard algorithm page and tables, use the algorithm to show the changes in the operator stack and RPN output
as each individual token is processed.
Assume an input of a correct, space separated, string of tokens representing an infix expression
Generate a space separated output string representing the RPN
Test with the input string:
3 + 4 * 2 / ( 1 - 5 ) ^ 2 ^ 3
print and display the output here.
Operator precedence is given in this table:
operator
precedence
associativity
operation
^
4
right
exponentiation
*
3
left
multiplication
/
3
left
division
+
2
left
addition
-
2
left
subtraction
Extra credit
Add extra text explaining the actions and an optional comment for the action on receipt of each token.
Note
The handling of functions and arguments is not required.
See also
Parsing/RPN calculator algorithm for a method of calculating a final value from this output RPN expression.
Parsing/RPN to infix conversion.
| #Fortran | Fortran | MODULE COMPILER !At least of arithmetic expressions.
INTEGER KBD,MSG !I/O units.
INTEGER ENUFF !How long s a piece of string?
PARAMETER (ENUFF = 66) !This long.
CHARACTER*(ENUFF) RP !Holds the Reverse Polish Notation.
INTEGER LR !And this is its length.
INTEGER OPSYMBOLS !Recognised operator symbols.
PARAMETER (OPSYMBOLS = 11) !There are also some associates.
TYPE SYMB !To recognise symbols and carry associated information.
CHARACTER*1 IS !Its text. Careful with the trailing space and comparisons.
INTEGER*1 PRECEDENCE !Controls the order of evaluation.
CHARACTER*48 USAGE !Description.
END TYPE SYMB !The cross-linkage of precedences is tricky.
TYPE(SYMB) SYMBOL(0:OPSYMBOLS) !Righto, I'll have some.
PARAMETER (SYMBOL =(/ !Note that "*" is not to be seen as a match to "**".
o SYMB(" ", 0,"Not recognised as an operator's symbol."),
1 SYMB(" ", 1,"separates symbols and aids legibility."),
2 SYMB(")", 4,"opened with ( to bracket a sub-expression."),
3 SYMB("]", 4,"opened with [ to bracket a sub-expression."),
4 SYMB("}", 4,"opened with { to bracket a sub-expression."),
5 SYMB("+",11,"addition, and unary + to no effect."),
6 SYMB("-",11,"subtraction, and unary - for neg. numbers."),
7 SYMB("*",12,"multiplication."),
8 SYMB("×",12,"multiplication, if you can find this."),
9 SYMB("/",12,"division."),
o SYMB("÷",12,"division for those with a fancy keyboard."),
C 13 is used so that stacked ^ will have lower priority than incoming ^, thus delivering right-to-left evaluation.
1 SYMB("^",14,"raise to power. Not recognised is **.")/))
CHARACTER*3 BRAOPEN,BRACLOSE !Three types are allowed.
PARAMETER (BRAOPEN = "([{", BRACLOSE = ")]}") !These.
INTEGER BRALEVEL !In and out, in and out. That's the game.
INTEGER PRBRA,PRPOW !Special double values.
PARAMETER (PRBRA = SYMBOL( 3).PRECEDENCE) !Bracketing
PARAMETER (PRPOW = SYMBOL(11).PRECEDENCE) !And powers refer leftwards.
CHARACTER*10 DIGIT !Numberish is a bit more complex.
PARAMETER (DIGIT = "0123456789") !But this will do for integers.
INTEGER STACKLIMIT !How high is a stack?
PARAMETER (STACKLIMIT = 66) !This should suffice.
TYPE DEFERRED !I need a siding for lower-precedence operations.
CHARACTER*1 OPC !The operation code.
INTEGER*1 PRECEDENCE !Its precedence in the siding may differ.
END TYPE DEFERRED !Anyway, that's enough.
TYPE(DEFERRED) OPSTACK(0:STACKLIMIT) !One siding, please.
INTEGER OSP !The operation stack pointer.
INTEGER INCOMING,TOKENTYPE,NOTHING,ANUMBER,OPENBRA,HUH !Some mnemonics.
PARAMETER (NOTHING = 0, ANUMBER = -1, OPENBRA = -2, HUH = -3) !The ordering is not arbitrary.
CONTAINS !Now to mess about.
SUBROUTINE EMIT(STUFF) !The objective is to produce some RPN text.
CHARACTER*(*) STUFF !The term of the moment.
INTEGER L !A length.
WRITE (MSG,1) STUFF !Announce.
1 FORMAT ("Emit ",A) !Whatever it is.
IF (STUFF.EQ."") RETURN !Ha ha.
L = LEN(STUFF) !So, how much is there to append?
IF (LR + L.GE.ENUFF) STOP "Too much RPN for RP!" !Perhaps too much.
IF (LR.GT.0) THEN !Is there existing stuff?
LR = LR + 1 !Yes. Advance one,
RP(LR:LR) = " " !And place a space.
END IF !So much for separators.
RP(LR + 1:LR + L) = STUFF !Place the stuff.
LR = LR + L !Count it in.
END SUBROUTINE EMIT !Simple enough, if a bit finicky.
SUBROUTINE STACKOP(C,P) !Push an item into the siding.
CHARACTER*1 C !The operation code.
INTEGER P !Its precedence.
OSP = OSP + 1 !Stacking up...
IF (OSP.GT.STACKLIMIT) STOP "OSP overtopped!" !Perhaps not.
OPSTACK(OSP).OPC = C !Righto,
OPSTACK(OSP).PRECEDENCE = P !The deed is simple.
WRITE (MSG,1) C,OPSTACK(1:OSP) !Announce.
1 FORMAT ("Stack ",A1,9X,",OpStk=",33(A1,I2:","))
END SUBROUTINE STACKOP !So this is more for mnemonic ease.
LOGICAL FUNCTION COMPILE(TEXT) !A compiler confronts a compiler!
CHARACTER*(*) TEXT !To be inspected.
INTEGER L1,L2 !Fingers for the scan.
CHARACTER*1 C !Character of the moment.
INTEGER HAPPY !Ah, shades of mood.
LR = 0 !No output yet.
OSP = 0 !Nothing stacked.
OPSTACK(0).OPC = "" !Prepare a bouncer.
OPSTACK(0).PRECEDENCE = 0 !So that loops won't go past.
BRALEVEL = 0 !None seen.
HAPPY = +1 !Nor any problems.
L2 = 1 !Syncopation: one past the end of the previous token.
Chew into an operand, possibly obstructed by an open bracket.
100 CALL FORASIGN !Find something to inspect.
IF (TOKENTYPE.EQ.NOTHING) THEN !Run off the end?
IF (OSP.GT.0) CALL GRUMP("Another operand or one of " !E.g. "1 +".
1 //BRAOPEN//" is expected.") !Give a hint, because stacked stuff awaits.
ELSE IF (TOKENTYPE.EQ.ANUMBER) THEN !If a number,
CALL EMIT(TEXT(L1:L2 - 1)) !Roll all its digits.
ELSE IF (TOKENTYPE.EQ.OPENBRA) THEN !Starting a sub-expression?
CALL STACKOP(C,PRBRA - 1) !Thus ( has less precedence than ).
GO TO 100 !And I still want an operand.
C ELSE IF (TOKENTYPE.EQ.ANAME) THEN !Name of something?
C CALL EMIT(TEXT(L1:L2 - 1)) !Roll it.
ELSE !No further options.
CALL GRUMP("Huh? Unexpected "//C) !Probably something like "1 + +"
END IF !Righto, finished with operands.
Chase after an operator, possibly interrupted by a close bracket,.
200 CALL FORASIGN !Find something to inspect.
IF (TOKENTYPE.LT.0) THEN !But, have I an operand-like token instead?
CALL GRUMP("Operator expected, not "//C) !It seems so.
ELSE !Normally, an operator is to hand. Possibly a NOTHING, though.
WRITE (MSG,201) C,INCOMING,OPSTACK(1:OSP) !Document it.
201 FORMAT ("Oprn=>",A1,"< Prec=",I2, !Try to align with other output.
1 ",OpStk=",33(A1,I2:",")) !So as not to clutter the display.
DO WHILE(OPSTACK(OSP).PRECEDENCE .GE. INCOMING) !Shunt higher-precedence stuff out.
IF (OPSTACK(OSP).PRECEDENCE .EQ. PRBRA - 1) !Only opening brackets have this precedence.
1 CALL GRUMP("Unbalanced "//OPSTACK(OSP).OPC) !And they vanish only when meeting their closing bracket.
CALL EMIT(OPSTACK(OSP).OPC) !Otherwise we have an operator.
OSP = OSP - 1 !It has gone forth.
END DO !On to the next.
IF (TOKENTYPE.GT.NOTHING) THEN !Now, only lower-precedence items are still in the stack.
IF (INCOMING.EQ.PRBRA) THEN !And this is a special arrival.
CALL BALANCEBRA(C) !It should match an earlier entry.
BRALEVEL = BRALEVEL - 1 !Count it out.
GO TO 200 !And I still haven't got an operator.
ELSE !All others are normal operators.
IF (C.EQ."^") INCOMING = PRPOW - 1 !Special trick to cause leftwards association of x^2^3.
CALL STACKOP(C,INCOMING) !Shunt aside, to await the next arrival.
END IF !So much for that operator.
END IF !Providing it was not just an end-of-input flusher.
END IF !And not a misplaced operand.
Carry on?
IF (HAPPY .GT. 0) GO TO 100 !No problems, and not a nothing from the end of the text.
Completed.
COMPILE = HAPPY.GE.0 !One hopes so.
CONTAINS !Now for some assistants.
SUBROUTINE GRUMP(GROWL) !There might be a problem.
CHARACTER*(*) GROWL !The fault.
WRITE (MSG,1) GROWL !Say it.
IF (L1.GT. 1) WRITE (MSG,1) "Tasty:",TEXT( 1:L1 - 1) !Now explain the context.
IF (L2.GT.L1) WRITE (MSG,1) "Nasty:",TEXT(L1:L2 - 1) !This is the token when trouble was found.
IF (L2.LE.LEN(TEXT)) WRITE (MSG,1) "Misty:",TEXT(L2:) !And this remains to be seen.
1 FORMAT (4X,A,1X,A) !A simple layout works nicely for reasonable-length texts.
HAPPY = -1 . !Just so.
END SUBROUTINE GRUMP !Enuogh said.
SUBROUTINE BALANCEBRA(B) !Perhaps a happy meeting.
CHARACTER*1 B !The closer.
CHARACTER*1 O !The putative opener.
INTEGER IT,L !Fingers.
CHARACTER*88 GROWL !A scratchpad.
O = OPSTACK(OSP).OPC !This should match B.
WRITE (MSG,1) O,B !Perhaps.
1 FORMAT ("Match ",2A) !Show what I've got, anyway.
IT = INDEX(BRAOPEN,O) !So, what sort did I save?
IF (IT .EQ. INDEX(BRACLOSE,B)) THEN !A friend?
OSP = OSP - 1 !Yes. They vanish together.
ELSE !Otherwise, something is out of place.
GROWL = "Unbalanced {[(...)]} bracketing! The closing " !Alas.
1 //B//" is unmatched." !So, a mess.
IF (IT.GT.0) GROWL(62:) = "A "//BRACLOSE(IT:IT) !Perhaps there had been no opening bracket.
1 //" would be better." !But if there had, this would be its friend.
CALL GRUMP(GROWL) !Take that!
END IF !So much for discrepancies.
END SUBROUTINE BALANCEBRA !But, hopefully, amity prevails.
SUBROUTINE FORASIGN !See what comes next.
INTEGER I !A stepper.
L1 = L2 !Pick up where the previous scan left off.
10 IF (L1.GT.LEN(TEXT)) THEN !Are we off the end yet?
C = "" !Yes. Scrub the marker.
L2 = L1 !TEXT(L1:L2 - 1) will be null.
TOKENTYPE = NOTHING !But this is to be checked first.
INCOMING = SYMBOL(1).PRECEDENCE !For flushing sidetracked operators.
HAPPY = 0 !Fading away.
ELSE !Otherwise, there is grist.
Check for spaces and move past them.
C = TEXT(L1:L1) !Grab the first character of the prospective token.
IF (C.LE." ") THEN !Boring?
L1 = L1 + 1 !Yes. Step past it.
GO TO 10 !And look afresh.
END IF !Otherwise, L1 now fingers the start.
Caught something to inspect.
L2 = L1 + 1 !This is one beyond. Just for digit strings.
IF (INDEX(DIGIT,C).GT.0) THEN !So, has one started?
TOKENTYPE = ANUMBER !Yep.
20 IF (L2.LE.LEN(TEXT)) THEN !Probe ahead.
IF (INDEX(DIGIT,TEXT(L2:L2)).GT.0) THEN !Another digit?
L2 = L2 + 1 !Yes. Leaving L1 fingering its start,
GO TO 20 !Chase its end.
END IF !So much for another digit.
END IF !And checking against the end.
C ELSE IF (INDEX(LETTERS,C).GT.0) THEN !Some sort of name?
C advance L2 while in NAMEISH.
ELSE IF (INDEX(BRAOPEN,C).GT.0) THEN !An open bracket?
TOKENTYPE = OPENBRA !Yep.
ELSE !Otherwise, anything else.
DO I = OPSYMBOLS,1,-1 !Scan backwards, to find ** before *, if present.
IF (SYMBOL(I).IS .EQ. C) EXIT !Found?
END DO !On to the next. A linear search will do.
IF (I.LE.0) THEN !Is it identified?
TOKENTYPE = HUH !No.
INCOMING = SYMBOL(0).PRECEDENCE !And this might provoke a flush.
ELSE !If it is identified,
TOKENTYPE = I !Then this is a positive number.
INCOMING = SYMBOL(I).PRECEDENCE !And this is of interest.
END IF !Righto, anything has been identified, possibly as HUH.
END IF !So much for classification.
END IF !If there is something to see.
WRITE (MSG,30) C,INCOMING,TOKENTYPE !Announce.
30 FORMAT ("Next=>",A1,"< Prec=",I2,",Ttype=",I2) !C might be blank.
END SUBROUTINE FORASIGN !I call for a sign, and I see what?
END FUNCTION COMPILE !That's the main activity.
END MODULE COMPILER !So, enough of this.
PROGRAM POKE
USE COMPILER
CHARACTER*66 TEXT
LOGICAL HIC
MSG = 6
KBD = 5
WRITE (MSG,1)
1 FORMAT ("Produce RPN from infix...",/)
10 WRITE (MSG,11)
11 FORMAT("Infix: ",$)
READ(KBD,12) TEXT
12 FORMAT (A)
IF (TEXT.EQ."") STOP "Enough."
HIC = COMPILE(TEXT)
WRITE (MSG,13) HIC,RP(1:LR)
13 FORMAT (L6," RPN: >",A,"<")
GO TO 10
END |
http://rosettacode.org/wiki/Paraffins | Paraffins |
This organic chemistry task is essentially to implement a tree enumeration algorithm.
Task
Enumerate, without repetitions and in order of increasing size, all possible paraffin molecules (also known as alkanes).
Paraffins are built up using only carbon atoms, which has four bonds, and hydrogen, which has one bond. All bonds for each atom must be used, so it is easiest to think of an alkane as linked carbon atoms forming the "backbone" structure, with adding hydrogen atoms linking the remaining unused bonds.
In a paraffin, one is allowed neither double bonds (two bonds between the same pair of atoms), nor cycles of linked carbons. So all paraffins with n carbon atoms share the empirical formula CnH2n+2
But for all n ≥ 4 there are several distinct molecules ("isomers") with the same formula but different structures.
The number of isomers rises rather rapidly when n increases.
In counting isomers it should be borne in mind that the four bond positions on a given carbon atom can be freely interchanged and bonds rotated (including 3-D "out of the paper" rotations when it's being observed on a flat diagram), so rotations or re-orientations of parts of the molecule (without breaking bonds) do not give different isomers. So what seem at first to be different molecules may in fact turn out to be different orientations of the same molecule.
Example
With n = 3 there is only one way of linking the carbons despite the different orientations the molecule can be drawn; and with n = 4 there are two configurations:
a straight chain: (CH3)(CH2)(CH2)(CH3)
a branched chain: (CH3)(CH(CH3))(CH3)
Due to bond rotations, it doesn't matter which direction the branch points in.
The phenomenon of "stereo-isomerism" (a molecule being different from its mirror image due to the actual 3-D arrangement of bonds) is ignored for the purpose of this task.
The input is the number n of carbon atoms of a molecule (for instance 17).
The output is how many different different paraffins there are with n carbon atoms (for instance 24,894 if n = 17).
The sequence of those results is visible in the OEIS entry:
oeis:A00602 number of n-node unrooted quartic trees; number of n-carbon alkanes C(n)H(2n+2) ignoring stereoisomers.
The sequence is (the index starts from zero, and represents the number of carbon atoms):
1, 1, 1, 1, 2, 3, 5, 9, 18, 35, 75, 159, 355, 802, 1858, 4347, 10359,
24894, 60523, 148284, 366319, 910726, 2278658, 5731580, 14490245,
36797588, 93839412, 240215803, 617105614, 1590507121, 4111846763,
10660307791, 27711253769, ...
Extra credit
Show the paraffins in some way.
A flat 1D representation, with arrays or lists is enough, for instance:
*Main> all_paraffins 1
[CCP H H H H]
*Main> all_paraffins 2
[BCP (C H H H) (C H H H)]
*Main> all_paraffins 3
[CCP H H (C H H H) (C H H H)]
*Main> all_paraffins 4
[BCP (C H H (C H H H)) (C H H (C H H H)),
CCP H (C H H H) (C H H H) (C H H H)]
*Main> all_paraffins 5
[CCP H H (C H H (C H H H)) (C H H (C H H H)),
CCP H (C H H H) (C H H H) (C H H (C H H H)),
CCP (C H H H) (C H H H) (C H H H) (C H H H)]
*Main> all_paraffins 6
[BCP (C H H (C H H (C H H H))) (C H H (C H H (C H H H))),
BCP (C H H (C H H (C H H H))) (C H (C H H H) (C H H H)),
BCP (C H (C H H H) (C H H H)) (C H (C H H H) (C H H H)),
CCP H (C H H H) (C H H (C H H H)) (C H H (C H H H)),
CCP (C H H H) (C H H H) (C H H H) (C H H (C H H H))]
Showing a basic 2D ASCII-art representation of the paraffins is better; for instance (molecule names aren't necessary):
methane ethane propane isobutane
H H H H H H H H H
│ │ │ │ │ │ │ │ │
H ─ C ─ H H ─ C ─ C ─ H H ─ C ─ C ─ C ─ H H ─ C ─ C ─ C ─ H
│ │ │ │ │ │ │ │ │
H H H H H H H │ H
│
H ─ C ─ H
│
H
Links
A paper that explains the problem and its solution in a functional language:
http://www.cs.wright.edu/~tkprasad/courses/cs776/paraffins-turner.pdf
A Haskell implementation:
https://github.com/ghc/nofib/blob/master/imaginary/paraffins/Main.hs
A Scheme implementation:
http://www.ccs.neu.edu/home/will/Twobit/src/paraffins.scm
A Fortress implementation: (this site has been closed)
http://java.net/projects/projectfortress/sources/sources/content/ProjectFortress/demos/turnersParaffins0.fss?rev=3005
| #Go | Go | package main
import (
"fmt"
"math/big"
)
const branches = 4
const nMax = 500
var rooted, unrooted [nMax + 1]big.Int
var c [branches]big.Int
var tmp = new(big.Int)
var one = big.NewInt(1)
func tree(br, n, l, sum int, cnt *big.Int) {
for b := br + 1; b <= branches; b++ {
sum += n
if sum > nMax {
return
}
if l*2 >= sum && b >= branches {
return
}
if b == br+1 {
c[br].Mul(&rooted[n], cnt)
} else {
tmp.Add(&rooted[n], tmp.SetInt64(int64(b-br-1)))
c[br].Mul(&c[br], tmp)
c[br].Div(&c[br], tmp.SetInt64(int64(b-br)))
}
if l*2 < sum {
unrooted[sum].Add(&unrooted[sum], &c[br])
}
if b < branches {
rooted[sum].Add(&rooted[sum], &c[br])
}
for m := n - 1; m > 0; m-- {
tree(b, m, l, sum, &c[br])
}
}
}
func bicenter(s int) {
if s&1 == 0 {
tmp.Rsh(tmp.Mul(&rooted[s/2], tmp.Add(&rooted[s/2], one)), 1)
unrooted[s].Add(&unrooted[s], tmp)
}
}
func main() {
rooted[0].SetInt64(1)
rooted[1].SetInt64(1)
unrooted[0].SetInt64(1)
unrooted[1].SetInt64(1)
for n := 1; n <= nMax; n++ {
tree(0, n, n, 1, big.NewInt(1))
bicenter(n)
fmt.Printf("%d: %d\n", n, &unrooted[n])
}
} |
http://rosettacode.org/wiki/Pangram_checker | Pangram checker | Pangram checker
You are encouraged to solve this task according to the task description, using any language you may know.
A pangram is a sentence that contains all the letters of the English alphabet at least once.
For example: The quick brown fox jumps over the lazy dog.
Task
Write a function or method to check a sentence to see if it is a pangram (or not) and show its use.
Related tasks
determine if a string has all the same characters
determine if a string has all unique characters
| #ALGOL_68 | ALGOL 68 | # init pangram: #
INT la = ABS "a", lz = ABS "z";
INT ua = ABS "A", uz = ABS "Z";
IF lz-la+1 > bits width THEN
put(stand error, "Exception: insufficient bits in word for task");
stop
FI;
PROC is a pangram = (STRING test)BOOL: (
BITS a2z := BIN(ABS(2r1 SHL (lz-la))-1); # assume: ASCII & Binary #
FOR i TO UPB test WHILE
INT c = ABS test[i];
IF la <= c AND c <= lz THEN
a2z := a2z AND NOT(2r1 SHL (c-la))
ELIF ua <= c AND c <= uz THEN
a2z := a2z AND NOT(2r1 SHL (c-ua))
FI;
# WHILE # a2z /= 2r0 DO
SKIP
OD;
a2z = 2r0
);
main:(
[]STRING test list = (
"Big fjiords vex quick waltz nymph",
"The quick brown fox jumps over a lazy dog",
"A quick brown fox jumps over a lazy dog"
);
FOR key TO UPB test list DO
STRING test = test list[key];
IF is a pangram(test) THEN
print(("""",test,""" is a pangram!", new line))
FI
OD
) |
http://rosettacode.org/wiki/Pascal_matrix_generation | Pascal matrix generation | A pascal matrix is a two-dimensional square matrix holding numbers from Pascal's triangle, also known as binomial coefficients and which can be shown as nCr.
Shown below are truncated 5-by-5 matrices M[i, j] for i,j in range 0..4.
A Pascal upper-triangular matrix that is populated with jCi:
[[1, 1, 1, 1, 1],
[0, 1, 2, 3, 4],
[0, 0, 1, 3, 6],
[0, 0, 0, 1, 4],
[0, 0, 0, 0, 1]]
A Pascal lower-triangular matrix that is populated with iCj (the transpose of the upper-triangular matrix):
[[1, 0, 0, 0, 0],
[1, 1, 0, 0, 0],
[1, 2, 1, 0, 0],
[1, 3, 3, 1, 0],
[1, 4, 6, 4, 1]]
A Pascal symmetric matrix that is populated with i+jCi:
[[1, 1, 1, 1, 1],
[1, 2, 3, 4, 5],
[1, 3, 6, 10, 15],
[1, 4, 10, 20, 35],
[1, 5, 15, 35, 70]]
Task
Write functions capable of generating each of the three forms of n-by-n matrices.
Use those functions to display upper, lower, and symmetric Pascal 5-by-5 matrices on this page.
The output should distinguish between different matrices and the rows of each matrix (no showing a list of 25 numbers assuming the reader should split it into rows).
Note
The Cholesky decomposition of a Pascal symmetric matrix is the Pascal lower-triangle matrix of the same size.
| #Delphi | Delphi | defmodule Pascal do
defp ij(n), do: for i <- 1..n, j <- 1..n, do: {i,j}
def upper_triangle(n) do
Enum.reduce(ij(n), Map.new, fn {i,j},acc ->
val = cond do
i==1 -> 1
j<i -> 0
true -> Map.get(acc, {i-1, j-1}) + Map.get(acc, {i, j-1})
end
Map.put(acc, {i,j}, val)
end) |> print(1..n)
end
def lower_triangle(n) do
Enum.reduce(ij(n), Map.new, fn {i,j},acc ->
val = cond do
j==1 -> 1
i<j -> 0
true -> Map.get(acc, {i-1, j-1}) + Map.get(acc, {i-1, j})
end
Map.put(acc, {i,j}, val)
end) |> print(1..n)
end
def symmetic_triangle(n) do
Enum.reduce(ij(n), Map.new, fn {i,j},acc ->
val = if i==1 or j==1, do: 1,
else: Map.get(acc, {i-1, j}) + Map.get(acc, {i, j-1})
Map.put(acc, {i,j}, val)
end) |> print(1..n)
end
def print(matrix, range) do
Enum.each(range, fn i ->
Enum.map(range, fn j -> Map.get(matrix, {i,j}) end) |> IO.inspect
end)
end
end
IO.puts "Pascal upper-triangular matrix:"
Pascal.upper_triangle(5)
IO.puts "Pascal lower-triangular matrix:"
Pascal.lower_triangle(5)
IO.puts "Pascal symmetric matrix:"
Pascal.symmetic_triangle(5) |
http://rosettacode.org/wiki/Parameterized_SQL_statement | Parameterized SQL statement | SQL injection
Using a SQL update statement like this one (spacing is optional):
UPDATE players
SET name = 'Smith, Steve', score = 42, active = TRUE
WHERE jerseyNum = 99
Non-parameterized SQL is the GoTo statement of database programming. Don't do it, and make sure your coworkers don't either. | #M2000_Interpreter | M2000 Interpreter |
Module Parametrized_Sql {
Base "rosetta" ' warning erase database if found it in current directory
Execute "rosetta", {create table players (name VarChar(64), score Float, active Integer, jerseyNum Integer);}
Append "rosetta", "players","name",0,FALSE,99
sql$={
UPDATE players
SET name = '{0}', score = {1}, active = {2}
WHERE jerseyNum = {3}
}
Execute "rosetta", format$(sql$,"Smith, Steve", 42,TRUE, 99)
Retrieve "rosetta","players",1,"jerseyNum",99
Read how_many
Read Name$,score, active,jerseynum
Print Name$="Smith, Steve", score=42, active=True, jerseynum=99 ' true true true true
}
Parametrized_Sql
|
http://rosettacode.org/wiki/Parameterized_SQL_statement | Parameterized SQL statement | SQL injection
Using a SQL update statement like this one (spacing is optional):
UPDATE players
SET name = 'Smith, Steve', score = 42, active = TRUE
WHERE jerseyNum = 99
Non-parameterized SQL is the GoTo statement of database programming. Don't do it, and make sure your coworkers don't either. | #Mathematica.2FWolfram_Language | Mathematica/Wolfram Language | Needs["DatabaseLink`"];
conn=OpenSQLConnection[JDBC["ODBC(DSN)","testdb"],"Username"->"John","Password"->"JohnsPassword"];
SQLExecute[conn,"UPDATE players SET name = `1`, score = `2`, active = `3` WHERE jerseyNum = `4`", {SQLArgument["Smith, Steve",42,True,99]}]
CloseSQLConnection[conn]; |
http://rosettacode.org/wiki/Parameterized_SQL_statement | Parameterized SQL statement | SQL injection
Using a SQL update statement like this one (spacing is optional):
UPDATE players
SET name = 'Smith, Steve', score = 42, active = TRUE
WHERE jerseyNum = 99
Non-parameterized SQL is the GoTo statement of database programming. Don't do it, and make sure your coworkers don't either. | #NetRexx | NetRexx | /* NetRexx */
options replace format comments java crossref symbols nobinary
import java.sql.
-- =============================================================================
class RParameterizedSQLSimple public
properties indirect
connexion = Connection
properties inheritable constant
DRIVER = "org.apache.derby.jdbc.EmbeddedDriver"
DBURL = "jdbc:derby:"
DBNAME = "/workspace/DB.DerbySamples/DB/TEAMS01"
DBMODE_CREATE = ";create=true"
DBMODE_NOCREATE = ";create=false"
DBMODE_SHUTDOWN = ";shutdown=true"
-- =============================================================================
method RParameterizedSQLSimple()
setConnexion(null)
return
-- =============================================================================
method createConnexion() inheritable returns Connection signals ClassNotFoundException, InstantiationException, IllegalAccessException
if getConnexion() = null then do
props = Properties()
props.put("user", "user1")
props.put("password", "user1")
xURL = String DBURL || DBNAME || DBMODE_CREATE
loadDriver(DRIVER)
setConnexion(DriverManager.getConnection(xURL, props))
end
return getConnexion()
-- =============================================================================
method shutdownConnexion() inheritable returns boolean signals SQLException
dbState = boolean
xURL = String DBURL || DBNAME || DBMODE_SHUTDOWN
do
DriverManager.getConnection(xURL)
dbState = isTrue
catch se = SQLException
if (se.getErrorCode() = 50000) & ("XJ015".equals(se.getSQLState())) then do
say "Derby shut down normally"
dbState = isTrue
end
else if (se.getErrorCode() = 45000) & ("08006".equals(se.getSQLState())) then do
say "Derby database shut down normally"
dbState = isTrue
end
else do
say "Derby did not shut down normally"
dbState = isFalse
signal se
end
end
return dbState
-- =============================================================================
method loadDriver(xdriver = String) inheritable static signals ClassNotFoundException, InstantiationException, IllegalAccessException
do
Class.forName(xdriver).newInstance()
say "Loaded the appropriate driver"
catch cnfe = ClassNotFoundException
say "Unable to load the JDBC driver" xdriver
say "Please check your CLASSPATH."
signal cnfe
catch ie = InstantiationException
say "Unable to instantiate the JDBC driver" xdriver
signal ie
catch iae = IllegalAccessException
say "Not allowed to access the JDBC driver" xdriver
signal iae
end
return
-- =============================================================================
method updatePlayer(jerseyNum = int, name = String, score = int, active = boolean) binary inheritable returns int signals SQLException
updateSQL = "" -
|| "UPDATE TEAM.PLAYERS" -
|| " SET NAME = ?, SCORE = ?, ACTIVE = ?" -
|| " WHERE JERSEYNUM = ?"
rowCt = int
ix = int 0
ps = getConnexion().prepareStatement(updateSQL)
ix = ix + 1; ps.setString(ix, name)
ix = ix + 1; ps.setInt(ix, score)
ix = ix + 1; ps.setBoolean(ix, active)
ix = ix + 1; ps.setInt(ix, jerseyNum)
rowCt = ps.executeUpdate()
return rowCt
-- =============================================================================
method main(args = String[]) public static
do
tda = RParameterizedSQLSimple()
tda.createConnexion()
if tda.getConnexion() \= null then do
updated = tda.updatePlayer(99, "Smith, Steve", 42, isTrue)
if updated > 0 then say "Update successful"
else say "Update failed"
finally
tda.shutdownConnexion()
end
catch ex = Exception
ex.printStackTrace
end
return
-- =============================================================================
method isTrue() public static returns boolean
return 1 == 1
-- =============================================================================
method isFalse() public static returns boolean
return \isTrue
|
http://rosettacode.org/wiki/Pascal%27s_triangle | Pascal's triangle | Pascal's triangle is an arithmetic and geometric figure often associated with the name of Blaise Pascal, but also studied centuries earlier in India, Persia, China and elsewhere.
Its first few rows look like this:
1
1 1
1 2 1
1 3 3 1
where each element of each row is either 1 or the sum of the two elements right above it.
For example, the next row of the triangle would be:
1 (since the first element of each row doesn't have two elements above it)
4 (1 + 3)
6 (3 + 3)
4 (3 + 1)
1 (since the last element of each row doesn't have two elements above it)
So the triangle now looks like this:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
Each row n (starting with row 0 at the top) shows the coefficients of the binomial expansion of (x + y)n.
Task
Write a function that prints out the first n rows of the triangle (with f(1) yielding the row consisting of only the element 1).
This can be done either by summing elements from the previous rows or using a binary coefficient or combination function.
Behavior for n ≤ 0 does not need to be uniform, but should be noted.
See also
Evaluate binomial coefficients
| #BQN | BQN | Pascal ← {(0⊸∾+∾⟜0)⍟(↕𝕩)⋈1}
•Show¨Pascal 6 |
http://rosettacode.org/wiki/Parse_an_IP_Address | Parse an IP Address | The purpose of this task is to demonstrate parsing of text-format IP addresses, using IPv4 and IPv6.
Taking the following as inputs:
127.0.0.1
The "localhost" IPv4 address
127.0.0.1:80
The "localhost" IPv4 address, with a specified port (80)
::1
The "localhost" IPv6 address
[::1]:80
The "localhost" IPv6 address, with a specified port (80)
2605:2700:0:3::4713:93e3
Rosetta Code's primary server's public IPv6 address
[2605:2700:0:3::4713:93e3]:80
Rosetta Code's primary server's public IPv6 address, with a specified port (80)
Task
Emit each described IP address as a hexadecimal integer representing the address, the address space, and the port number specified, if any.
In languages where variant result types are clumsy, the result should be ipv4 or ipv6 address number, something which says which address space was represented, port number and something that says if the port was specified.
Example
127.0.0.1 has the address number 7F000001 (2130706433 decimal)
in the ipv4 address space.
::ffff:127.0.0.1 represents the same address in the ipv6 address space where it has the
address number FFFF7F000001 (281472812449793 decimal).
::1 has address number 1 and serves the same purpose in the ipv6 address
space that 127.0.0.1 serves in the ipv4 address space.
| #Phix | Phix | with javascript_semantics
function parse_ip(string txt)
sequence r
integer dot = find('.',txt),
colon = find(':',txt),
openbr = find('[',txt),
closebr = find(']',txt)
if (colon=5 and txt[1..4] = "http")
or (colon=6 and txt[1..5] = "https") then
txt = trim(txt[colon+1..$],"/")
return parse_ip(txt)
end if
bool ipv6 = openbr or dot=0 or (colon and colon<dot)
if ipv6 then
sequence res = repeat(0,8)
if openbr then
if openbr!=1 or closebr<openbr then return 0 end if
r = scanf(txt[closebr+1..$],":%d")
if length(r)!=1 then return 0 end if
atom port = r[1][1]
if port<0 or port>65535 then return 0 end if
res &= port
txt = txt[2..closebr-1]
end if
if dot then
colon = rfind(':',txt)
if colon>dot then return 0 end if
object r4 = parse_ip(txt[colon+1..$])
if not sequence(r4) or length(r4)!=4 then return 0 end if
res[7] = r4[1]*#100+r4[2]
res[8] = r4[3]*#100+r4[4]
txt = txt[1..colon-1+(txt[colon-1]=':')]
dot = 2
end if
sequence r8 = {}
integer ip = 0
while length(txt) do
colon = find(':',txt)
if colon=1 then
if ip then return 0 end if
ip = length(r8)+1
txt = txt[2+(length(r8)=0)..$]
else
r = scanf(txt[1..colon-1],"%x")
if length(r)!=1 then return 0 end if
atom r11 = r[1][1]
if r11<0 or r11>#FFFF then return 0 end if
r8 &= r11
if colon=0 then exit end if
txt = txt[colon+1..$]
end if
end while
if ip then
if length(r8)>=(8-dot) then return 0 end if
r8[ip..ip-1] = repeat(0,(8-dot)-length(r8))
else
if length(r8)!=8-dot then return 0 end if
end if
res[1..8-dot] = r8
return res
end if
-- ipv4:
if dot=0 or (colon and colon<dot) then return 0 end if
string d4 = substitute(txt[1..colon-1],'.',' ')
r = scanf(d4,"%d %d %d %d")
if length(r)!=1 then return 0 end if
r = r[1]
for i=1 to length(r) do
if r[i]<0 or r[i]>255 then return 0 end if
end for
if colon then
sequence r2 = scanf(txt[colon+1..$],"%d")
if length(r2)!=1 then return 0 end if
atom port = r2[1][1]
if port<0 or port>65535 then return 0 end if
r &= port
end if
return r
end function
constant tests = {{"127.0.0.1",{127,0,0,1}},
{"127.0.0.1:80",{127,0,0,1,80}},
{"::1",{0,0,0,0,0,0,0,1}},
{"[::1]:80",{0,0,0,0,0,0,0,1,80}},
{"2605:2700:0:3::4713:93e3",{#2605,#2700,0,3,0,0,#4713,#93e3}},
{"[2605:2700:0:3::4713:93e3]:80",{#2605,#2700,0,3,0,0,#4713,#93e3,80}},
{"::ffff:192.168.173.22",{0,0,0,0,0,#ffff,#c0a8,#ad16}},
{"[::ffff:192.168.173.22]:80",{0,0,0,0,0,#ffff,#c0a8,#ad16,80}},
{"1::",{1,0,0,0,0,0,0,0}},
{"[1::]:80",{1,0,0,0,0,0,0,0,80}},
{"::",{0,0,0,0,0,0,0,0}},
{"[::]:80",{0,0,0,0,0,0,0,0,80}},
{"192.168.0.1",{192,168,0,1}},
{"2001:db8:85a3:0:0:8a2e:370:7334",{#2001,#db8,#85a3,0,0,#8a2e,#370,#7334}},
{"2001:db8:85a3::8a2e:370:7334",{#2001,#db8,#85a3,0,0,#8a2e,#370,#7334}},
{"[2001:db8:85a3:8d3:1319:8a2e:370:7334]:443",{#2001,#db8,#85a3,#8d3,#1319,#8a2e,#370,#7334,443}},
{"256.0.0.0",0},
{"g::1",0},
{"::ffff:127.0.0.0.1",0},
{"a::b::1",0},
{"0000",0},
{"0000:0000",0},
{"0000:0000:0000:0000:0000:0000:0000:0000",{0,0,0,0,0,0,0,0}},
{"0000:0000:0000::0000:0000",{0,0,0,0,0,0,0,0}},
{"0000::0000::0000:0000",0},
{"ffff:ffff:ffff:ffff:ffff:ffff:ffff:ffff",{#ffff,#ffff,#ffff,#ffff,#ffff,#ffff,#ffff,#ffff}},
{"ffff:ffff:ffff:fffg:ffff:ffff:ffff:ffff",0},
{"fff:ffff:ffff:ffff:ffff:ffff:ffff:ffff",{#fff,#ffff,#ffff,#ffff,#ffff,#ffff,#ffff,#ffff}},
{"fff:ffff:0:ffff:ffff:ffff:ffff:ffff",{#fff,#ffff,0,#ffff,#ffff,#ffff,#ffff,#ffff}},
{"2001:0db8:0:0:0:0:1428:57ab",{#2001,#0db8,0,0,0,0,#1428,#57ab}},
{"2001:0db8::1428:57ab",{#2001,#0db8,0,0,0,0,#1428,#57ab}},
{"2001:0db8:0:0:8d3:0:0:0",{#2001,#0db8,0,0,#8d3,0,0,0}},
{"2001:0db8:0:0:8d3::",{#2001,#0db8,0,0,#8d3,0,0,0}},
{"2001:0db8::8d3:0:0:0",{#2001,#0db8,0,0,#8d3,0,0,0}},
{"http://127.0.0.1/",{127,0,0,1}},
{"https://127.0.0.1/",{127,0,0,1}},
{"http:",0},
{"http:/2001:db8:3:4::127.0.2.0",{#2001,#db8,3,4,0,0,#7F00,#200}},
{"::192.168.0.1",{0,0,0,0,0,0,#C0A8,1}},
{"::ffff:0:255.255.255.255",{0,0,0,0,#ffff,0,#ffff,#ffff}},
{"",0},
{"ffffffffffffffffffffffffffffffff::",0},
{"[1::]:9999999999999999999999999999",0},
{"I think that's enough tests for now",0}}
for i=1 to length(tests) do
{string ip, object expected} = tests[i]
object actual = parse_ip(ip)
if actual!=expected then
?{ip,actual,expected}
?9/0
end if
string addr
if actual=0 then
addr = "**not a valid ip address**"
else
if remainder(length(actual),2) then
actual[$] = sprintf(", port %d",actual[$])
else
actual = append(actual,"")
end if
if length(actual)=5 then
addr = sprintf("ivp4 address: %02x%02x%02x%02x%s",actual)
elsif length(actual)=9 then
addr = sprintf("ivp6 address: %04x%04x%04x%04x%04x%04x%04x%04x%s",actual)
else
?9/0
end if
end if
printf(1,"%45s %s\n",{ip,addr})
end for
|
http://rosettacode.org/wiki/Parametric_polymorphism | Parametric polymorphism | Parametric Polymorphism
type variables
Task
Write a small example for a type declaration that is parametric over another type, together with a short bit of code (and its type signature) that uses it.
A good example is a container type, let's say a binary tree, together with some function that traverses the tree, say, a map-function that operates on every element of the tree.
This language feature only applies to statically-typed languages.
| #OCaml | OCaml | type 'a tree = Empty | Node of 'a * 'a tree * 'a tree
(** val map_tree : ('a -> 'b) -> 'a tree -> 'b tree *)
let rec map_tree f = function
| Empty -> Empty
| Node (x,l,r) -> Node (f x, map_tree f l, map_tree f r) |
http://rosettacode.org/wiki/Parametric_polymorphism | Parametric polymorphism | Parametric Polymorphism
type variables
Task
Write a small example for a type declaration that is parametric over another type, together with a short bit of code (and its type signature) that uses it.
A good example is a container type, let's say a binary tree, together with some function that traverses the tree, say, a map-function that operates on every element of the tree.
This language feature only applies to statically-typed languages.
| #Phix | Phix | <-------- object --------->
| |
+-atom +-sequence
| |
+-integer +-string
|
http://rosettacode.org/wiki/Parametric_polymorphism | Parametric polymorphism | Parametric Polymorphism
type variables
Task
Write a small example for a type declaration that is parametric over another type, together with a short bit of code (and its type signature) that uses it.
A good example is a container type, let's say a binary tree, together with some function that traverses the tree, say, a map-function that operates on every element of the tree.
This language feature only applies to statically-typed languages.
| #PicoLisp | PicoLisp | (de mapTree (Tree Fun)
(set Tree (Fun (car Tree)))
(and (cadr Tree) (mapTree @ Fun))
(and (cddr Tree) (mapTree @ Fun)) ) |
http://rosettacode.org/wiki/Parametric_polymorphism | Parametric polymorphism | Parametric Polymorphism
type variables
Task
Write a small example for a type declaration that is parametric over another type, together with a short bit of code (and its type signature) that uses it.
A good example is a container type, let's say a binary tree, together with some function that traverses the tree, say, a map-function that operates on every element of the tree.
This language feature only applies to statically-typed languages.
| #Racket | Racket |
#lang typed/racket
(define-type (Tree A) (U False (Node A)))
(struct: (A) Node
([val : A] [left : (Tree A)] [right : (Tree A)])
#:transparent)
(: tree-map (All (A B) (A -> B) (Tree A) -> (Tree B)))
(define (tree-map f tree)
(match tree
[#f #f]
[(Node val left right)
(Node (f val) (tree-map f left) (tree-map f right))]))
;; unit tests
(require typed/rackunit)
(check-equal?
(tree-map add1 (Node 5 (Node 3 #f #f) #f))
(Node 6 (Node 4 #f #f) #f))
|
http://rosettacode.org/wiki/Parsing/RPN_to_infix_conversion | Parsing/RPN to infix conversion | Parsing/RPN to infix conversion
You are encouraged to solve this task according to the task description, using any language you may know.
Task
Create a program that takes an RPN representation of an expression formatted as a space separated sequence of tokens and generates the equivalent expression in infix notation.
Assume an input of a correct, space separated, string of tokens
Generate a space separated output string representing the same expression in infix notation
Show how the major datastructure of your algorithm changes with each new token parsed.
Test with the following input RPN strings then print and display the output here.
RPN input
sample output
3 4 2 * 1 5 - 2 3 ^ ^ / +
3 + 4 * 2 / ( 1 - 5 ) ^ 2 ^ 3
1 2 + 3 4 + ^ 5 6 + ^
( ( 1 + 2 ) ^ ( 3 + 4 ) ) ^ ( 5 + 6 )
Operator precedence and operator associativity is given in this table:
operator
precedence
associativity
operation
^
4
right
exponentiation
*
3
left
multiplication
/
3
left
division
+
2
left
addition
-
2
left
subtraction
See also
Parsing/Shunting-yard algorithm for a method of generating an RPN from an infix expression.
Parsing/RPN calculator algorithm for a method of calculating a final value from this output RPN expression.
Postfix to infix from the RubyQuiz site.
| #Groovy | Groovy | class PostfixToInfix {
static class Expression {
final static String ops = "-+/*^"
String op, ex
int precedence = 3
Expression(String e) {
ex = e
}
Expression(String e1, String e2, String o) {
ex = String.format "%s %s %s", e1, o, e2
op = o
precedence = (ops.indexOf(o) / 2) as int
}
@Override
String toString() {
return ex
}
}
static String postfixToInfix(final String postfix) {
Stack<Expression> expr = new Stack<>()
for (String token in postfix.split("\\s+")) {
char c = token.charAt(0)
int idx = Expression.ops.indexOf(c as int)
if (idx != -1 && token.length() == 1) {
Expression r = expr.pop()
Expression l = expr.pop()
int opPrecedence = (idx / 2) as int
if (l.precedence < opPrecedence || (l.precedence == opPrecedence && c == '^' as char))
l.ex = '(' + l.ex + ')'
if (r.precedence < opPrecedence || (r.precedence == opPrecedence && c != '^' as char))
r.ex = '(' + r.ex + ')'
expr << new Expression(l.ex, r.ex, token)
} else {
expr << new Expression(token)
}
printf "%s -> %s%n", token, expr
}
expr.peek().ex
}
static void main(String[] args) {
(["3 4 2 * 1 5 - 2 3 ^ ^ / +", "1 2 + 3 4 + ^ 5 6 + ^"]).each { String e ->
printf "Postfix : %s%n", e
printf "Infix : %s%n", postfixToInfix(e)
println()
}
}
} |
http://rosettacode.org/wiki/Partial_function_application | Partial function application | Partial function application is the ability to take a function of many
parameters and apply arguments to some of the parameters to create a new
function that needs only the application of the remaining arguments to
produce the equivalent of applying all arguments to the original function.
E.g:
Given values v1, v2
Given f(param1, param2)
Then partial(f, param1=v1) returns f'(param2)
And f(param1=v1, param2=v2) == f'(param2=v2) (for any value v2)
Note that in the partial application of a parameter, (in the above case param1), other parameters are not explicitly mentioned. This is a recurring feature of partial function application.
Task
Create a function fs( f, s ) that takes a function, f( n ), of one value and a sequence of values s.
Function fs should return an ordered sequence of the result of applying function f to every value of s in turn.
Create function f1 that takes a value and returns it multiplied by 2.
Create function f2 that takes a value and returns it squared.
Partially apply f1 to fs to form function fsf1( s )
Partially apply f2 to fs to form function fsf2( s )
Test fsf1 and fsf2 by evaluating them with s being the sequence of integers from 0 to 3 inclusive and then the sequence of even integers from 2 to 8 inclusive.
Notes
In partially applying the functions f1 or f2 to fs, there should be no explicit mention of any other parameters to fs, although introspection of fs within the partial applicator to find its parameters is allowed.
This task is more about how results are generated rather than just getting results.
| #JavaScript | JavaScript | var f1 = function (x) { return x * 2; },
f2 = function (x) { return x * x; },
fs = function (f, s) {
return function (s) {
return s.map(f);
}
},
fsf1 = fs(f1),
fsf2 = fs(f2);
// Test
[
fsf1([0, 1, 2, 3]),
fsf2([0, 1, 2, 3]),
fsf1([2, 4, 6, 8]),
fsf2([2, 4, 6, 8])
] |
http://rosettacode.org/wiki/Partition_an_integer_x_into_n_primes | Partition an integer x into n primes | Task
Partition a positive integer X into N distinct primes.
Or, to put it in another way:
Find N unique primes such that they add up to X.
Show in the output section the sum X and the N primes in ascending order separated by plus (+) signs:
• partition 99809 with 1 prime.
• partition 18 with 2 primes.
• partition 19 with 3 primes.
• partition 20 with 4 primes.
• partition 2017 with 24 primes.
• partition 22699 with 1, 2, 3, and 4 primes.
• partition 40355 with 3 primes.
The output could/should be shown in a format such as:
Partitioned 19 with 3 primes: 3+5+11
Use any spacing that may be appropriate for the display.
You need not validate the input(s).
Use the lowest primes possible; use 18 = 5+13, not 18 = 7+11.
You only need to show one solution.
This task is similar to factoring an integer.
Related tasks
Count in factors
Prime decomposition
Factors of an integer
Sieve of Eratosthenes
Primality by trial division
Factors of a Mersenne number
Factors of a Mersenne number
Sequence of primes by trial division
| #REXX | REXX | /*REXX program partitions integer(s) (greater than unity) into N primes. */
parse arg what /*obtain an optional list from the C.L.*/
do until what=='' /*possibly process a series of integers*/
parse var what x n what; parse var x x '-' y /*get possible range and # partitions.*/
parse var n n '-' m /* " " " " " " */
if x=='' | x=="," then x= 19 /*Not specified? Then use the default.*/
if y=='' | y=="," then y= x /* " " " " " " */
if n=='' | n=="," then n= 3 /* " " " " " " */
if m=='' | m=="," then m= n /* " " " " " " */
call genP y /*generate Y number of primes. */
do g=x to y /*partition X ───► Y into partitions.*/
do q=n to m; call part /* " G into Q primes. */
end /*q*/
end /*g*/
end /*until*/
exit 0 /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
genP: arg high; @.1= 2; @.2= 3; #= 2 /*get highest prime, assign some vars. */
do j=@.#+2 by 2 until @.#>high /*only find odd primes from here on. */
do k=2 while k*k<=j /*divide by some known low odd primes. */
if j // @.k==0 then iterate j /*Is J divisible by P? Then ¬ prime.*/
end /*k*/ /* [↓] a prime (J) has been found. */
#= # + 1; @.#= j /*bump prime count; assign prime to @.*/
end /*j*/; return
/*──────────────────────────────────────────────────────────────────────────────────────*/
getP: procedure expose i. p. @.; parse arg z /*bump the prime in the partition list.*/
if i.z==0 then do; _= z - 1; i.z= i._; end
i.z= i.z + 1; _= i.z; p.z= @._; return 0
/*──────────────────────────────────────────────────────────────────────────────────────*/
list: _= p.1; if $==g then do j=2 to q; _= _ p.j
end /*j*/
else _= '__(not_possible)'
return 'prime' || word("s", 1 + (q==1)) translate(_, '+ ', " _") /*plural? */
/*──────────────────────────────────────────────────────────────────────────────────────*/
part: i.= 0; do j=1 for q; call getP j
end /*j*/
do !=0 by 0; $= 0 /*!: a DO variable for LEAVE & ITERATE*/
do s=1 for q; $= $ + p.s /* [↓] is sum of the primes too large?*/
if $>g then do; if s==1 then leave ! /*perform a quick exit?*/
do k=s to q; i.k= 0; end /*k*/
do r=s-1 to q; call getP r; end /*r*/
iterate !
end
end /*s*/
if $==g then leave /*is sum of the primes exactly right ? */
if $ <g then do; call getP q; iterate; end
end /*!*/ /* [↑] Is sum too low? Bump a prime.*/
say 'partitioned' center(g,9) "into" center(q, 5) list()
return |
http://rosettacode.org/wiki/Pascal%27s_triangle/Puzzle | Pascal's triangle/Puzzle | This puzzle involves a Pascals Triangle, also known as a Pyramid of Numbers.
[ 151]
[ ][ ]
[40][ ][ ]
[ ][ ][ ][ ]
[ X][11][ Y][ 4][ Z]
Each brick of the pyramid is the sum of the two bricks situated below it.
Of the three missing numbers at the base of the pyramid,
the middle one is the sum of the other two (that is, Y = X + Z).
Task
Write a program to find a solution to this puzzle.
| #Python | Python | # Pyramid solver
# [151]
# [ ] [ ]
# [ 40] [ ] [ ]
# [ ] [ ] [ ] [ ]
#[ X ] [ 11] [ Y ] [ 4 ] [ Z ]
# X -Y + Z = 0
def combine( snl, snr ):
cl = {}
if isinstance(snl, int):
cl['1'] = snl
elif isinstance(snl, string):
cl[snl] = 1
else:
cl.update( snl)
if isinstance(snr, int):
n = cl.get('1', 0)
cl['1'] = n + snr
elif isinstance(snr, string):
n = cl.get(snr, 0)
cl[snr] = n + 1
else:
for k,v in snr.items():
n = cl.get(k, 0)
cl[k] = n+v
return cl
def constrain(nsum, vn ):
nn = {}
nn.update(vn)
n = nn.get('1', 0)
nn['1'] = n - nsum
return nn
def makeMatrix( constraints ):
vmap = set()
for c in constraints:
vmap.update( c.keys())
vmap.remove('1')
nvars = len(vmap)
vmap = sorted(vmap) # sort here so output is in sorted order
mtx = []
for c in constraints:
row = []
for vv in vmap:
row.append(float(c.get(vv, 0)))
row.append(-float(c.get('1',0)))
mtx.append(row)
if len(constraints) == nvars:
print 'System appears solvable'
elif len(constraints) < nvars:
print 'System is not solvable - needs more constraints.'
return mtx, vmap
def SolvePyramid( vl, cnstr ):
vl.reverse()
constraints = [cnstr]
lvls = len(vl)
for lvln in range(1,lvls):
lvd = vl[lvln]
for k in range(lvls - lvln):
sn = lvd[k]
ll = vl[lvln-1]
vn = combine(ll[k], ll[k+1])
if sn is None:
lvd[k] = vn
else:
constraints.append(constrain( sn, vn ))
print 'Constraint Equations:'
for cstr in constraints:
fset = ('%d*%s'%(v,k) for k,v in cstr.items() )
print ' + '.join(fset), ' = 0'
mtx,vmap = makeMatrix(constraints)
MtxSolve(mtx)
d = len(vmap)
for j in range(d):
print vmap[j],'=', mtx[j][d]
def MtxSolve(mtx):
# Simple Matrix solver...
mDim = len(mtx) # dimension---
for j in range(mDim):
rw0= mtx[j]
f = 1.0/rw0[j]
for k in range(j, mDim+1):
rw0[k] *= f
for l in range(1+j,mDim):
rwl = mtx[l]
f = -rwl[j]
for k in range(j, mDim+1):
rwl[k] += f * rw0[k]
# backsolve part ---
for j1 in range(1,mDim):
j = mDim - j1
rw0= mtx[j]
for l in range(0, j):
rwl = mtx[l]
f = -rwl[j]
rwl[j] += f * rw0[j]
rwl[mDim] += f * rw0[mDim]
return mtx
p = [ [151], [None,None], [40,None,None], [None,None,None,None], ['X', 11, 'Y', 4, 'Z'] ]
addlConstraint = { 'X':1, 'Y':-1, 'Z':1, '1':0 }
SolvePyramid( p, addlConstraint) |
http://rosettacode.org/wiki/Password_generator | Password generator | Create a password generation program which will generate passwords containing random ASCII characters from the following groups:
lower-case letters: a ──► z
upper-case letters: A ──► Z
digits: 0 ──► 9
other printable characters: !"#$%&'()*+,-./:;<=>?@[]^_{|}~
(the above character list excludes white-space, backslash and grave)
The generated password(s) must include at least one (of each of the four groups):
lower-case letter,
upper-case letter,
digit (numeral), and
one "other" character.
The user must be able to specify the password length and the number of passwords to generate.
The passwords should be displayed or written to a file, one per line.
The randomness should be from a system source or library.
The program should implement a help option or button which should describe the program and options when invoked.
You may also allow the user to specify a seed value, and give the option of excluding visually similar characters.
For example: Il1 O0 5S 2Z where the characters are:
capital eye, lowercase ell, the digit one
capital oh, the digit zero
the digit five, capital ess
the digit two, capital zee
| #OCaml | OCaml | let lower = "abcdefghijklmnopqrstuvwxyz"
let upper = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
let digit = "0123456789"
let other = "!\034#$%&'()*+,-./:;<=>?@[]^_{|}~"
let visually_similar = "Il1O05S2Z"
let mk_pwd len readable =
let get_char i =
match i mod 4 with
| 0 -> lower.[Random.int (String.length lower)]
| 1 -> upper.[Random.int (String.length upper)]
| 2 -> digit.[Random.int (String.length digit)]
| 3 -> other.[Random.int (String.length other)]
| _ -> raise Exit
in
let f =
if readable then
(fun i ->
let rec aux () =
let c = get_char i in
if String.contains visually_similar c
then aux ()
else (String.make 1 c)
in
aux ()
)
else
(fun i ->
let c = get_char i in
(String.make 1 c)
)
in
let r = Array.init len f in
Array.sort (fun _ _ -> (Random.int 3) - 1) r;
(String.concat "" (Array.to_list r))
let () =
Random.self_init ();
let num = ref 1 in
let len = ref 8 in
let readable = ref false in
let speclist = [
"-n", Arg.Set_int num, "number of passwords";
"-c", Arg.Set_int len, "number of characters";
"--readable", Arg.Set readable, "readable";
"--rand-init", Arg.String (fun s ->
Random.full_init
(Array.map int_of_char (Array.of_seq (String.to_seq s)))
), "initialise the random generator with a string";
] in
Arg.parse speclist (fun s -> invalid_arg s) "Password generator";
for i = 1 to !num do
print_endline (mk_pwd !len !readable)
done |
http://rosettacode.org/wiki/Permutations | Permutations | Task
Write a program that generates all permutations of n different objects. (Practically numerals!)
Related tasks
Find the missing permutation
Permutations/Derangements
The number of samples of size k from n objects.
With combinations and permutations generation tasks.
Order Unimportant
Order Important
Without replacement
(
n
k
)
=
n
C
k
=
n
(
n
−
1
)
…
(
n
−
k
+
1
)
k
(
k
−
1
)
…
1
{\displaystyle {\binom {n}{k}}=^{n}\operatorname {C} _{k}={\frac {n(n-1)\ldots (n-k+1)}{k(k-1)\dots 1}}}
n
P
k
=
n
⋅
(
n
−
1
)
⋅
(
n
−
2
)
⋯
(
n
−
k
+
1
)
{\displaystyle ^{n}\operatorname {P} _{k}=n\cdot (n-1)\cdot (n-2)\cdots (n-k+1)}
Task: Combinations
Task: Permutations
With replacement
(
n
+
k
−
1
k
)
=
n
+
k
−
1
C
k
=
(
n
+
k
−
1
)
!
(
n
−
1
)
!
k
!
{\displaystyle {\binom {n+k-1}{k}}=^{n+k-1}\operatorname {C} _{k}={(n+k-1)! \over (n-1)!k!}}
n
k
{\displaystyle n^{k}}
Task: Combinations with repetitions
Task: Permutations with repetitions
| #VBScript | VBScript |
'permutation ,recursive
a=array("Hello",1,True,3.141592)
cnt=0
perm a,0
wscript.echo vbcrlf &"Count " & cnt
sub print(a)
s=""
for i=0 to ubound(a):
s=s &" " & a(i):
next:
wscript.echo s :
cnt=cnt+1 :
end sub
sub swap(a,b) t=a: a=b :b=t: end sub
sub perm(byval a,i)
if i=ubound(a) then print a: exit sub
for j= i to ubound(a)
swap a(i),a(j)
perm a,i+1
swap a(i),a(j)
next
end sub
|
http://rosettacode.org/wiki/Pathological_floating_point_problems | Pathological floating point problems | Most programmers are familiar with the inexactness of floating point calculations in a binary processor.
The classic example being:
0.1 + 0.2 = 0.30000000000000004
In many situations the amount of error in such calculations is very small and can be overlooked or eliminated with rounding.
There are pathological problems however, where seemingly simple, straight-forward calculations are extremely sensitive to even tiny amounts of imprecision.
This task's purpose is to show how your language deals with such classes of problems.
A sequence that seems to converge to a wrong limit.
Consider the sequence:
v1 = 2
v2 = -4
vn = 111 - 1130 / vn-1 + 3000 / (vn-1 * vn-2)
As n grows larger, the series should converge to 6 but small amounts of error will cause it to approach 100.
Task 1
Display the values of the sequence where n = 3, 4, 5, 6, 7, 8, 20, 30, 50 & 100 to at least 16 decimal places.
n = 3 18.5
n = 4 9.378378
n = 5 7.801153
n = 6 7.154414
n = 7 6.806785
n = 8 6.5926328
n = 20 6.0435521101892689
n = 30 6.006786093031205758530554
n = 50 6.0001758466271871889456140207471954695237
n = 100 6.000000019319477929104086803403585715024350675436952458072592750856521767230266
Task 2
The Chaotic Bank Society is offering a new investment account to their customers.
You first deposit $e - 1 where e is 2.7182818... the base of natural logarithms.
After each year, your account balance will be multiplied by the number of years that have passed, and $1 in service charges will be removed.
So ...
after 1 year, your balance will be multiplied by 1 and $1 will be removed for service charges.
after 2 years your balance will be doubled and $1 removed.
after 3 years your balance will be tripled and $1 removed.
...
after 10 years, multiplied by 10 and $1 removed, and so on.
What will your balance be after 25 years?
Starting balance: $e-1
Balance = (Balance * year) - 1 for 25 years
Balance after 25 years: $0.0399387296732302
Task 3, extra credit
Siegfried Rump's example. Consider the following function, designed by Siegfried Rump in 1988.
f(a,b) = 333.75b6 + a2( 11a2b2 - b6 - 121b4 - 2 ) + 5.5b8 + a/(2b)
compute f(a,b) where a=77617.0 and b=33096.0
f(77617.0, 33096.0) = -0.827396059946821
Demonstrate how to solve at least one of the first two problems, or both, and the third if you're feeling particularly jaunty.
See also;
Floating-Point Arithmetic Section 1.3.2 Difficult problems.
| #Wren | Wren | import "/big" for BigRat
import "/fmt" for Fmt
var LIMIT = 100
var bigE = BigRat.fromDecimal("2.71828182845904523536028747135266249775724709369995")
// v(n) sequence task
var c1 = BigRat.new(111)
var c2 = BigRat.new(1130)
var c3 = BigRat.new(3000)
var v1 = BigRat.two
var v2 = BigRat.new(-4)
for (i in 3..LIMIT) {
var v3 = c1 - c2/v2 + c3/(v2*v1)
Fmt.print("$3d : $19s", i, v3.toDecimal(16, true, true))
v1 = v2
v2 = v3
}
// Chaotic Building Society task
var balance = bigE - 1
for (year in 1..25) balance = balance * year - 1
System.print("\nBalance after 25 years is %(balance.toDecimal(16))")
// Siegfried Rump task
var a = BigRat.new(77617)
var b = BigRat.new(33096)
var c4 = BigRat.new(33375, 100)
var c5 = BigRat.new(11)
var c6 = BigRat.new(121)
var c7 = BigRat.new(11, 2)
var f = c4 * b.pow(6) + c7 * b.pow(8) + a/(b*2)
var c8 = c5 * a.pow(2) * b.pow(2) - b.pow(6) - c6 * b.pow(4) - 2
f = f + c8 * a.pow(2)
System.print("\nf(77617.0, 33096.0) is %(f.toDecimal(16))") |
http://rosettacode.org/wiki/Parallel_brute_force | Parallel brute force | Task
Find, through brute force, the five-letter passwords corresponding with the following SHA-256 hashes:
1. 1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad
2. 3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b
3. 74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f
Your program should naively iterate through all possible passwords consisting only of five lower-case ASCII English letters. It should use concurrent or parallel processing, if your language supports that feature. You may calculate SHA-256 hashes by calling a library or through a custom implementation. Print each matching password, along with its SHA-256 hash.
Related task: SHA-256
| #Erlang | Erlang |
#! /usr/bin/escript
-mode(compile).
-export([cracker/4, supervisor/3]).
hexdigit(N) when (N >= 0) and (N =< 9) -> N + $0;
hexdigit(N) when (N >= 10) and (N < 16) -> N - 10 + $a.
hexbyte(N) -> [hexdigit(N bsr 4), hexdigit(N band 15)].
match(Key, Hash) ->
B = crypto:hash(sha256, Key),
Hash == lists:append([hexbyte(X) || <<X:8/integer>> <= B]).
cracker(Prefixes, Rest, Hashes, Boss) ->
Results = [[[P|Q], Hash]
|| P <- Prefixes, Q <- Rest, Hash <- Hashes, match([P|Q], Hash)],
Boss ! {done, Results}.
supervisor(0, Results, Caller) -> Caller ! {done, Results};
supervisor(Tasks, Results, Caller) ->
receive
{done, Cracked} -> supervisor(Tasks - 1, Cracked ++ Results, Caller)
end.
main(_) ->
Tasks = ["abc", "def", "ghi", "jkl", "mno", "pqr", "stuv", "wxyz"],
Letter = lists:seq($a, $z),
Rest = [[B, C, D, E] || B <- Letter, C <- Letter, D <- Letter, E <- Letter],
Hashes = [
"1115dd800feaacefdf481f1f9070374a2a81e27880f187396db67958b207cbad",
"3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b",
"74e1bb62f8dabb8125a58852b63bdf6eaef667cb56ac7f7cdba6d7305c50a22f"
],
Boss = spawn(?MODULE, supervisor, [length(Tasks), [], self()]),
[spawn(?MODULE, cracker, [Prefixes, Rest, Hashes, Boss])
|| Prefixes <- Tasks],
receive
{done, Results} -> Results
end,
[io:format("~s: ~s~n", Result) || Result <- Results].
|
http://rosettacode.org/wiki/Parallel_calculations | Parallel calculations | Many programming languages allow you to specify computations to be run in parallel.
While Concurrent computing is focused on concurrency,
the purpose of this task is to distribute time-consuming calculations
on as many CPUs as possible.
Assume we have a collection of numbers, and want to find the one
with the largest minimal prime factor
(that is, the one that contains relatively large factors).
To speed up the search, the factorization should be done
in parallel using separate threads or processes,
to take advantage of multi-core CPUs.
Show how this can be formulated in your language.
Parallelize the factorization of those numbers,
then search the returned list of numbers and factors
for the largest minimal factor,
and return that number and its prime factors.
For the prime number decomposition
you may use the solution of the Prime decomposition task.
| #Fortran | Fortran |
program Primes
use ISO_FORTRAN_ENV
implicit none
integer(int64), dimension(7) :: data = (/2099726827, 15780709, 1122725370, 15808973, 576460741, 12878611, 12757923/)
integer(int64), dimension(100) :: outprimes
integer(int64) :: largest_factor = 0, largest = 0, minim = 0, val = 0
integer(int16) :: count = 0, OMP_GET_THREAD_NUM
call omp_set_num_threads(4);
!$omp parallel do private(val,outprimes,count) shared(data,largest_factor,largest)
do val = 1, 7
outprimes = 0
call find_factors(data(val), outprimes, count)
minim = minval(outprimes(1:count))
if (minim > largest_factor) then
largest_factor = minim
largest = data(val)
end if
write(*, fmt = '(A7,i0,A2,i12,100i12)') 'Thread ', OMP_GET_THREAD_NUM(), ': ', data(val), outprimes(1:count)
end do
!$omp end parallel do
write(*, fmt = '(i0,A26,i0)') largest, ' have the Largest factor: ', largest_factor
return
contains
subroutine find_factors(n, d, count)
integer(int64), intent(in) :: n
integer(int64), dimension(:), intent(out) :: d
integer(int16), intent(out) :: count
integer(int16) :: i
integer(int64) :: div, next, rest
i = 1
div = 2; next = 3; rest = n
do while (rest /= 1)
do while (mod(rest, div) == 0)
d(i) = div
i = i + 1
rest = rest / div
end do
div = next
next = next + 2
end do
count = i - 1
end subroutine find_factors
end program Primes
|
http://rosettacode.org/wiki/Parsing/RPN_calculator_algorithm | Parsing/RPN calculator algorithm | Task
Create a stack-based evaluator for an expression in reverse Polish notation (RPN) that also shows the changes in the stack as each individual token is processed as a table.
Assume an input of a correct, space separated, string of tokens of an RPN expression
Test with the RPN expression generated from the Parsing/Shunting-yard algorithm task:
3 4 2 * 1 5 - 2 3 ^ ^ / +
Print or display the output here
Notes
^ means exponentiation in the expression above.
/ means division.
See also
Parsing/Shunting-yard algorithm for a method of generating an RPN from an infix expression.
Several solutions to 24 game/Solve make use of RPN evaluators (although tracing how they work is not a part of that task).
Parsing/RPN to infix conversion.
Arithmetic evaluation.
| #Ceylon | Ceylon | import ceylon.collection {
ArrayList
}
shared void run() {
value ops = map {
"+" -> plus<Float>,
"*" -> times<Float>,
"-" -> ((Float a, Float b) => a - b),
"/" -> ((Float a, Float b) => a / b),
"^" -> ((Float a, Float b) => a ^ b)
};
void printTableRow(String|Float token, String description, {Float*} stack) {
print("``token.string.padTrailing(8)````description.padTrailing(30)````stack``");
}
function calculate(String input) {
value stack = ArrayList<Float>();
value tokens = input.split().map((String element)
=> if(ops.keys.contains(element)) then element else parseFloat(element));
print("Token Operation Stack");
for(token in tokens.coalesced) {
if(is Float token) {
stack.push(token);
printTableRow(token, "push", stack);
} else if(exists op = ops[token], exists first = stack.pop(), exists second = stack.pop()) {
value result = op(second, first);
stack.push(result);
printTableRow(token, "perform ``token`` on ``formatFloat(second, 1, 1)`` and ``formatFloat(first, 1, 1)``", stack);
} else {
throw Exception("bad syntax");
}
}
return stack.pop();
}
print(calculate("3 4 2 * 1 5 - 2 3 ^ ^ / +"));
} |
http://rosettacode.org/wiki/Pancake_numbers | Pancake numbers | Adrian Monk has problems and an assistant, Sharona Fleming. Sharona can deal with most of Adrian's problems except his lack of punctuality paying her remuneration. 2 pay checks down and she prepares him pancakes for breakfast. Knowing that he will be unable to eat them unless they are stacked in ascending order of size she leaves him only a skillet which he can insert at any point in the pile and flip all the above pancakes, repeating until the pile is sorted. Sharona has left the pile of n pancakes such that the maximum number of flips is required. Adrian is determined to do this in as few flips as possible. This sequence n->p(n) is known as the Pancake numbers.
The task is to determine p(n) for n = 1 to 9, and for each show an example requiring p(n) flips.
Sorting_algorithms/Pancake_sort actually performs the sort some giving the number of flips used. How do these compare with p(n)?
Few people know p(20), generously I shall award an extra credit for anyone doing more than p(16).
References
Bill Gates and the pancake problem
A058986
| #Phix | Phix | with javascript_semantics
function pancake(integer n)
integer gap = 2, sum_gaps = gap, adj = -1
while sum_gaps<n do
adj += 1
gap = gap*2-1
sum_gaps += gap
end while
n += adj
return n
end function
sequence t = tagset(20),
r = columnize({t,apply(t,pancake)}),
p = apply(true,sprintf,{{"p(%2d) = %2d"},r})
printf(1,"%s\n",join_by(p,1,5))
|
http://rosettacode.org/wiki/Pancake_numbers | Pancake numbers | Adrian Monk has problems and an assistant, Sharona Fleming. Sharona can deal with most of Adrian's problems except his lack of punctuality paying her remuneration. 2 pay checks down and she prepares him pancakes for breakfast. Knowing that he will be unable to eat them unless they are stacked in ascending order of size she leaves him only a skillet which he can insert at any point in the pile and flip all the above pancakes, repeating until the pile is sorted. Sharona has left the pile of n pancakes such that the maximum number of flips is required. Adrian is determined to do this in as few flips as possible. This sequence n->p(n) is known as the Pancake numbers.
The task is to determine p(n) for n = 1 to 9, and for each show an example requiring p(n) flips.
Sorting_algorithms/Pancake_sort actually performs the sort some giving the number of flips used. How do these compare with p(n)?
Few people know p(20), generously I shall award an extra credit for anyone doing more than p(16).
References
Bill Gates and the pancake problem
A058986
| #Python | Python | """Pancake numbers. Requires Python>=3.7."""
import time
from collections import deque
from operator import itemgetter
from typing import Tuple
Pancakes = Tuple[int, ...]
def flip(pancakes: Pancakes, position: int) -> Pancakes:
"""Flip the stack of pancakes at the given position."""
return tuple([*reversed(pancakes[:position]), *pancakes[position:]])
def pancake(n: int) -> Tuple[Pancakes, int]:
"""Return the nth pancake number."""
init_stack = tuple(range(1, n + 1))
stack_flips = {init_stack: 0}
queue = deque([init_stack])
while queue:
stack = queue.popleft()
flips = stack_flips[stack] + 1
for i in range(2, n + 1):
flipped = flip(stack, i)
if flipped not in stack_flips:
stack_flips[flipped] = flips
queue.append(flipped)
return max(stack_flips.items(), key=itemgetter(1))
if __name__ == "__main__":
start = time.time()
for n in range(1, 10):
pancakes, p = pancake(n)
print(f"pancake({n}) = {p:>2}. Example: {list(pancakes)}")
print(f"\nTook {time.time() - start:.3} seconds.")
|
http://rosettacode.org/wiki/Palindromic_gapful_numbers | Palindromic gapful numbers | Palindromic gapful numbers
You are encouraged to solve this task according to the task description, using any language you may know.
Numbers (positive integers expressed in base ten) that are (evenly) divisible by the number formed by the
first and last digit are known as gapful numbers.
Evenly divisible means divisible with no remainder.
All one─ and two─digit numbers have this property and are trivially excluded. Only
numbers ≥ 100 will be considered for this Rosetta Code task.
Example
1037 is a gapful number because it is evenly divisible by the
number 17 which is formed by the first and last decimal digits
of 1037.
A palindromic number is (for this task, a positive integer expressed in base ten), when the number is
reversed, is the same as the original number.
Task
Show (nine sets) the first 20 palindromic gapful numbers that end with:
the digit 1
the digit 2
the digit 3
the digit 4
the digit 5
the digit 6
the digit 7
the digit 8
the digit 9
Show (nine sets, like above) of palindromic gapful numbers:
the last 15 palindromic gapful numbers (out of 100)
the last 10 palindromic gapful numbers (out of 1,000) {optional}
For other ways of expressing the (above) requirements, see the discussion page.
Note
All palindromic gapful numbers are divisible by eleven.
Related tasks
palindrome detection.
gapful numbers.
Also see
The OEIS entry: A108343 gapful numbers.
| #C | C | #include <stdbool.h>
#include <stdio.h>
#include <stdint.h>
typedef uint64_t integer;
integer reverse(integer n) {
integer rev = 0;
while (n > 0) {
rev = rev * 10 + (n % 10);
n /= 10;
}
return rev;
}
typedef struct palgen_tag {
integer power;
integer next;
int digit;
bool even;
} palgen_t;
void init_palgen(palgen_t* palgen, int digit) {
palgen->power = 10;
palgen->next = digit * palgen->power - 1;
palgen->digit = digit;
palgen->even = false;
}
integer next_palindrome(palgen_t* p) {
++p->next;
if (p->next == p->power * (p->digit + 1)) {
if (p->even)
p->power *= 10;
p->next = p->digit * p->power;
p->even = !p->even;
}
return p->next * (p->even ? 10 * p->power : p->power)
+ reverse(p->even ? p->next : p->next/10);
}
bool gapful(integer n) {
integer m = n;
while (m >= 10)
m /= 10;
return n % (n % 10 + 10 * m) == 0;
}
void print(int len, integer array[][len]) {
for (int digit = 1; digit < 10; ++digit) {
printf("%d: ", digit);
for (int i = 0; i < len; ++i)
printf(" %llu", array[digit - 1][i]);
printf("\n");
}
}
int main() {
const int n1 = 20, n2 = 15, n3 = 10;
const int m1 = 100, m2 = 1000;
integer pg1[9][n1];
integer pg2[9][n2];
integer pg3[9][n3];
for (int digit = 1; digit < 10; ++digit) {
palgen_t pgen;
init_palgen(&pgen, digit);
for (int i = 0; i < m2; ) {
integer n = next_palindrome(&pgen);
if (!gapful(n))
continue;
if (i < n1)
pg1[digit - 1][i] = n;
else if (i < m1 && i >= m1 - n2)
pg2[digit - 1][i - (m1 - n2)] = n;
else if (i >= m2 - n3)
pg3[digit - 1][i - (m2 - n3)] = n;
++i;
}
}
printf("First %d palindromic gapful numbers ending in:\n", n1);
print(n1, pg1);
printf("\nLast %d of first %d palindromic gapful numbers ending in:\n", n2, m1);
print(n2, pg2);
printf("\nLast %d of first %d palindromic gapful numbers ending in:\n", n3, m2);
print(n3, pg3);
return 0;
} |
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