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http://rosettacode.org/wiki/Fibonacci_sequence | Fibonacci sequence | The Fibonacci sequence is a sequence Fn of natural numbers defined recursively:
F0 = 0
F1 = 1
Fn = Fn-1 + Fn-2, if n>1
Task
Write a function to generate the nth Fibonacci number.
Solutions can be iterative or recursive (though recursive solutions are generally considered too slow and are mostly used as an exercise in recursion).
The sequence is sometimes extended into negative numbers by using a straightforward inverse of the positive definition:
Fn = Fn+2 - Fn+1, if n<0
support for negative n in the solution is optional.
Related tasks
Fibonacci n-step number sequences
Leonardo numbers
References
Wikipedia, Fibonacci number
Wikipedia, Lucas number
MathWorld, Fibonacci Number
Some identities for r-Fibonacci numbers
OEIS Fibonacci numbers
OEIS Lucas numbers
| #VBScript | VBScript | class generator
dim t1
dim t2
dim tn
dim cur_overflow
Private Sub Class_Initialize
cur_overflow = false
t1 = ccur(0)
t2 = ccur(1)
tn = ccur(t1 + t2)
end sub
public default property get generated
on error resume next
generated = ccur(tn)
if err.number <> 0 then
generated = cdbl(tn)
cur_overflow = true
end if
t1 = ccur(t2)
if err.number <> 0 then
t1 = cdbl(t2)
cur_overflow = true
end if
t2 = ccur(tn)
if err.number <> 0 then
t2 = cdbl(tn)
cur_overflow = true
end if
tn = ccur(t1+ t2)
if err.number <> 0 then
tn = cdbl(t1) + cdbl(t2)
cur_overflow = true
end if
on error goto 0
end property
public property get overflow
overflow = cur_overflow
end property
end class |
http://rosettacode.org/wiki/Equal_prime_and_composite_sums | Equal prime and composite sums | Suppose we have a sequence of prime sums, where each term Pn is the sum of the first n primes.
P = (2), (2 + 3), (2 + 3 + 5), (2 + 3 + 5 + 7), (2 + 3 + 5 + 7 + 11), ...
P = 2, 5, 10, 17, 28, etc.
Further; suppose we have a sequence of composite sums, where each term Cm is the sum of the first m composites.
C = (4), (4 + 6), (4 + 6 + 8), (4 + 6 + 8 + 9), (4 + 6 + 8 + 9 + 10), ...
C = 4, 10, 18, 27, 37, etc.
Notice that the third term of P; P3 (10) is equal to the second term of C; C2 (10);
Task
Find and display the indices (n, m) and value of at least the first 6 terms of the sequence of numbers that are both the sum of the first n primes and the first m composites.
See also
OEIS:A007504 - Sum of the first n primes
OEIS:A053767 - Sum of first n composite numbers
OEIS:A294174 - Numbers that can be expressed both as the sum of first primes and as the sum of first composites
| #XPL0 | XPL0 | func IsPrime(N); \Return 'true' if N is prime
int N, I;
[if N <= 2 then return N = 2;
if (N&1) = 0 then \even >2\ return false;
for I:= 3 to sqrt(N) do
[if rem(N/I) = 0 then return false;
I:= I+1;
];
return true;
];
int Cnt, N, M, SumP, SumC, NumP, NumC;
[Cnt:= 0;
N:= 1; M:= 1;
NumP:= 2; NumC:= 4;
SumP:= 2; SumC:= 4;
Format(8, 0);
Text(0, " sum prime composit
");
loop [if SumC > SumP then
[repeat NumP:= NumP+1 until IsPrime(NumP);
SumP:= SumP + NumP;
N:= N+1;
];
if SumP > SumC then
[repeat NumC:= NumC+1 until not IsPrime(NumC);
SumC:= SumC + NumC;
M:= M+1;
];
if SumP = SumC then
[RlOut(0, float(SumP));
RlOut(0, float(N));
RlOut(0, float(M)); CrLf(0);
Cnt:= Cnt+1;
if Cnt >= 6 then quit;
repeat NumC:= NumC+1 until not IsPrime(NumC);
SumC:= SumC + NumC;
M:= M+1;
];
];
] |
http://rosettacode.org/wiki/Entropy/Narcissist | Entropy/Narcissist |
Task
Write a computer program that computes and shows its own entropy.
Related Tasks
Fibonacci_word
Entropy
| #Ada | Ada | with Ada.Text_Io;
with Ada.Command_Line;
with Ada.Numerics.Elementary_Functions;
procedure Entropy is
use Ada.Text_Io;
type Hist_Type is array (Character) of Natural;
function Log_2 (V : Float) return Float is
use Ada.Numerics.Elementary_Functions;
begin
return Log (V) / Log (2.0);
end Log_2;
procedure Read_File (Name : String; Hist : out Hist_Type) is
File : File_Type;
Char : Character;
begin
Hist := (others => 0);
Open (File, In_File, Name);
while not End_Of_File (File) loop
Get (File, Char);
Hist (Char) := Hist (Char) + 1;
end loop;
Close (File);
end Read_File;
function Length_Of (Hist : Hist_Type) return Natural is
Sum : Natural := 0;
begin
for V of Hist loop
Sum := Sum + V;
end loop;
return Sum;
end Length_Of;
function Entropy_Of (Hist : Hist_Type) return Float is
Length : constant Float := Float (Length_Of (Hist));
Sum : Float := 0.0;
begin
for V of Hist loop
if V > 0 then
Sum := Sum + Float (V) / Length * Log_2 (Float (V) / Length);
end if;
end loop;
return -Sum;
end Entropy_Of;
package Float_Io is new Ada.Text_Io.Float_Io (Float);
Name : constant String := Ada.Command_Line.Argument (1);
Hist : Hist_Type;
Entr : Float;
begin
Float_Io.Default_Exp := 0;
Float_Io.Default_Aft := 6;
Read_File (Name, Hist);
Entr := Entropy_Of (Hist);
Put ("Entropy of '");
Put (Name);
Put ("' is ");
Float_Io.Put (Entr);
New_Line;
end Entropy; |
http://rosettacode.org/wiki/Ethiopian_multiplication | Ethiopian multiplication | Ethiopian multiplication is a method of multiplying integers using only addition, doubling, and halving.
Method:
Take two numbers to be multiplied and write them down at the top of two columns.
In the left-hand column repeatedly halve the last number, discarding any remainders, and write the result below the last in the same column, until you write a value of 1.
In the right-hand column repeatedly double the last number and write the result below. stop when you add a result in the same row as where the left hand column shows 1.
Examine the table produced and discard any row where the value in the left column is even.
Sum the values in the right-hand column that remain to produce the result of multiplying the original two numbers together
For example: 17 × 34
17 34
Halving the first column:
17 34
8
4
2
1
Doubling the second column:
17 34
8 68
4 136
2 272
1 544
Strike-out rows whose first cell is even:
17 34
8 68
4 136
2 272
1 544
Sum the remaining numbers in the right-hand column:
17 34
8 --
4 ---
2 ---
1 544
====
578
So 17 multiplied by 34, by the Ethiopian method is 578.
Task
The task is to define three named functions/methods/procedures/subroutines:
one to halve an integer,
one to double an integer, and
one to state if an integer is even.
Use these functions to create a function that does Ethiopian multiplication.
References
Ethiopian multiplication explained (BBC Video clip)
A Night Of Numbers - Go Forth And Multiply (Video)
Russian Peasant Multiplication
Programming Praxis: Russian Peasant Multiplication
| #ActionScript | ActionScript | function Divide(a:Number):Number {
return ((a-(a%2))/2);
}
function Multiply(a:Number):Number {
return (a *= 2);
}
function isEven(a:Number):Boolean {
if (a%2 == 0) {
return (true);
} else {
return (false);
}
}
function Ethiopian(left:Number, right:Number) {
var r:Number = 0;
trace(left+" "+right);
while (left != 1) {
var State:String = "Keep";
if (isEven(Divide(left))) {
State = "Strike";
}
trace(Divide(left)+" "+Multiply(right)+" "+State);
left = Divide(left);
right = Multiply(right);
if (State == "Keep") {
r += right;
}
}
trace("="+" "+r);
}
} |
http://rosettacode.org/wiki/Equilibrium_index | Equilibrium index | An equilibrium index of a sequence is an index into the sequence such that the sum of elements at lower indices is equal to the sum of elements at higher indices.
For example, in a sequence
A
{\displaystyle A}
:
A
0
=
−
7
{\displaystyle A_{0}=-7}
A
1
=
1
{\displaystyle A_{1}=1}
A
2
=
5
{\displaystyle A_{2}=5}
A
3
=
2
{\displaystyle A_{3}=2}
A
4
=
−
4
{\displaystyle A_{4}=-4}
A
5
=
3
{\displaystyle A_{5}=3}
A
6
=
0
{\displaystyle A_{6}=0}
3 is an equilibrium index, because:
A
0
+
A
1
+
A
2
=
A
4
+
A
5
+
A
6
{\displaystyle A_{0}+A_{1}+A_{2}=A_{4}+A_{5}+A_{6}}
6 is also an equilibrium index, because:
A
0
+
A
1
+
A
2
+
A
3
+
A
4
+
A
5
=
0
{\displaystyle A_{0}+A_{1}+A_{2}+A_{3}+A_{4}+A_{5}=0}
(sum of zero elements is zero)
7 is not an equilibrium index, because it is not a valid index of sequence
A
{\displaystyle A}
.
Task;
Write a function that, given a sequence, returns its equilibrium indices (if any).
Assume that the sequence may be very long.
| #ABAP | ABAP | REPORT equilibrium_index.
TYPES: y_i TYPE STANDARD TABLE OF i WITH EMPTY KEY.
cl_demo_output=>display( REDUCE y_i( LET sequences = VALUE y_i( ( -7 ) ( 1 ) ( 5 ) ( 2 ) ( -4 ) ( 3 ) ( 0 ) )
total_sum = REDUCE #( INIT sum = 0
FOR sequence IN sequences
NEXT sum = sum + ( sequence ) ) IN
INIT x = VALUE y_i( )
y = 0
FOR i = 1 UNTIL i > lines( sequences )
LET z = sequences[ i ] IN
NEXT x = COND #( WHEN y = ( total_sum - y - z ) THEN VALUE y_i( BASE x ( i - 1 ) ) ELSE x )
y = y + z ) ). |
http://rosettacode.org/wiki/Environment_variables | Environment variables | Task
Show how to get one of your process's environment variables.
The available variables vary by system; some of the common ones available on Unix include:
PATH
HOME
USER
| #Ada | Ada | with Ada.Environment_Variables; use Ada.Environment_Variables;
with Ada.Text_Io; use Ada.Text_Io;
procedure Print_Path is
begin
Put_Line("Path : " & Value("PATH"));
end Print_Path; |
http://rosettacode.org/wiki/Environment_variables | Environment variables | Task
Show how to get one of your process's environment variables.
The available variables vary by system; some of the common ones available on Unix include:
PATH
HOME
USER
| #ALGOL_68 | ALGOL 68 | print((getenv("HOME"), new line)) |
http://rosettacode.org/wiki/Environment_variables | Environment variables | Task
Show how to get one of your process's environment variables.
The available variables vary by system; some of the common ones available on Unix include:
PATH
HOME
USER
| #APL | APL |
⎕ENV 'HOME'
HOME /home/russtopia
|
http://rosettacode.org/wiki/Esthetic_numbers | Esthetic numbers | An esthetic number is a positive integer where every adjacent digit differs from its neighbour by 1.
E.G.
12 is an esthetic number. One and two differ by 1.
5654 is an esthetic number. Each digit is exactly 1 away from its neighbour.
890 is not an esthetic number. Nine and zero differ by 9.
These examples are nominally in base 10 but the concept extends easily to numbers in other bases. Traditionally, single digit numbers are included in esthetic numbers; zero may or may not be. For our purposes, for this task, do not include zero (0) as an esthetic number. Do not include numbers with leading zeros.
Esthetic numbers are also sometimes referred to as stepping numbers.
Task
Write a routine (function, procedure, whatever) to find esthetic numbers in a given base.
Use that routine to find esthetic numbers in bases 2 through 16 and display, here on this page, the esthectic numbers from index (base × 4) through index (base × 6), inclusive. (E.G. for base 2: 8th through 12th, for base 6: 24th through 36th, etc.)
Find and display, here on this page, the base 10 esthetic numbers with a magnitude between 1000 and 9999.
Stretch: Find and display, here on this page, the base 10 esthetic numbers with a magnitude between 1.0e8 and 1.3e8.
Related task
numbers with equal rises and falls
See also
OEIS A033075 - Positive numbers n such that all pairs of consecutive decimal digits differ by 1
Numbers Aplenty - Esthetic numbers
Geeks for Geeks - Stepping numbers
| #J | J | isesthetic=: 10&$: :(1 */ .=2 |@-/\ #.inv)"0
gen=: {{r=.$k=.1 while.y>#r do. r=.r,k#~u k
k=.1+({:k)+i.2*#k end.y{.r}} |
http://rosettacode.org/wiki/Euler%27s_sum_of_powers_conjecture | Euler's sum of powers conjecture | There is a conjecture in mathematics that held for over two hundred years before it was disproved by the finding of a counterexample in 1966 by Lander and Parkin.
Euler's (disproved) sum of powers conjecture
At least k positive kth powers are required to sum to a kth power,
except for the trivial case of one kth power: yk = yk
In 1966, Leon J. Lander and Thomas R. Parkin used a brute-force search on a CDC 6600 computer restricting numbers to those less than 250.
Task
Write a program to search for an integer solution for:
x05 + x15 + x25 + x35 == y5
Where all xi's and y are distinct integers between 0 and 250 (exclusive).
Show an answer here.
Related tasks
Pythagorean quadruples.
Pythagorean triples.
| #AWK | AWK |
# syntax: GAWK -f EULERS_SUM_OF_POWERS_CONJECTURE.AWK
BEGIN {
start_int = systime()
main()
printf("%d seconds\n",systime()-start_int)
exit(0)
}
function main( sum,s1,x0,x1,x2,x3) {
for (x0=1; x0<=250; x0++) {
for (x1=1; x1<=x0; x1++) {
for (x2=1; x2<=x1; x2++) {
for (x3=1; x3<=x2; x3++) {
sum = (x0^5) + (x1^5) + (x2^5) + (x3^5)
s1 = int(sum ^ 0.2)
if (sum == s1^5) {
printf("%d^5 + %d^5 + %d^5 + %d^5 = %d^5\n",x0,x1,x2,x3,s1)
return
}
}
}
}
}
}
|
http://rosettacode.org/wiki/Factorial | Factorial | Definitions
The factorial of 0 (zero) is defined as being 1 (unity).
The Factorial Function of a positive integer, n, is defined as the product of the sequence:
n, n-1, n-2, ... 1
Task
Write a function to return the factorial of a number.
Solutions can be iterative or recursive.
Support for trapping negative n errors is optional.
Related task
Primorial numbers
| #Microsoft_Small_Basic | Microsoft Small Basic | 'Factorial - smallbasic - 05/01/2019
For n = 1 To 25
f = 1
For i = 1 To n
f = f * i
EndFor
TextWindow.WriteLine("Factorial(" + n + ")=" + f)
EndFor |
http://rosettacode.org/wiki/Even_or_odd | Even or odd | Task
Test whether an integer is even or odd.
There is more than one way to solve this task:
Use the even and odd predicates, if the language provides them.
Check the least significant digit. With binary integers, i bitwise-and 1 equals 0 iff i is even, or equals 1 iff i is odd.
Divide i by 2. The remainder equals 0 iff i is even. The remainder equals +1 or -1 iff i is odd.
Use modular congruences:
i ≡ 0 (mod 2) iff i is even.
i ≡ 1 (mod 2) iff i is odd.
| #ALGOL-M | ALGOL-M |
BEGIN
% RETURN 1 IF EVEN, OTHERWISE 0 %
INTEGER FUNCTION EVEN(I);
INTEGER I;
BEGIN
EVEN := 1 - (I - 2 * (I / 2));
END;
% TEST THE ROUTINE %
INTEGER K;
FOR K := 1 STEP 3 UNTIL 10 DO
WRITE(K," IS ", IF EVEN(K) = 1 THEN "EVEN" ELSE "ODD");
END |
http://rosettacode.org/wiki/Even_or_odd | Even or odd | Task
Test whether an integer is even or odd.
There is more than one way to solve this task:
Use the even and odd predicates, if the language provides them.
Check the least significant digit. With binary integers, i bitwise-and 1 equals 0 iff i is even, or equals 1 iff i is odd.
Divide i by 2. The remainder equals 0 iff i is even. The remainder equals +1 or -1 iff i is odd.
Use modular congruences:
i ≡ 0 (mod 2) iff i is even.
i ≡ 1 (mod 2) iff i is odd.
| #ALGOL_W | ALGOL W | begin
% the Algol W standard procedure odd returns true if its integer %
% parameter is odd, false if it is even %
for i := 1, 1702, 23, -26
do begin
write( i, " is ", if odd( i ) then "odd" else "even" )
end for_i
end. |
http://rosettacode.org/wiki/Euler_method | Euler method | Euler's method numerically approximates solutions of first-order ordinary differential equations (ODEs) with a given initial value. It is an explicit method for solving initial value problems (IVPs), as described in the wikipedia page.
The ODE has to be provided in the following form:
d
y
(
t
)
d
t
=
f
(
t
,
y
(
t
)
)
{\displaystyle {\frac {dy(t)}{dt}}=f(t,y(t))}
with an initial value
y
(
t
0
)
=
y
0
{\displaystyle y(t_{0})=y_{0}}
To get a numeric solution, we replace the derivative on the LHS with a finite difference approximation:
d
y
(
t
)
d
t
≈
y
(
t
+
h
)
−
y
(
t
)
h
{\displaystyle {\frac {dy(t)}{dt}}\approx {\frac {y(t+h)-y(t)}{h}}}
then solve for
y
(
t
+
h
)
{\displaystyle y(t+h)}
:
y
(
t
+
h
)
≈
y
(
t
)
+
h
d
y
(
t
)
d
t
{\displaystyle y(t+h)\approx y(t)+h\,{\frac {dy(t)}{dt}}}
which is the same as
y
(
t
+
h
)
≈
y
(
t
)
+
h
f
(
t
,
y
(
t
)
)
{\displaystyle y(t+h)\approx y(t)+h\,f(t,y(t))}
The iterative solution rule is then:
y
n
+
1
=
y
n
+
h
f
(
t
n
,
y
n
)
{\displaystyle y_{n+1}=y_{n}+h\,f(t_{n},y_{n})}
where
h
{\displaystyle h}
is the step size, the most relevant parameter for accuracy of the solution. A smaller step size increases accuracy but also the computation cost, so it has always has to be hand-picked according to the problem at hand.
Example: Newton's Cooling Law
Newton's cooling law describes how an object of initial temperature
T
(
t
0
)
=
T
0
{\displaystyle T(t_{0})=T_{0}}
cools down in an environment of temperature
T
R
{\displaystyle T_{R}}
:
d
T
(
t
)
d
t
=
−
k
Δ
T
{\displaystyle {\frac {dT(t)}{dt}}=-k\,\Delta T}
or
d
T
(
t
)
d
t
=
−
k
(
T
(
t
)
−
T
R
)
{\displaystyle {\frac {dT(t)}{dt}}=-k\,(T(t)-T_{R})}
It says that the cooling rate
d
T
(
t
)
d
t
{\displaystyle {\frac {dT(t)}{dt}}}
of the object is proportional to the current temperature difference
Δ
T
=
(
T
(
t
)
−
T
R
)
{\displaystyle \Delta T=(T(t)-T_{R})}
to the surrounding environment.
The analytical solution, which we will compare to the numerical approximation, is
T
(
t
)
=
T
R
+
(
T
0
−
T
R
)
e
−
k
t
{\displaystyle T(t)=T_{R}+(T_{0}-T_{R})\;e^{-kt}}
Task
Implement a routine of Euler's method and then to use it to solve the given example of Newton's cooling law with it for three different step sizes of:
2 s
5 s and
10 s
and to compare with the analytical solution.
Initial values
initial temperature
T
0
{\displaystyle T_{0}}
shall be 100 °C
room temperature
T
R
{\displaystyle T_{R}}
shall be 20 °C
cooling constant
k
{\displaystyle k}
shall be 0.07
time interval to calculate shall be from 0 s ──► 100 s
A reference solution (Common Lisp) can be seen below. We see that bigger step sizes lead to reduced approximation accuracy.
| #J | J | NB.*euler a Approximates Y(t) in Y'(t)=f(t,Y) with Y(a)=Y0 and t=a..b and step size h.
euler=: adverb define
'Y0 a b h'=. 4{. y
t=. i.@>:&.(%&h) b - a
Y=. (+ h * u)^:(<#t) Y0
t,.Y
)
ncl=: _0.07 * -&20 NB. Newton's Cooling Law
|
http://rosettacode.org/wiki/Euler_method | Euler method | Euler's method numerically approximates solutions of first-order ordinary differential equations (ODEs) with a given initial value. It is an explicit method for solving initial value problems (IVPs), as described in the wikipedia page.
The ODE has to be provided in the following form:
d
y
(
t
)
d
t
=
f
(
t
,
y
(
t
)
)
{\displaystyle {\frac {dy(t)}{dt}}=f(t,y(t))}
with an initial value
y
(
t
0
)
=
y
0
{\displaystyle y(t_{0})=y_{0}}
To get a numeric solution, we replace the derivative on the LHS with a finite difference approximation:
d
y
(
t
)
d
t
≈
y
(
t
+
h
)
−
y
(
t
)
h
{\displaystyle {\frac {dy(t)}{dt}}\approx {\frac {y(t+h)-y(t)}{h}}}
then solve for
y
(
t
+
h
)
{\displaystyle y(t+h)}
:
y
(
t
+
h
)
≈
y
(
t
)
+
h
d
y
(
t
)
d
t
{\displaystyle y(t+h)\approx y(t)+h\,{\frac {dy(t)}{dt}}}
which is the same as
y
(
t
+
h
)
≈
y
(
t
)
+
h
f
(
t
,
y
(
t
)
)
{\displaystyle y(t+h)\approx y(t)+h\,f(t,y(t))}
The iterative solution rule is then:
y
n
+
1
=
y
n
+
h
f
(
t
n
,
y
n
)
{\displaystyle y_{n+1}=y_{n}+h\,f(t_{n},y_{n})}
where
h
{\displaystyle h}
is the step size, the most relevant parameter for accuracy of the solution. A smaller step size increases accuracy but also the computation cost, so it has always has to be hand-picked according to the problem at hand.
Example: Newton's Cooling Law
Newton's cooling law describes how an object of initial temperature
T
(
t
0
)
=
T
0
{\displaystyle T(t_{0})=T_{0}}
cools down in an environment of temperature
T
R
{\displaystyle T_{R}}
:
d
T
(
t
)
d
t
=
−
k
Δ
T
{\displaystyle {\frac {dT(t)}{dt}}=-k\,\Delta T}
or
d
T
(
t
)
d
t
=
−
k
(
T
(
t
)
−
T
R
)
{\displaystyle {\frac {dT(t)}{dt}}=-k\,(T(t)-T_{R})}
It says that the cooling rate
d
T
(
t
)
d
t
{\displaystyle {\frac {dT(t)}{dt}}}
of the object is proportional to the current temperature difference
Δ
T
=
(
T
(
t
)
−
T
R
)
{\displaystyle \Delta T=(T(t)-T_{R})}
to the surrounding environment.
The analytical solution, which we will compare to the numerical approximation, is
T
(
t
)
=
T
R
+
(
T
0
−
T
R
)
e
−
k
t
{\displaystyle T(t)=T_{R}+(T_{0}-T_{R})\;e^{-kt}}
Task
Implement a routine of Euler's method and then to use it to solve the given example of Newton's cooling law with it for three different step sizes of:
2 s
5 s and
10 s
and to compare with the analytical solution.
Initial values
initial temperature
T
0
{\displaystyle T_{0}}
shall be 100 °C
room temperature
T
R
{\displaystyle T_{R}}
shall be 20 °C
cooling constant
k
{\displaystyle k}
shall be 0.07
time interval to calculate shall be from 0 s ──► 100 s
A reference solution (Common Lisp) can be seen below. We see that bigger step sizes lead to reduced approximation accuracy.
| #Java | Java |
public class Euler {
private static void euler (Callable f, double y0, int a, int b, int h) {
int t = a;
double y = y0;
while (t < b) {
System.out.println ("" + t + " " + y);
t += h;
y += h * f.compute (t, y);
}
System.out.println ("DONE");
}
public static void main (String[] args) {
Callable cooling = new Cooling ();
int[] steps = {2, 5, 10};
for (int stepSize : steps) {
System.out.println ("Step size: " + stepSize);
euler (cooling, 100.0, 0, 100, stepSize);
}
}
}
// interface used so we can plug in alternative functions to Euler
interface Callable {
public double compute (int time, double t);
}
// class to implement the newton cooling equation
class Cooling implements Callable {
public double compute (int time, double t) {
return -0.07 * (t - 20);
}
}
|
http://rosettacode.org/wiki/Evaluate_binomial_coefficients | Evaluate binomial coefficients | This programming task, is to calculate ANY binomial coefficient.
However, it has to be able to output
(
5
3
)
{\displaystyle {\binom {5}{3}}}
, which is 10.
This formula is recommended:
(
n
k
)
=
n
!
(
n
−
k
)
!
k
!
=
n
(
n
−
1
)
(
n
−
2
)
…
(
n
−
k
+
1
)
k
(
k
−
1
)
(
k
−
2
)
…
1
{\displaystyle {\binom {n}{k}}={\frac {n!}{(n-k)!k!}}={\frac {n(n-1)(n-2)\ldots (n-k+1)}{k(k-1)(k-2)\ldots 1}}}
See Also:
Combinations and permutations
Pascal's triangle
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
| #C.23 | C# | using System;
namespace BinomialCoefficients
{
class Program
{
static void Main(string[] args)
{
ulong n = 1000000, k = 3;
ulong result = biCoefficient(n, k);
Console.WriteLine("The Binomial Coefficient of {0}, and {1}, is equal to: {2}", n, k, result);
Console.ReadLine();
}
static int fact(int n)
{
if (n == 0) return 1;
else return n * fact(n - 1);
}
static ulong biCoefficient(ulong n, ulong k)
{
if (k > n - k)
{
k = n - k;
}
ulong c = 1;
for (uint i = 0; i < k; i++)
{
c = c * (n - i);
c = c / (i + 1);
}
return c;
}
}
} |
http://rosettacode.org/wiki/Fibonacci_sequence | Fibonacci sequence | The Fibonacci sequence is a sequence Fn of natural numbers defined recursively:
F0 = 0
F1 = 1
Fn = Fn-1 + Fn-2, if n>1
Task
Write a function to generate the nth Fibonacci number.
Solutions can be iterative or recursive (though recursive solutions are generally considered too slow and are mostly used as an exercise in recursion).
The sequence is sometimes extended into negative numbers by using a straightforward inverse of the positive definition:
Fn = Fn+2 - Fn+1, if n<0
support for negative n in the solution is optional.
Related tasks
Fibonacci n-step number sequences
Leonardo numbers
References
Wikipedia, Fibonacci number
Wikipedia, Lucas number
MathWorld, Fibonacci Number
Some identities for r-Fibonacci numbers
OEIS Fibonacci numbers
OEIS Lucas numbers
| #Vedit_macro_language | Vedit macro language | :FIBONACCI:
#11 = 0
#12 = 1
Repeat(#1) {
#10 = #11 + #12
#11 = #12
#12 = #10
}
Return(#11) |
http://rosettacode.org/wiki/Entropy/Narcissist | Entropy/Narcissist |
Task
Write a computer program that computes and shows its own entropy.
Related Tasks
Fibonacci_word
Entropy
| #ALGOL_68 | ALGOL 68 | BEGIN
# calculate the shannon entropy of a string #
PROC shannon entropy = ( STRING s )REAL:
BEGIN
INT string length = ( UPB s - LWB s ) + 1;
# count the occurances of each character #
[ 0 : max abs char ]INT char count;
FOR char pos FROM LWB char count TO UPB char count DO
char count[ char pos ] := 0
OD;
FOR char pos FROM LWB s TO UPB s DO
char count[ ABS s[ char pos ] ] +:= 1
OD;
# calculate the entropy, we use log base 10 and then convert #
# to log base 2 after calculating the sum #
REAL entropy := 0;
FOR char pos FROM LWB char count TO UPB char count DO
IF char count[ char pos ] /= 0
THEN
# have a character that occurs in the string #
REAL probability = char count[ char pos ] / string length;
entropy -:= probability * log( probability )
FI
OD;
entropy / log( 2 )
END; # shannon entropy #
IF FILE input file;
STRING file name = "entropyNarcissist.a68";
open( input file, file name, stand in channel ) /= 0
THEN
# failed to open the file #
print( ( "Unable to open """ + file name + """", newline ) )
ELSE
# file opened OK #
BOOL at eof := FALSE;
# set the EOF handler for the file #
on logical file end( input file
, ( REF FILE f )BOOL:
BEGIN
# note that we reached EOF on the latest read #
at eof := TRUE;
# return TRUE so processing can continue #
TRUE
END
);
# construct a string containing the whole file #
STRING file contents := "";
WHILE STRING line;
get( input file, ( line, newline ) );
NOT at eof
DO
file contents +:= line + REPR 12
OD;
close( input file );
# show the entropy of the file cotents #
print( ( shannon entropy( file contents ), newline ) )
FI
END |
http://rosettacode.org/wiki/Ethiopian_multiplication | Ethiopian multiplication | Ethiopian multiplication is a method of multiplying integers using only addition, doubling, and halving.
Method:
Take two numbers to be multiplied and write them down at the top of two columns.
In the left-hand column repeatedly halve the last number, discarding any remainders, and write the result below the last in the same column, until you write a value of 1.
In the right-hand column repeatedly double the last number and write the result below. stop when you add a result in the same row as where the left hand column shows 1.
Examine the table produced and discard any row where the value in the left column is even.
Sum the values in the right-hand column that remain to produce the result of multiplying the original two numbers together
For example: 17 × 34
17 34
Halving the first column:
17 34
8
4
2
1
Doubling the second column:
17 34
8 68
4 136
2 272
1 544
Strike-out rows whose first cell is even:
17 34
8 68
4 136
2 272
1 544
Sum the remaining numbers in the right-hand column:
17 34
8 --
4 ---
2 ---
1 544
====
578
So 17 multiplied by 34, by the Ethiopian method is 578.
Task
The task is to define three named functions/methods/procedures/subroutines:
one to halve an integer,
one to double an integer, and
one to state if an integer is even.
Use these functions to create a function that does Ethiopian multiplication.
References
Ethiopian multiplication explained (BBC Video clip)
A Night Of Numbers - Go Forth And Multiply (Video)
Russian Peasant Multiplication
Programming Praxis: Russian Peasant Multiplication
| #Ada | Ada |
with ada.text_io;use ada.text_io;
procedure ethiopian is
function double (n : Natural) return Natural is (2*n);
function halve (n : Natural) return Natural is (n/2);
function is_even (n : Natural) return Boolean is (n mod 2 = 0);
function mul (l, r : Natural) return Natural is
(if l = 0 then 0 elsif l = 1 then r elsif is_even (l) then mul (halve (l),double (r))
else r + double (mul (halve (l), r)));
begin
put_line (mul (17,34)'img);
end ethiopian; |
http://rosettacode.org/wiki/Equilibrium_index | Equilibrium index | An equilibrium index of a sequence is an index into the sequence such that the sum of elements at lower indices is equal to the sum of elements at higher indices.
For example, in a sequence
A
{\displaystyle A}
:
A
0
=
−
7
{\displaystyle A_{0}=-7}
A
1
=
1
{\displaystyle A_{1}=1}
A
2
=
5
{\displaystyle A_{2}=5}
A
3
=
2
{\displaystyle A_{3}=2}
A
4
=
−
4
{\displaystyle A_{4}=-4}
A
5
=
3
{\displaystyle A_{5}=3}
A
6
=
0
{\displaystyle A_{6}=0}
3 is an equilibrium index, because:
A
0
+
A
1
+
A
2
=
A
4
+
A
5
+
A
6
{\displaystyle A_{0}+A_{1}+A_{2}=A_{4}+A_{5}+A_{6}}
6 is also an equilibrium index, because:
A
0
+
A
1
+
A
2
+
A
3
+
A
4
+
A
5
=
0
{\displaystyle A_{0}+A_{1}+A_{2}+A_{3}+A_{4}+A_{5}=0}
(sum of zero elements is zero)
7 is not an equilibrium index, because it is not a valid index of sequence
A
{\displaystyle A}
.
Task;
Write a function that, given a sequence, returns its equilibrium indices (if any).
Assume that the sequence may be very long.
| #Action.21 | Action! | PROC PrintArray(INT ARRAY a INT size)
INT i
Put('[)
FOR i=0 TO size-1
DO
IF i>0 THEN Put(' ) FI
PrintI(a(i))
OD
Put(']) PutE()
RETURN
INT FUNC SumRange(INT ARRAY a INT first,last)
INT sum
INT i
sum=0
FOR i=first TO last
DO
sum==+a(i)
OD
RETURN(sum)
PROC EquilibriumIndices(INT ARRAY a INT size
INT ARRAY indices INT POINTER indSize)
INT i,left,right
indSize^=0
FOR i=0 TO size-1
DO
left=SumRange(a,0,i-1)
right=SumRange(a,i+1,size-1)
IF left=right THEN
indices(indSize^)=i
indSize^==+1
FI
OD
RETURN
PROC Test(INT ARRAY a INT size)
INT ARRAY indices(100)
INT indSize
EquilibriumIndices(a,size,indices,@indSize)
Print("Array=") PrintArray(a,size)
Print("Equilibrium indices=") PrintArray(indices,indSize)
PutE()
RETURN
PROC Main()
INT ARRAY a=[65529 1 5 2 65532 3 0]
INT ARRAY b=[65535 1 65535 1 65535 1 65535]
INT ARRAY c=[1 2 3 4 5 6 7 8 9]
INT ARRAY d=[0]
Test(a,7)
Test(b,7)
Test(c,9)
Test(d,1)
RETURN |
http://rosettacode.org/wiki/Equilibrium_index | Equilibrium index | An equilibrium index of a sequence is an index into the sequence such that the sum of elements at lower indices is equal to the sum of elements at higher indices.
For example, in a sequence
A
{\displaystyle A}
:
A
0
=
−
7
{\displaystyle A_{0}=-7}
A
1
=
1
{\displaystyle A_{1}=1}
A
2
=
5
{\displaystyle A_{2}=5}
A
3
=
2
{\displaystyle A_{3}=2}
A
4
=
−
4
{\displaystyle A_{4}=-4}
A
5
=
3
{\displaystyle A_{5}=3}
A
6
=
0
{\displaystyle A_{6}=0}
3 is an equilibrium index, because:
A
0
+
A
1
+
A
2
=
A
4
+
A
5
+
A
6
{\displaystyle A_{0}+A_{1}+A_{2}=A_{4}+A_{5}+A_{6}}
6 is also an equilibrium index, because:
A
0
+
A
1
+
A
2
+
A
3
+
A
4
+
A
5
=
0
{\displaystyle A_{0}+A_{1}+A_{2}+A_{3}+A_{4}+A_{5}=0}
(sum of zero elements is zero)
7 is not an equilibrium index, because it is not a valid index of sequence
A
{\displaystyle A}
.
Task;
Write a function that, given a sequence, returns its equilibrium indices (if any).
Assume that the sequence may be very long.
| #Ada | Ada | with Ada.Containers.Vectors;
generic
type Index_Type is range <>;
type Element_Type is private;
Zero : Element_Type;
with function "+" (Left, Right : Element_Type) return Element_Type is <>;
with function "-" (Left, Right : Element_Type) return Element_Type is <>;
with function "=" (Left, Right : Element_Type) return Boolean is <>;
type Array_Type is private;
with function Element (From : Array_Type; Key : Index_Type) return Element_Type is <>;
package Equilibrium is
package Index_Vectors is new Ada.Containers.Vectors
(Index_Type => Positive, Element_Type => Index_Type);
function Get_Indices (From : Array_Type) return Index_Vectors.Vector;
end Equilibrium; |
http://rosettacode.org/wiki/Environment_variables | Environment variables | Task
Show how to get one of your process's environment variables.
The available variables vary by system; some of the common ones available on Unix include:
PATH
HOME
USER
| #AppleScript | AppleScript |
tell application "Finder" to get name of home
|
http://rosettacode.org/wiki/Environment_variables | Environment variables | Task
Show how to get one of your process's environment variables.
The available variables vary by system; some of the common ones available on Unix include:
PATH
HOME
USER
| #Arturo | Arturo | print ["path:" env\PATH]
print ["user:" env\USER]
print ["home:" env\HOME] |
http://rosettacode.org/wiki/Environment_variables | Environment variables | Task
Show how to get one of your process's environment variables.
The available variables vary by system; some of the common ones available on Unix include:
PATH
HOME
USER
| #AutoHotkey | AutoHotkey | EnvGet, OutputVar, Path
MsgBox, %OutputVar% |
http://rosettacode.org/wiki/Esthetic_numbers | Esthetic numbers | An esthetic number is a positive integer where every adjacent digit differs from its neighbour by 1.
E.G.
12 is an esthetic number. One and two differ by 1.
5654 is an esthetic number. Each digit is exactly 1 away from its neighbour.
890 is not an esthetic number. Nine and zero differ by 9.
These examples are nominally in base 10 but the concept extends easily to numbers in other bases. Traditionally, single digit numbers are included in esthetic numbers; zero may or may not be. For our purposes, for this task, do not include zero (0) as an esthetic number. Do not include numbers with leading zeros.
Esthetic numbers are also sometimes referred to as stepping numbers.
Task
Write a routine (function, procedure, whatever) to find esthetic numbers in a given base.
Use that routine to find esthetic numbers in bases 2 through 16 and display, here on this page, the esthectic numbers from index (base × 4) through index (base × 6), inclusive. (E.G. for base 2: 8th through 12th, for base 6: 24th through 36th, etc.)
Find and display, here on this page, the base 10 esthetic numbers with a magnitude between 1000 and 9999.
Stretch: Find and display, here on this page, the base 10 esthetic numbers with a magnitude between 1.0e8 and 1.3e8.
Related task
numbers with equal rises and falls
See also
OEIS A033075 - Positive numbers n such that all pairs of consecutive decimal digits differ by 1
Numbers Aplenty - Esthetic numbers
Geeks for Geeks - Stepping numbers
| #Java | Java | import java.util.ArrayList;
import java.util.stream.IntStream;
import java.util.stream.LongStream;
public class EstheticNumbers {
interface RecTriConsumer<A, B, C> {
void accept(RecTriConsumer<A, B, C> f, A a, B b, C c);
}
private static boolean isEsthetic(long n, long b) {
if (n == 0) {
return false;
}
var i = n % b;
var n2 = n / b;
while (n2 > 0) {
var j = n2 % b;
if (Math.abs(i - j) != 1) {
return false;
}
n2 /= b;
i = j;
}
return true;
}
private static void listEsths(long n, long n2, long m, long m2, int perLine, boolean all) {
var esths = new ArrayList<Long>();
var dfs = new RecTriConsumer<Long, Long, Long>() {
public void accept(Long n, Long m, Long i) {
accept(this, n, m, i);
}
@Override
public void accept(RecTriConsumer<Long, Long, Long> f, Long n, Long m, Long i) {
if (n <= i && i <= m) {
esths.add(i);
}
if (i == 0 || i > m) {
return;
}
var d = i % 10;
var i1 = i * 10 + d - 1;
var i2 = i1 + 2;
if (d == 0) {
f.accept(f, n, m, i2);
} else if (d == 9) {
f.accept(f, n, m, i1);
} else {
f.accept(f, n, m, i1);
f.accept(f, n, m, i2);
}
}
};
LongStream.range(0, 10).forEach(i -> dfs.accept(n2, m2, i));
var le = esths.size();
System.out.printf("Base 10: %d esthetic numbers between %d and %d:%n", le, n, m);
if (all) {
for (int i = 0; i < esths.size(); i++) {
System.out.printf("%d ", esths.get(i));
if ((i + 1) % perLine == 0) {
System.out.println();
}
}
} else {
for (int i = 0; i < perLine; i++) {
System.out.printf("%d ", esths.get(i));
}
System.out.println();
System.out.println("............");
for (int i = le - perLine; i < le; i++) {
System.out.printf("%d ", esths.get(i));
}
}
System.out.println();
System.out.println();
}
public static void main(String[] args) {
IntStream.rangeClosed(2, 16).forEach(b -> {
System.out.printf("Base %d: %dth to %dth esthetic numbers:%n", b, 4 * b, 6 * b);
var n = 1L;
var c = 0L;
while (c < 6 * b) {
if (isEsthetic(n, b)) {
c++;
if (c >= 4 * b) {
System.out.printf("%s ", Long.toString(n, b));
}
}
n++;
}
System.out.println();
});
System.out.println();
// the following all use the obvious range limitations for the numbers in question
listEsths(1000, 1010, 9999, 9898, 16, true);
listEsths((long) 1e8, 101_010_101, 13 * (long) 1e7, 123_456_789, 9, true);
listEsths((long) 1e11, 101_010_101_010L, 13 * (long) 1e10, 123_456_789_898L, 7, false);
listEsths((long) 1e14, 101_010_101_010_101L, 13 * (long) 1e13, 123_456_789_898_989L, 5, false);
listEsths((long) 1e17, 101_010_101_010_101_010L, 13 * (long) 1e16, 123_456_789_898_989_898L, 4, false);
}
} |
http://rosettacode.org/wiki/Euler%27s_sum_of_powers_conjecture | Euler's sum of powers conjecture | There is a conjecture in mathematics that held for over two hundred years before it was disproved by the finding of a counterexample in 1966 by Lander and Parkin.
Euler's (disproved) sum of powers conjecture
At least k positive kth powers are required to sum to a kth power,
except for the trivial case of one kth power: yk = yk
In 1966, Leon J. Lander and Thomas R. Parkin used a brute-force search on a CDC 6600 computer restricting numbers to those less than 250.
Task
Write a program to search for an integer solution for:
x05 + x15 + x25 + x35 == y5
Where all xi's and y are distinct integers between 0 and 250 (exclusive).
Show an answer here.
Related tasks
Pythagorean quadruples.
Pythagorean triples.
| #BCPL | BCPL |
GET "libhdr"
LET solve() BE {
LET pow5 = VEC 249
LET sum = ?
FOR i = 1 TO 249
pow5!i := i * i * i * i * i
FOR w = 4 TO 249
FOR x = 3 TO w - 1
FOR y = 2 TO x - 1
FOR z = 1 TO y - 1 {
sum := pow5!w + pow5!x + pow5!y + pow5!z
FOR a = w + 1 TO 249
IF pow5!a = sum {
writef("solution found: %d %d %d %d %d *n", w, x, y, z, a)
RETURN
}
}
writef("Sorry, no solution found.*n")
}
LET start() = VALOF {
solve()
RESULTIS 0
}
|
http://rosettacode.org/wiki/Factorial | Factorial | Definitions
The factorial of 0 (zero) is defined as being 1 (unity).
The Factorial Function of a positive integer, n, is defined as the product of the sequence:
n, n-1, n-2, ... 1
Task
Write a function to return the factorial of a number.
Solutions can be iterative or recursive.
Support for trapping negative n errors is optional.
Related task
Primorial numbers
| #min | min | ((dup 0 ==) 'succ (dup pred) '* linrec) :factorial |
http://rosettacode.org/wiki/Even_or_odd | Even or odd | Task
Test whether an integer is even or odd.
There is more than one way to solve this task:
Use the even and odd predicates, if the language provides them.
Check the least significant digit. With binary integers, i bitwise-and 1 equals 0 iff i is even, or equals 1 iff i is odd.
Divide i by 2. The remainder equals 0 iff i is even. The remainder equals +1 or -1 iff i is odd.
Use modular congruences:
i ≡ 0 (mod 2) iff i is even.
i ≡ 1 (mod 2) iff i is odd.
| #AntLang | AntLang | odd: {x mod 2}
even: {1 - x mod 2} |
http://rosettacode.org/wiki/Even_or_odd | Even or odd | Task
Test whether an integer is even or odd.
There is more than one way to solve this task:
Use the even and odd predicates, if the language provides them.
Check the least significant digit. With binary integers, i bitwise-and 1 equals 0 iff i is even, or equals 1 iff i is odd.
Divide i by 2. The remainder equals 0 iff i is even. The remainder equals +1 or -1 iff i is odd.
Use modular congruences:
i ≡ 0 (mod 2) iff i is even.
i ≡ 1 (mod 2) iff i is odd.
| #APL | APL | 2|28
0
2|37
1 |
http://rosettacode.org/wiki/Euler_method | Euler method | Euler's method numerically approximates solutions of first-order ordinary differential equations (ODEs) with a given initial value. It is an explicit method for solving initial value problems (IVPs), as described in the wikipedia page.
The ODE has to be provided in the following form:
d
y
(
t
)
d
t
=
f
(
t
,
y
(
t
)
)
{\displaystyle {\frac {dy(t)}{dt}}=f(t,y(t))}
with an initial value
y
(
t
0
)
=
y
0
{\displaystyle y(t_{0})=y_{0}}
To get a numeric solution, we replace the derivative on the LHS with a finite difference approximation:
d
y
(
t
)
d
t
≈
y
(
t
+
h
)
−
y
(
t
)
h
{\displaystyle {\frac {dy(t)}{dt}}\approx {\frac {y(t+h)-y(t)}{h}}}
then solve for
y
(
t
+
h
)
{\displaystyle y(t+h)}
:
y
(
t
+
h
)
≈
y
(
t
)
+
h
d
y
(
t
)
d
t
{\displaystyle y(t+h)\approx y(t)+h\,{\frac {dy(t)}{dt}}}
which is the same as
y
(
t
+
h
)
≈
y
(
t
)
+
h
f
(
t
,
y
(
t
)
)
{\displaystyle y(t+h)\approx y(t)+h\,f(t,y(t))}
The iterative solution rule is then:
y
n
+
1
=
y
n
+
h
f
(
t
n
,
y
n
)
{\displaystyle y_{n+1}=y_{n}+h\,f(t_{n},y_{n})}
where
h
{\displaystyle h}
is the step size, the most relevant parameter for accuracy of the solution. A smaller step size increases accuracy but also the computation cost, so it has always has to be hand-picked according to the problem at hand.
Example: Newton's Cooling Law
Newton's cooling law describes how an object of initial temperature
T
(
t
0
)
=
T
0
{\displaystyle T(t_{0})=T_{0}}
cools down in an environment of temperature
T
R
{\displaystyle T_{R}}
:
d
T
(
t
)
d
t
=
−
k
Δ
T
{\displaystyle {\frac {dT(t)}{dt}}=-k\,\Delta T}
or
d
T
(
t
)
d
t
=
−
k
(
T
(
t
)
−
T
R
)
{\displaystyle {\frac {dT(t)}{dt}}=-k\,(T(t)-T_{R})}
It says that the cooling rate
d
T
(
t
)
d
t
{\displaystyle {\frac {dT(t)}{dt}}}
of the object is proportional to the current temperature difference
Δ
T
=
(
T
(
t
)
−
T
R
)
{\displaystyle \Delta T=(T(t)-T_{R})}
to the surrounding environment.
The analytical solution, which we will compare to the numerical approximation, is
T
(
t
)
=
T
R
+
(
T
0
−
T
R
)
e
−
k
t
{\displaystyle T(t)=T_{R}+(T_{0}-T_{R})\;e^{-kt}}
Task
Implement a routine of Euler's method and then to use it to solve the given example of Newton's cooling law with it for three different step sizes of:
2 s
5 s and
10 s
and to compare with the analytical solution.
Initial values
initial temperature
T
0
{\displaystyle T_{0}}
shall be 100 °C
room temperature
T
R
{\displaystyle T_{R}}
shall be 20 °C
cooling constant
k
{\displaystyle k}
shall be 0.07
time interval to calculate shall be from 0 s ──► 100 s
A reference solution (Common Lisp) can be seen below. We see that bigger step sizes lead to reduced approximation accuracy.
| #JavaScript | JavaScript |
// Function that takes differential-equation, initial condition,
// ending x, and step size as parameters
function eulersMethod(f, x1, y1, x2, h) {
// Header
console.log("\tX\t|\tY\t");
console.log("------------------------------------");
// Initial Variables
var x=x1, y=y1;
// While we're not done yet
// Both sides of the OR let you do Euler's Method backwards
while ((x<x2 && x1<x2) || (x>x2 && x1>x2)) {
// Print what we have
console.log("\t" + x + "\t|\t" + y);
// Calculate the next values
y += h*f(x, y)
x += h;
}
return y;
}
function cooling(x, y) {
return -0.07 * (y-20);
}
eulersMethod(cooling, 0, 100, 100, 10);
|
http://rosettacode.org/wiki/Evaluate_binomial_coefficients | Evaluate binomial coefficients | This programming task, is to calculate ANY binomial coefficient.
However, it has to be able to output
(
5
3
)
{\displaystyle {\binom {5}{3}}}
, which is 10.
This formula is recommended:
(
n
k
)
=
n
!
(
n
−
k
)
!
k
!
=
n
(
n
−
1
)
(
n
−
2
)
…
(
n
−
k
+
1
)
k
(
k
−
1
)
(
k
−
2
)
…
1
{\displaystyle {\binom {n}{k}}={\frac {n!}{(n-k)!k!}}={\frac {n(n-1)(n-2)\ldots (n-k+1)}{k(k-1)(k-2)\ldots 1}}}
See Also:
Combinations and permutations
Pascal's triangle
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
| #C.2B.2B | C++ | double Factorial(double nValue)
{
double result = nValue;
double result_next;
double pc = nValue;
do
{
result_next = result*(pc-1);
result = result_next;
pc--;
}while(pc>2);
nValue = result;
return nValue;
}
double binomialCoefficient(double n, double k)
{
if (abs(n - k) < 1e-7 || k < 1e-7) return 1.0;
if( abs(k-1.0) < 1e-7 || abs(k - (n-1)) < 1e-7)return n;
return Factorial(n) /(Factorial(k)*Factorial((n - k)));
}
|
http://rosettacode.org/wiki/Fibonacci_sequence | Fibonacci sequence | The Fibonacci sequence is a sequence Fn of natural numbers defined recursively:
F0 = 0
F1 = 1
Fn = Fn-1 + Fn-2, if n>1
Task
Write a function to generate the nth Fibonacci number.
Solutions can be iterative or recursive (though recursive solutions are generally considered too slow and are mostly used as an exercise in recursion).
The sequence is sometimes extended into negative numbers by using a straightforward inverse of the positive definition:
Fn = Fn+2 - Fn+1, if n<0
support for negative n in the solution is optional.
Related tasks
Fibonacci n-step number sequences
Leonardo numbers
References
Wikipedia, Fibonacci number
Wikipedia, Lucas number
MathWorld, Fibonacci Number
Some identities for r-Fibonacci numbers
OEIS Fibonacci numbers
OEIS Lucas numbers
| #Visual_Basic | Visual Basic | Sub fibonacci()
Const n = 139
Dim i As Integer
Dim f1 As Variant, f2 As Variant, f3 As Variant 'for Decimal
f1 = CDec(0): f2 = CDec(1) 'for Decimal setting
Debug.Print "fibo("; 0; ")="; f1
Debug.Print "fibo("; 1; ")="; f2
For i = 2 To n
f3 = f1 + f2
Debug.Print "fibo("; i; ")="; f3
f1 = f2
f2 = f3
Next i
End Sub 'fibonacci |
http://rosettacode.org/wiki/Entropy/Narcissist | Entropy/Narcissist |
Task
Write a computer program that computes and shows its own entropy.
Related Tasks
Fibonacci_word
Entropy
| #AutoHotkey | AutoHotkey | FileRead, var, *C %A_ScriptFullPath%
MsgBox, % Entropy(var)
Entropy(n) {
a := [], len := StrLen(n), m := n
while StrLen(m) {
s := SubStr(m, 1, 1)
m := RegExReplace(m, s, "", c)
a[s] := c
}
for key, val in a {
m := Log(p := val / len)
e -= p * m / Log(2)
}
return, e
} |
http://rosettacode.org/wiki/Entropy/Narcissist | Entropy/Narcissist |
Task
Write a computer program that computes and shows its own entropy.
Related Tasks
Fibonacci_word
Entropy
| #AWK | AWK |
BEGIN{FS=""
RS="\x04"#EOF
getline<"entropy.awk"
for(i=1;i<=NF;i++)H[$i]++
for(i in H)E-=(h=H[i]/NF)*log(h)
print "bytes ",NF," entropy ",E/log(2)
exit} |
http://rosettacode.org/wiki/Enumerations | Enumerations | Task
Create an enumeration of constants with and without explicit values.
| #11l | 11l | T.enum TokenCategory
NAME
KEYWORD
CONSTANT
TEST_CATEGORY = 10 |
http://rosettacode.org/wiki/Ethiopian_multiplication | Ethiopian multiplication | Ethiopian multiplication is a method of multiplying integers using only addition, doubling, and halving.
Method:
Take two numbers to be multiplied and write them down at the top of two columns.
In the left-hand column repeatedly halve the last number, discarding any remainders, and write the result below the last in the same column, until you write a value of 1.
In the right-hand column repeatedly double the last number and write the result below. stop when you add a result in the same row as where the left hand column shows 1.
Examine the table produced and discard any row where the value in the left column is even.
Sum the values in the right-hand column that remain to produce the result of multiplying the original two numbers together
For example: 17 × 34
17 34
Halving the first column:
17 34
8
4
2
1
Doubling the second column:
17 34
8 68
4 136
2 272
1 544
Strike-out rows whose first cell is even:
17 34
8 68
4 136
2 272
1 544
Sum the remaining numbers in the right-hand column:
17 34
8 --
4 ---
2 ---
1 544
====
578
So 17 multiplied by 34, by the Ethiopian method is 578.
Task
The task is to define three named functions/methods/procedures/subroutines:
one to halve an integer,
one to double an integer, and
one to state if an integer is even.
Use these functions to create a function that does Ethiopian multiplication.
References
Ethiopian multiplication explained (BBC Video clip)
A Night Of Numbers - Go Forth And Multiply (Video)
Russian Peasant Multiplication
Programming Praxis: Russian Peasant Multiplication
| #Aime | Aime | void
halve(integer &x)
{
x >>= 1;
}
void
double(integer &x)
{
x <<= 1;
}
integer
iseven(integer x)
{
return (x & 1) == 0;
}
integer
ethiopian(integer plier, integer plicand, integer tutor)
{
integer result;
result = 0;
if (tutor) {
o_form("ethiopian multiplication of ~ by ~\n", plier, plicand);
}
while (plier >= 1) {
if (iseven(plier)) {
if (tutor) {
o_form("/w4/ /w6/ struck\n", plier, plicand);
}
} else {
if (tutor) {
o_form("/w4/ /w6/ kept\n", plier, plicand);
}
result += plicand;
}
halve(plier);
double(plicand);
}
return result;
}
integer
main(void)
{
o_integer(ethiopian(17, 34, 1));
o_byte('\n');
return 0;
} |
http://rosettacode.org/wiki/Equilibrium_index | Equilibrium index | An equilibrium index of a sequence is an index into the sequence such that the sum of elements at lower indices is equal to the sum of elements at higher indices.
For example, in a sequence
A
{\displaystyle A}
:
A
0
=
−
7
{\displaystyle A_{0}=-7}
A
1
=
1
{\displaystyle A_{1}=1}
A
2
=
5
{\displaystyle A_{2}=5}
A
3
=
2
{\displaystyle A_{3}=2}
A
4
=
−
4
{\displaystyle A_{4}=-4}
A
5
=
3
{\displaystyle A_{5}=3}
A
6
=
0
{\displaystyle A_{6}=0}
3 is an equilibrium index, because:
A
0
+
A
1
+
A
2
=
A
4
+
A
5
+
A
6
{\displaystyle A_{0}+A_{1}+A_{2}=A_{4}+A_{5}+A_{6}}
6 is also an equilibrium index, because:
A
0
+
A
1
+
A
2
+
A
3
+
A
4
+
A
5
=
0
{\displaystyle A_{0}+A_{1}+A_{2}+A_{3}+A_{4}+A_{5}=0}
(sum of zero elements is zero)
7 is not an equilibrium index, because it is not a valid index of sequence
A
{\displaystyle A}
.
Task;
Write a function that, given a sequence, returns its equilibrium indices (if any).
Assume that the sequence may be very long.
| #Aime | Aime | list
eqindex(list l)
{
integer e, i, s, sum;
list x;
s = sum = 0;
l.ucall(add_i, 1, sum);
for (i, e in l) {
if (s * 2 + e == sum) {
x.append(i);
}
s += e;
}
x;
}
integer
main(void)
{
list(-7, 1, 5, 2, -4, 3, 0).eqindex.ucall(o_, 0, "\n");
0;
} |
http://rosettacode.org/wiki/Equilibrium_index | Equilibrium index | An equilibrium index of a sequence is an index into the sequence such that the sum of elements at lower indices is equal to the sum of elements at higher indices.
For example, in a sequence
A
{\displaystyle A}
:
A
0
=
−
7
{\displaystyle A_{0}=-7}
A
1
=
1
{\displaystyle A_{1}=1}
A
2
=
5
{\displaystyle A_{2}=5}
A
3
=
2
{\displaystyle A_{3}=2}
A
4
=
−
4
{\displaystyle A_{4}=-4}
A
5
=
3
{\displaystyle A_{5}=3}
A
6
=
0
{\displaystyle A_{6}=0}
3 is an equilibrium index, because:
A
0
+
A
1
+
A
2
=
A
4
+
A
5
+
A
6
{\displaystyle A_{0}+A_{1}+A_{2}=A_{4}+A_{5}+A_{6}}
6 is also an equilibrium index, because:
A
0
+
A
1
+
A
2
+
A
3
+
A
4
+
A
5
=
0
{\displaystyle A_{0}+A_{1}+A_{2}+A_{3}+A_{4}+A_{5}=0}
(sum of zero elements is zero)
7 is not an equilibrium index, because it is not a valid index of sequence
A
{\displaystyle A}
.
Task;
Write a function that, given a sequence, returns its equilibrium indices (if any).
Assume that the sequence may be very long.
| #ALGOL_68 | ALGOL 68 | MODE YIELDINT = PROC(INT)VOID;
PROC gen equilibrium index = ([]INT arr, YIELDINT yield)VOID:
(
INT sum := 0;
FOR i FROM LWB arr TO UPB arr DO
sum +:= arr[i]
OD;
INT left:=0, right:=sum;
FOR i FROM LWB arr TO UPB arr DO
right -:= arr[i];
IF left = right THEN yield(i) FI;
left +:= arr[i]
OD
);
test:(
[]INT arr = []INT(-7, 1, 5, 2, -4, 3, 0)[@0];
# FOR INT index IN # gen equilibrium index(arr, # ) DO ( #
## (INT index)VOID:
print(index)
# OD # );
print(new line)
) |
http://rosettacode.org/wiki/Environment_variables | Environment variables | Task
Show how to get one of your process's environment variables.
The available variables vary by system; some of the common ones available on Unix include:
PATH
HOME
USER
| #AutoIt | AutoIt | ConsoleWrite("# Environment:" & @CRLF)
Local $sEnvVar = EnvGet("LANG")
ConsoleWrite("LANG : " & $sEnvVar & @CRLF)
ShowEnv("SystemDrive")
ShowEnv("USERNAME")
Func ShowEnv($N)
ConsoleWrite( StringFormat("%-12s : %s\n", $N, EnvGet($N)) )
EndFunc ;==>ShowEnv |
http://rosettacode.org/wiki/Environment_variables | Environment variables | Task
Show how to get one of your process's environment variables.
The available variables vary by system; some of the common ones available on Unix include:
PATH
HOME
USER
| #AWK | AWK | $ awk 'BEGIN{print "HOME:"ENVIRON["HOME"],"USER:"ENVIRON["USER"]}' |
http://rosettacode.org/wiki/Environment_variables | Environment variables | Task
Show how to get one of your process's environment variables.
The available variables vary by system; some of the common ones available on Unix include:
PATH
HOME
USER
| #BASIC | BASIC | x$ = ENVIRON$("path")
PRINT x$ |
http://rosettacode.org/wiki/Esthetic_numbers | Esthetic numbers | An esthetic number is a positive integer where every adjacent digit differs from its neighbour by 1.
E.G.
12 is an esthetic number. One and two differ by 1.
5654 is an esthetic number. Each digit is exactly 1 away from its neighbour.
890 is not an esthetic number. Nine and zero differ by 9.
These examples are nominally in base 10 but the concept extends easily to numbers in other bases. Traditionally, single digit numbers are included in esthetic numbers; zero may or may not be. For our purposes, for this task, do not include zero (0) as an esthetic number. Do not include numbers with leading zeros.
Esthetic numbers are also sometimes referred to as stepping numbers.
Task
Write a routine (function, procedure, whatever) to find esthetic numbers in a given base.
Use that routine to find esthetic numbers in bases 2 through 16 and display, here on this page, the esthectic numbers from index (base × 4) through index (base × 6), inclusive. (E.G. for base 2: 8th through 12th, for base 6: 24th through 36th, etc.)
Find and display, here on this page, the base 10 esthetic numbers with a magnitude between 1000 and 9999.
Stretch: Find and display, here on this page, the base 10 esthetic numbers with a magnitude between 1.0e8 and 1.3e8.
Related task
numbers with equal rises and falls
See also
OEIS A033075 - Positive numbers n such that all pairs of consecutive decimal digits differ by 1
Numbers Aplenty - Esthetic numbers
Geeks for Geeks - Stepping numbers
| #JavaScript | JavaScript | function isEsthetic(inp, base = 10) {
let arr = inp.toString(base).split('');
if (arr.length == 1) return false;
for (let i = 0; i < arr.length; i++)
arr[i] = parseInt(arr[i], base);
for (i = 0; i < arr.length-1; i++)
if (Math.abs(arr[i]-arr[i+1]) !== 1) return false;
return true;
}
function collectEsthetics(base, range) {
let out = [], x;
if (range) {
for (x = range[0]; x < range[1]; x++)
if (isEsthetic(x)) out.push(x);
return out;
} else {
x = 1;
while (out.length < base*6) {
s = x.toString(base);
if (isEsthetic(s, base)) out.push(s.toUpperCase());
x++;
}
return out.slice(base*4);
}
}
// main
let d = new Date();
for (let x = 2; x <= 36; x++) { // we put b17 .. b36 on top, because we can
console.log(`${x}:`);
console.log( collectEsthetics(x),
(new Date() - d) / 1000 + ' s');
}
console.log( collectEsthetics(10, [1000, 9999]),
(new Date() - d) / 1000 + ' s' );
console.log( collectEsthetics(10, [1e8, 1.3e8]),
(new Date() - d) / 1000 + ' s' ); |
http://rosettacode.org/wiki/Euler%27s_sum_of_powers_conjecture | Euler's sum of powers conjecture | There is a conjecture in mathematics that held for over two hundred years before it was disproved by the finding of a counterexample in 1966 by Lander and Parkin.
Euler's (disproved) sum of powers conjecture
At least k positive kth powers are required to sum to a kth power,
except for the trivial case of one kth power: yk = yk
In 1966, Leon J. Lander and Thomas R. Parkin used a brute-force search on a CDC 6600 computer restricting numbers to those less than 250.
Task
Write a program to search for an integer solution for:
x05 + x15 + x25 + x35 == y5
Where all xi's and y are distinct integers between 0 and 250 (exclusive).
Show an answer here.
Related tasks
Pythagorean quadruples.
Pythagorean triples.
| #Bracmat | Bracmat | 0:?x0
& whl
' ( 1+!x0:<250:?x0
& out$(x0 !x0)
& 0:?x1
& whl
' ( 1+!x1:~>!x0:?x1
& out$(x0 !x0 x1 !x1)
& 0:?x2
& whl
' ( 1+!x2:~>!x1:?x2
& 0:?x3
& whl
' ( 1+!x3:~>!x2:?x3
& (!x0^5+!x1^5+!x2^5+!x3^5)^1/5
: ( #?y
& out$(x0 !x0 x1 !x1 x2 !x2 x3 !x3 y !y)
& 250:?x0:?x1:?x2:?x3
| ?
)
)
)
)
) |
http://rosettacode.org/wiki/Factorial | Factorial | Definitions
The factorial of 0 (zero) is defined as being 1 (unity).
The Factorial Function of a positive integer, n, is defined as the product of the sequence:
n, n-1, n-2, ... 1
Task
Write a function to return the factorial of a number.
Solutions can be iterative or recursive.
Support for trapping negative n errors is optional.
Related task
Primorial numbers
| #MiniScript | MiniScript | factorial = function(n)
result = 1
for i in range(2,n)
result = result * i
end for
return result
end function
print factorial(10) |
http://rosettacode.org/wiki/Even_or_odd | Even or odd | Task
Test whether an integer is even or odd.
There is more than one way to solve this task:
Use the even and odd predicates, if the language provides them.
Check the least significant digit. With binary integers, i bitwise-and 1 equals 0 iff i is even, or equals 1 iff i is odd.
Divide i by 2. The remainder equals 0 iff i is even. The remainder equals +1 or -1 iff i is odd.
Use modular congruences:
i ≡ 0 (mod 2) iff i is even.
i ≡ 1 (mod 2) iff i is odd.
| #AppleScript | AppleScript | set L to {3, 2, 1, 0, -1, -2, -3}
set evens to {}
set odds to {}
repeat with x in L
if (x mod 2 = 0) then
set the end of evens to x's contents
else
set the end of odds to x's contents
end if
end repeat
return {even:evens, odd:odds} |
http://rosettacode.org/wiki/Euler_method | Euler method | Euler's method numerically approximates solutions of first-order ordinary differential equations (ODEs) with a given initial value. It is an explicit method for solving initial value problems (IVPs), as described in the wikipedia page.
The ODE has to be provided in the following form:
d
y
(
t
)
d
t
=
f
(
t
,
y
(
t
)
)
{\displaystyle {\frac {dy(t)}{dt}}=f(t,y(t))}
with an initial value
y
(
t
0
)
=
y
0
{\displaystyle y(t_{0})=y_{0}}
To get a numeric solution, we replace the derivative on the LHS with a finite difference approximation:
d
y
(
t
)
d
t
≈
y
(
t
+
h
)
−
y
(
t
)
h
{\displaystyle {\frac {dy(t)}{dt}}\approx {\frac {y(t+h)-y(t)}{h}}}
then solve for
y
(
t
+
h
)
{\displaystyle y(t+h)}
:
y
(
t
+
h
)
≈
y
(
t
)
+
h
d
y
(
t
)
d
t
{\displaystyle y(t+h)\approx y(t)+h\,{\frac {dy(t)}{dt}}}
which is the same as
y
(
t
+
h
)
≈
y
(
t
)
+
h
f
(
t
,
y
(
t
)
)
{\displaystyle y(t+h)\approx y(t)+h\,f(t,y(t))}
The iterative solution rule is then:
y
n
+
1
=
y
n
+
h
f
(
t
n
,
y
n
)
{\displaystyle y_{n+1}=y_{n}+h\,f(t_{n},y_{n})}
where
h
{\displaystyle h}
is the step size, the most relevant parameter for accuracy of the solution. A smaller step size increases accuracy but also the computation cost, so it has always has to be hand-picked according to the problem at hand.
Example: Newton's Cooling Law
Newton's cooling law describes how an object of initial temperature
T
(
t
0
)
=
T
0
{\displaystyle T(t_{0})=T_{0}}
cools down in an environment of temperature
T
R
{\displaystyle T_{R}}
:
d
T
(
t
)
d
t
=
−
k
Δ
T
{\displaystyle {\frac {dT(t)}{dt}}=-k\,\Delta T}
or
d
T
(
t
)
d
t
=
−
k
(
T
(
t
)
−
T
R
)
{\displaystyle {\frac {dT(t)}{dt}}=-k\,(T(t)-T_{R})}
It says that the cooling rate
d
T
(
t
)
d
t
{\displaystyle {\frac {dT(t)}{dt}}}
of the object is proportional to the current temperature difference
Δ
T
=
(
T
(
t
)
−
T
R
)
{\displaystyle \Delta T=(T(t)-T_{R})}
to the surrounding environment.
The analytical solution, which we will compare to the numerical approximation, is
T
(
t
)
=
T
R
+
(
T
0
−
T
R
)
e
−
k
t
{\displaystyle T(t)=T_{R}+(T_{0}-T_{R})\;e^{-kt}}
Task
Implement a routine of Euler's method and then to use it to solve the given example of Newton's cooling law with it for three different step sizes of:
2 s
5 s and
10 s
and to compare with the analytical solution.
Initial values
initial temperature
T
0
{\displaystyle T_{0}}
shall be 100 °C
room temperature
T
R
{\displaystyle T_{R}}
shall be 20 °C
cooling constant
k
{\displaystyle k}
shall be 0.07
time interval to calculate shall be from 0 s ──► 100 s
A reference solution (Common Lisp) can be seen below. We see that bigger step sizes lead to reduced approximation accuracy.
| #jq | jq | # euler_method takes a filter (df), initial condition
# (x1,y1), ending x (x2), and step size as parameters;
# it emits the y values at each iteration.
# df must take [x,y] as its input.
def euler_method(df; x1; y1; x2; h):
h as $h
| [x1, y1]
| recurse( if ((.[0] < x2 and x1 < x2) or
(.[0] > x2 and x1 > x2)) then
[ (.[0] + $h), (.[1] + $h*df) ]
else empty
end )
| .[1] ;
# We could now solve the task by writing for each step-size, $h
# euler_method(-0.07 * (.[1]-20); 0; 100; 100; $h)
# but for clarity, we shall define a function named "cooling":
# [x,y] is input
def cooling: -0.07 * (.[1]-20);
# The following solves the task:
# (2,5,10) | [., [ euler_method(cooling; 0; 100; 100; .) ] ]
|
http://rosettacode.org/wiki/Euler_method | Euler method | Euler's method numerically approximates solutions of first-order ordinary differential equations (ODEs) with a given initial value. It is an explicit method for solving initial value problems (IVPs), as described in the wikipedia page.
The ODE has to be provided in the following form:
d
y
(
t
)
d
t
=
f
(
t
,
y
(
t
)
)
{\displaystyle {\frac {dy(t)}{dt}}=f(t,y(t))}
with an initial value
y
(
t
0
)
=
y
0
{\displaystyle y(t_{0})=y_{0}}
To get a numeric solution, we replace the derivative on the LHS with a finite difference approximation:
d
y
(
t
)
d
t
≈
y
(
t
+
h
)
−
y
(
t
)
h
{\displaystyle {\frac {dy(t)}{dt}}\approx {\frac {y(t+h)-y(t)}{h}}}
then solve for
y
(
t
+
h
)
{\displaystyle y(t+h)}
:
y
(
t
+
h
)
≈
y
(
t
)
+
h
d
y
(
t
)
d
t
{\displaystyle y(t+h)\approx y(t)+h\,{\frac {dy(t)}{dt}}}
which is the same as
y
(
t
+
h
)
≈
y
(
t
)
+
h
f
(
t
,
y
(
t
)
)
{\displaystyle y(t+h)\approx y(t)+h\,f(t,y(t))}
The iterative solution rule is then:
y
n
+
1
=
y
n
+
h
f
(
t
n
,
y
n
)
{\displaystyle y_{n+1}=y_{n}+h\,f(t_{n},y_{n})}
where
h
{\displaystyle h}
is the step size, the most relevant parameter for accuracy of the solution. A smaller step size increases accuracy but also the computation cost, so it has always has to be hand-picked according to the problem at hand.
Example: Newton's Cooling Law
Newton's cooling law describes how an object of initial temperature
T
(
t
0
)
=
T
0
{\displaystyle T(t_{0})=T_{0}}
cools down in an environment of temperature
T
R
{\displaystyle T_{R}}
:
d
T
(
t
)
d
t
=
−
k
Δ
T
{\displaystyle {\frac {dT(t)}{dt}}=-k\,\Delta T}
or
d
T
(
t
)
d
t
=
−
k
(
T
(
t
)
−
T
R
)
{\displaystyle {\frac {dT(t)}{dt}}=-k\,(T(t)-T_{R})}
It says that the cooling rate
d
T
(
t
)
d
t
{\displaystyle {\frac {dT(t)}{dt}}}
of the object is proportional to the current temperature difference
Δ
T
=
(
T
(
t
)
−
T
R
)
{\displaystyle \Delta T=(T(t)-T_{R})}
to the surrounding environment.
The analytical solution, which we will compare to the numerical approximation, is
T
(
t
)
=
T
R
+
(
T
0
−
T
R
)
e
−
k
t
{\displaystyle T(t)=T_{R}+(T_{0}-T_{R})\;e^{-kt}}
Task
Implement a routine of Euler's method and then to use it to solve the given example of Newton's cooling law with it for three different step sizes of:
2 s
5 s and
10 s
and to compare with the analytical solution.
Initial values
initial temperature
T
0
{\displaystyle T_{0}}
shall be 100 °C
room temperature
T
R
{\displaystyle T_{R}}
shall be 20 °C
cooling constant
k
{\displaystyle k}
shall be 0.07
time interval to calculate shall be from 0 s ──► 100 s
A reference solution (Common Lisp) can be seen below. We see that bigger step sizes lead to reduced approximation accuracy.
| #Julia | Julia | euler(f::Function, T::Number, t0::Int, t1::Int, h::Int) = collect(begin T += h * f(T); T end for t in t0:h:t1)
# Prints a series of arbitrary values in a tabular form, left aligned in cells with a given width
tabular(width, cells...) = println(join(map(s -> rpad(s, width), cells)))
# prints the table according to the task description for h=5 and 10 sec
for h in (5, 10)
print("Step $h:\n\n")
tabular(15, "Time", "Euler", "Analytic")
t = 0
for T in euler(y -> -0.07 * (y - 20.0), 100.0, 0, 100, h)
tabular(15, t, round(T,digits=6), round(20.0 + 80.0 * exp(-0.07t), digits=6))
t += h
end
println()
end |
http://rosettacode.org/wiki/Evaluate_binomial_coefficients | Evaluate binomial coefficients | This programming task, is to calculate ANY binomial coefficient.
However, it has to be able to output
(
5
3
)
{\displaystyle {\binom {5}{3}}}
, which is 10.
This formula is recommended:
(
n
k
)
=
n
!
(
n
−
k
)
!
k
!
=
n
(
n
−
1
)
(
n
−
2
)
…
(
n
−
k
+
1
)
k
(
k
−
1
)
(
k
−
2
)
…
1
{\displaystyle {\binom {n}{k}}={\frac {n!}{(n-k)!k!}}={\frac {n(n-1)(n-2)\ldots (n-k+1)}{k(k-1)(k-2)\ldots 1}}}
See Also:
Combinations and permutations
Pascal's triangle
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
| #Clojure | Clojure | (defn binomial-coefficient [n k]
(let [rprod (fn [a b] (reduce * (range a (inc b))))]
(/ (rprod (- n k -1) n) (rprod 1 k)))) |
http://rosettacode.org/wiki/Evaluate_binomial_coefficients | Evaluate binomial coefficients | This programming task, is to calculate ANY binomial coefficient.
However, it has to be able to output
(
5
3
)
{\displaystyle {\binom {5}{3}}}
, which is 10.
This formula is recommended:
(
n
k
)
=
n
!
(
n
−
k
)
!
k
!
=
n
(
n
−
1
)
(
n
−
2
)
…
(
n
−
k
+
1
)
k
(
k
−
1
)
(
k
−
2
)
…
1
{\displaystyle {\binom {n}{k}}={\frac {n!}{(n-k)!k!}}={\frac {n(n-1)(n-2)\ldots (n-k+1)}{k(k-1)(k-2)\ldots 1}}}
See Also:
Combinations and permutations
Pascal's triangle
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
| #CoffeeScript | CoffeeScript |
binomial_coefficient = (n, k) ->
result = 1
for i in [0...k]
result *= (n - i) / (i + 1)
result
n = 5
for k in [0..n]
console.log "binomial_coefficient(#{n}, #{k}) = #{binomial_coefficient(n,k)}"
|
http://rosettacode.org/wiki/Fibonacci_sequence | Fibonacci sequence | The Fibonacci sequence is a sequence Fn of natural numbers defined recursively:
F0 = 0
F1 = 1
Fn = Fn-1 + Fn-2, if n>1
Task
Write a function to generate the nth Fibonacci number.
Solutions can be iterative or recursive (though recursive solutions are generally considered too slow and are mostly used as an exercise in recursion).
The sequence is sometimes extended into negative numbers by using a straightforward inverse of the positive definition:
Fn = Fn+2 - Fn+1, if n<0
support for negative n in the solution is optional.
Related tasks
Fibonacci n-step number sequences
Leonardo numbers
References
Wikipedia, Fibonacci number
Wikipedia, Lucas number
MathWorld, Fibonacci Number
Some identities for r-Fibonacci numbers
OEIS Fibonacci numbers
OEIS Lucas numbers
| #Visual_Basic_.NET | Visual Basic .NET | Function Fib(ByVal n As Integer) As Decimal
Dim fib0, fib1, sum As Decimal
Dim i As Integer
fib0 = 0
fib1 = 1
For i = 1 To n
sum = fib0 + fib1
fib0 = fib1
fib1 = sum
Next
Fib = fib0
End Function |
http://rosettacode.org/wiki/Entropy/Narcissist | Entropy/Narcissist |
Task
Write a computer program that computes and shows its own entropy.
Related Tasks
Fibonacci_word
Entropy
| #C | C | #include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <string.h>
#include <math.h>
#define MAXLEN 961 //maximum string length
int makehist(char *S,int *hist,int len){
int wherechar[256];
int i,histlen;
histlen=0;
for(i=0;i<256;i++)wherechar[i]=-1;
for(i=0;i<len;i++){
if(wherechar[(int)S[i]]==-1){
wherechar[(int)S[i]]=histlen;
histlen++;
}
hist[wherechar[(int)S[i]]]++;
}
return histlen;
}
double entropy(int *hist,int histlen,int len){
int i;
double H;
H=0;
for(i=0;i<histlen;i++){
H-=(double)hist[i]/len*log2((double)hist[i]/len);
}
return H;
}
int main(void){
char S[MAXLEN];
int len,*hist,histlen;
double H;
FILE *f;
f=fopen("entropy.c","r");
for(len=0;!feof(f);len++)S[len]=fgetc(f);
S[--len]='\0';
hist=(int*)calloc(len,sizeof(int));
histlen=makehist(S,hist,len);
//hist now has no order (known to the program) but that doesn't matter
H=entropy(hist,histlen,len);
printf("%lf\n",H);
return 0;
} |
http://rosettacode.org/wiki/Enumerations | Enumerations | Task
Create an enumeration of constants with and without explicit values.
| #6502_Assembly | 6502 Assembly | Sunday equ 0
Monday equ 1
Tuesday equ 2
Wednesday equ 3
Thursday equ 4
Friday equ 5
Saturday equ 6 |
http://rosettacode.org/wiki/Enumerations | Enumerations | Task
Create an enumeration of constants with and without explicit values.
| #68000_Assembly | 68000 Assembly | Sunday equ 0
Monday equ 1
Tuesday equ 2
Wednesday equ 3
Thursday equ 4
Friday equ 5
Saturday equ 6 |
http://rosettacode.org/wiki/Enumerations | Enumerations | Task
Create an enumeration of constants with and without explicit values.
| #8086_Assembly | 8086 Assembly | Sunday equ 0
Monday equ 1
Tuesday equ 2
Wednesday equ 3
Thursday equ 4
Friday equ 5
Saturday equ 6
Sunday equ 7 |
http://rosettacode.org/wiki/Ethiopian_multiplication | Ethiopian multiplication | Ethiopian multiplication is a method of multiplying integers using only addition, doubling, and halving.
Method:
Take two numbers to be multiplied and write them down at the top of two columns.
In the left-hand column repeatedly halve the last number, discarding any remainders, and write the result below the last in the same column, until you write a value of 1.
In the right-hand column repeatedly double the last number and write the result below. stop when you add a result in the same row as where the left hand column shows 1.
Examine the table produced and discard any row where the value in the left column is even.
Sum the values in the right-hand column that remain to produce the result of multiplying the original two numbers together
For example: 17 × 34
17 34
Halving the first column:
17 34
8
4
2
1
Doubling the second column:
17 34
8 68
4 136
2 272
1 544
Strike-out rows whose first cell is even:
17 34
8 68
4 136
2 272
1 544
Sum the remaining numbers in the right-hand column:
17 34
8 --
4 ---
2 ---
1 544
====
578
So 17 multiplied by 34, by the Ethiopian method is 578.
Task
The task is to define three named functions/methods/procedures/subroutines:
one to halve an integer,
one to double an integer, and
one to state if an integer is even.
Use these functions to create a function that does Ethiopian multiplication.
References
Ethiopian multiplication explained (BBC Video clip)
A Night Of Numbers - Go Forth And Multiply (Video)
Russian Peasant Multiplication
Programming Praxis: Russian Peasant Multiplication
| #ALGOL_68 | ALGOL 68 | PROC halve = (REF INT x)VOID: x := ABS(BIN x SHR 1);
PROC doublit = (REF INT x)VOID: x := ABS(BIN x SHL 1);
PROC iseven = (#CONST# INT x)BOOL: NOT ODD x;
PROC ethiopian = (INT in plier,
INT in plicand, #CONST# BOOL tutor)INT:
(
INT plier := in plier, plicand := in plicand;
INT result:=0;
IF tutor THEN
printf(($"ethiopian multiplication of "g(0)," by "g(0)l$, plier, plicand)) FI;
WHILE plier >= 1 DO
IF iseven(plier) THEN
IF tutor THEN printf(($" "4d," "6d" struck"l$, plier, plicand)) FI
ELSE
IF tutor THEN printf(($" "4d," "6d" kept"l$, plier, plicand)) FI;
result +:= plicand
FI;
halve(plier); doublit(plicand)
OD;
result
);
main:
(
printf(($g(0)l$, ethiopian(17, 34, TRUE)))
) |
http://rosettacode.org/wiki/Equilibrium_index | Equilibrium index | An equilibrium index of a sequence is an index into the sequence such that the sum of elements at lower indices is equal to the sum of elements at higher indices.
For example, in a sequence
A
{\displaystyle A}
:
A
0
=
−
7
{\displaystyle A_{0}=-7}
A
1
=
1
{\displaystyle A_{1}=1}
A
2
=
5
{\displaystyle A_{2}=5}
A
3
=
2
{\displaystyle A_{3}=2}
A
4
=
−
4
{\displaystyle A_{4}=-4}
A
5
=
3
{\displaystyle A_{5}=3}
A
6
=
0
{\displaystyle A_{6}=0}
3 is an equilibrium index, because:
A
0
+
A
1
+
A
2
=
A
4
+
A
5
+
A
6
{\displaystyle A_{0}+A_{1}+A_{2}=A_{4}+A_{5}+A_{6}}
6 is also an equilibrium index, because:
A
0
+
A
1
+
A
2
+
A
3
+
A
4
+
A
5
=
0
{\displaystyle A_{0}+A_{1}+A_{2}+A_{3}+A_{4}+A_{5}=0}
(sum of zero elements is zero)
7 is not an equilibrium index, because it is not a valid index of sequence
A
{\displaystyle A}
.
Task;
Write a function that, given a sequence, returns its equilibrium indices (if any).
Assume that the sequence may be very long.
| #AppleScript | AppleScript | -- equilibriumIndices :: [Int] -> [Int]
on equilibriumIndices(xs)
script balancedPair
on |λ|(a, pair, i)
set {x, y} to pair
if x = y then
{i - 1} & a
else
a
end if
end |λ|
end script
script plus
on |λ|(a, b)
a + b
end |λ|
end script
-- Fold over zipped pairs of sums from left
-- and sums from right
foldr(balancedPair, {}, ¬
zip(scanl1(plus, xs), scanr1(plus, xs)))
end equilibriumIndices
-- TEST -----------------------------------------------------------------------
on run
map(equilibriumIndices, {¬
{-7, 1, 5, 2, -4, 3, 0}, ¬
{2, 4, 6}, ¬
{2, 9, 2}, ¬
{1, -1, 1, -1, 1, -1, 1}, ¬
{1}, ¬
{}})
--> {{3, 6}, {}, {1}, {0, 1, 2, 3, 4, 5, 6}, {0}, {}}
end run
-- GENERIC FUNCTIONS ----------------------------------------------------------
-- foldl :: (a -> b -> a) -> a -> [b] -> a
on foldl(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from 1 to lng
set v to |λ|(v, item i of xs, i, xs)
end repeat
return v
end tell
end foldl
-- foldr :: (a -> b -> a) -> a -> [b] -> a
on foldr(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from lng to 1 by -1
set v to |λ|(v, item i of xs, i, xs)
end repeat
return v
end tell
end foldr
-- init :: [a] -> [a]
on init(xs)
set lng to length of xs
if lng > 1 then
items 1 thru -2 of xs
else if lng > 0 then
{}
else
missing value
end if
end init
-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map
-- min :: Ord a => a -> a -> a
on min(x, y)
if y < x then
y
else
x
end if
end min
-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: Handler -> Script
on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn
-- scanl :: (b -> a -> b) -> b -> [a] -> [b]
on scanl(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
set lst to {startValue}
repeat with i from 1 to lng
set v to |λ|(v, item i of xs, i, xs)
set end of lst to v
end repeat
return lst
end tell
end scanl
-- scanl1 :: (a -> a -> a) -> [a] -> [a]
on scanl1(f, xs)
if length of xs > 0 then
scanl(f, item 1 of xs, tail(xs))
else
{}
end if
end scanl1
-- scanr :: (b -> a -> b) -> b -> [a] -> [b]
on scanr(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
set lst to {startValue}
repeat with i from lng to 1 by -1
set v to |λ|(v, item i of xs, i, xs)
set end of lst to v
end repeat
return reverse of lst
end tell
end scanr
-- scanr1 :: (a -> a -> a) -> [a] -> [a]
on scanr1(f, xs)
if length of xs > 0 then
scanr(f, item -1 of xs, init(xs))
else
{}
end if
end scanr1
-- tail :: [a] -> [a]
on tail(xs)
if length of xs > 1 then
items 2 thru -1 of xs
else
{}
end if
end tail
-- zip :: [a] -> [b] -> [(a, b)]
on zip(xs, ys)
set lng to min(length of xs, length of ys)
set lst to {}
repeat with i from 1 to lng
set end of lst to {item i of xs, item i of ys}
end repeat
return lst
end zip |
http://rosettacode.org/wiki/Environment_variables | Environment variables | Task
Show how to get one of your process's environment variables.
The available variables vary by system; some of the common ones available on Unix include:
PATH
HOME
USER
| #Batch_File | Batch File | echo %Foo% |
http://rosettacode.org/wiki/Environment_variables | Environment variables | Task
Show how to get one of your process's environment variables.
The available variables vary by system; some of the common ones available on Unix include:
PATH
HOME
USER
| #BBC_BASIC | BBC BASIC | PRINT FNenvironment("PATH")
PRINT FNenvironment("USERNAME")
END
DEF FNenvironment(envar$)
LOCAL buffer%, size%
SYS "GetEnvironmentVariable", envar$, 0, 0 TO size%
DIM buffer% LOCAL size%
SYS "GetEnvironmentVariable", envar$, buffer%, size%+1
= $$buffer% |
http://rosettacode.org/wiki/Environment_variables | Environment variables | Task
Show how to get one of your process's environment variables.
The available variables vary by system; some of the common ones available on Unix include:
PATH
HOME
USER
| #C | C | #include <stdlib.h>
#include <stdio.h>
int main() {
puts(getenv("HOME"));
puts(getenv("PATH"));
puts(getenv("USER"));
return 0;
}
|
http://rosettacode.org/wiki/Esthetic_numbers | Esthetic numbers | An esthetic number is a positive integer where every adjacent digit differs from its neighbour by 1.
E.G.
12 is an esthetic number. One and two differ by 1.
5654 is an esthetic number. Each digit is exactly 1 away from its neighbour.
890 is not an esthetic number. Nine and zero differ by 9.
These examples are nominally in base 10 but the concept extends easily to numbers in other bases. Traditionally, single digit numbers are included in esthetic numbers; zero may or may not be. For our purposes, for this task, do not include zero (0) as an esthetic number. Do not include numbers with leading zeros.
Esthetic numbers are also sometimes referred to as stepping numbers.
Task
Write a routine (function, procedure, whatever) to find esthetic numbers in a given base.
Use that routine to find esthetic numbers in bases 2 through 16 and display, here on this page, the esthectic numbers from index (base × 4) through index (base × 6), inclusive. (E.G. for base 2: 8th through 12th, for base 6: 24th through 36th, etc.)
Find and display, here on this page, the base 10 esthetic numbers with a magnitude between 1000 and 9999.
Stretch: Find and display, here on this page, the base 10 esthetic numbers with a magnitude between 1.0e8 and 1.3e8.
Related task
numbers with equal rises and falls
See also
OEIS A033075 - Positive numbers n such that all pairs of consecutive decimal digits differ by 1
Numbers Aplenty - Esthetic numbers
Geeks for Geeks - Stepping numbers
| #Julia | Julia | using Formatting
import Base.iterate, Base.IteratorSize, Base.IteratorEltype
"""
struct Esthetic
Used for iteration of esthetic numbers
"""
struct Esthetic{T}
lowerlimit::T where T <: Integer
base::T
upperlimit::T
Esthetic{T}(n, bas, m=typemax(T)) where T = new{T}(nextesthetic(n, bas), bas, m)
end
Base.IteratorSize(n::Esthetic) = Base.IsInfinite()
Base.IteratorEltype(n::Esthetic) = Integer
function Base.iterate(es::Esthetic, state=typeof(es.lowerlimit)[])
state = isempty(state) ? digits(es.lowerlimit, base=es.base) : increment!(state, es.base, 1)
n = toInt(state, es.base)
return n <= es.upperlimit ? (n, state) : nothing
end
isesthetic(n, b) = (d = digits(n, base=b); all(i -> abs(d[i] - d[i + 1]) == 1, 1:length(d)-1))
toInt(dig, bas) = foldr((i, j) -> i + bas * j, dig)
nextesthetic(n, b) = for i in n:typemax(typeof(n)) if isesthetic(i, b) return i end; end
""" Fill the digits below pos in vector with the least esthetic number fitting there """
function filldecreasing!(vec, pos)
if pos > 1
n = vec[pos]
for i in pos-1:-1:1
n = (n == 0) ? 1 : n - 1
vec[i] = n
end
end
return vec
end
""" Get the next esthetic number's digits from the previous number's digits """
function increment!(vec, bas, startpos = 1)
len = length(vec)
if len == 1
if vec[1] < bas - 1
vec[1] += 1
else
vec[1] = 0
push!(vec, 1)
end
else
pos = findfirst(i -> vec[i] < vec[i + 1], startpos:len-1)
if pos == nothing
if vec[end] >= bas - 1
push!(vec, 1)
filldecreasing!(vec, len + 1)
else
vec[end] += 1
filldecreasing!(vec, len)
end
else
for i in pos:len
if i == len
if vec[i] < bas - 1
vec[i] += 1
filldecreasing!(vec, i)
else
push!(vec, 1)
filldecreasing!(vec, len + 1)
end
elseif vec[i] < vec[i + 1] && vec[i] < bas - 2
vec[i] += 2
filldecreasing!(vec, i)
break
end
end
end
end
return vec
end
for b in 2:16
println("For base $b, the esthetic numbers indexed from $(4b) to $(6b) are:")
printed = 0
for (i, n) in enumerate(Iterators.take(Esthetic{Int}(1, b), 6b))
if i >= 4b
printed += 1
print(string(n, base=b), printed % 21 == 20 ? "\n" : " ")
end
end
println("\n")
end
for (bottom, top, cols, T) in [[1000, 9999, 16, Int], [100_000_000, 130_000_000, 8, Int],
[101_010_000_000, 130_000_000_000, 6, Int], [101_010_101_010_000_000, 130_000_000_000_000_000, 4, Int],
[101_010_101_010_101_000_000, 130_000_000_000_000_000_000, 4, Int128]]
esth, count = Esthetic{T}(bottom, 10, top), 0
println("\nBase 10 esthetic numbers between $(format(bottom, commas=true)) and $(format(top, commas=true)):")
for n in esth
count += 1
if count == 64
println(" ...")
elseif count < 64
print(format(n, commas=true), count % cols == 0 ? "\n" : " ")
end
end
println("\nTotal esthetic numbers in interval: $count")
end
|
http://rosettacode.org/wiki/Euler%27s_sum_of_powers_conjecture | Euler's sum of powers conjecture | There is a conjecture in mathematics that held for over two hundred years before it was disproved by the finding of a counterexample in 1966 by Lander and Parkin.
Euler's (disproved) sum of powers conjecture
At least k positive kth powers are required to sum to a kth power,
except for the trivial case of one kth power: yk = yk
In 1966, Leon J. Lander and Thomas R. Parkin used a brute-force search on a CDC 6600 computer restricting numbers to those less than 250.
Task
Write a program to search for an integer solution for:
x05 + x15 + x25 + x35 == y5
Where all xi's and y are distinct integers between 0 and 250 (exclusive).
Show an answer here.
Related tasks
Pythagorean quadruples.
Pythagorean triples.
| #C | C | // Alexander Maximov, July 2nd, 2015
#include <stdio.h>
#include <time.h>
typedef long long mylong;
void compute(int N, char find_only_one_solution)
{ const int M = 30; /* x^5 == x modulo M=2*3*5 */
int a, b, c, d, e;
mylong s, t, max, *p5 = (mylong*)malloc(sizeof(mylong)*(N+M));
for(s=0; s < N; ++s)
p5[s] = s * s, p5[s] *= p5[s] * s;
for(max = p5[N - 1]; s < (N + M); p5[s++] = max + 1);
for(a = 1; a < N; ++a)
for(b = a + 1; b < N; ++b)
for(c = b + 1; c < N; ++c)
for(d = c + 1, e = d + ((t = p5[a] + p5[b] + p5[c]) % M); ((s = t + p5[d]) <= max); ++d, ++e)
{ for(e -= M; p5[e + M] <= s; e += M); /* jump over M=30 values for e>d */
if(p5[e] == s)
{ printf("%d %d %d %d %d\r\n", a, b, c, d, e);
if(find_only_one_solution) goto onexit;
}
}
onexit:
free(p5);
}
int main(void)
{
int tm = clock();
compute(250, 0);
printf("time=%d milliseconds\r\n", (int)((clock() - tm) * 1000 / CLOCKS_PER_SEC));
return 0;
} |
http://rosettacode.org/wiki/Factorial | Factorial | Definitions
The factorial of 0 (zero) is defined as being 1 (unity).
The Factorial Function of a positive integer, n, is defined as the product of the sequence:
n, n-1, n-2, ... 1
Task
Write a function to return the factorial of a number.
Solutions can be iterative or recursive.
Support for trapping negative n errors is optional.
Related task
Primorial numbers
| #MiniZinc | MiniZinc | var int: factorial(int: n) =
let {
array[0..n] of var int: factorial;
constraint forall(a in 0..n)(
factorial[a] == if (a == 0) then
1
else
a*factorial[a-1]
endif
)} in factorial[n];
var int: fac = factorial(6);
solve satisfy;
output [show(fac),"\n"]; |
http://rosettacode.org/wiki/Even_or_odd | Even or odd | Task
Test whether an integer is even or odd.
There is more than one way to solve this task:
Use the even and odd predicates, if the language provides them.
Check the least significant digit. With binary integers, i bitwise-and 1 equals 0 iff i is even, or equals 1 iff i is odd.
Divide i by 2. The remainder equals 0 iff i is even. The remainder equals +1 or -1 iff i is odd.
Use modular congruences:
i ≡ 0 (mod 2) iff i is even.
i ≡ 1 (mod 2) iff i is odd.
| #Arendelle | Arendelle | ( input , "Please enter a number: " )
{ @input % 2 = 0 ,
"| @input | is even!"
,
"| @input | is odd!"
} |
http://rosettacode.org/wiki/Even_or_odd | Even or odd | Task
Test whether an integer is even or odd.
There is more than one way to solve this task:
Use the even and odd predicates, if the language provides them.
Check the least significant digit. With binary integers, i bitwise-and 1 equals 0 iff i is even, or equals 1 iff i is odd.
Divide i by 2. The remainder equals 0 iff i is even. The remainder equals +1 or -1 iff i is odd.
Use modular congruences:
i ≡ 0 (mod 2) iff i is even.
i ≡ 1 (mod 2) iff i is odd.
| #ARM_Assembly | ARM Assembly |
/* ARM assembly Raspberry PI or android 32 bits */
/* program oddEven.s */
/* REMARK 1 : this program use routines in a include file
see task Include a file language arm assembly
for the routine affichageMess conversion10
see at end of this program the instruction include */
/* for constantes see task include a file in arm assembly */
/************************************/
/* Constantes */
/************************************/
.include "../constantes.inc"
/*********************************/
/* Initialized data */
/*********************************/
.data
sMessResultOdd: .asciz " @ is odd (impair) \n"
sMessResultEven: .asciz " @ is even (pair) \n"
szCarriageReturn: .asciz "\n"
/*********************************/
/* UnInitialized data */
/*********************************/
.bss
sZoneConv: .skip 24
/*********************************/
/* code section */
/*********************************/
.text
.global main
main: @ entry of program
mov r0,#5
bl testOddEven
mov r0,#12
bl testOddEven
mov r0,#2021
bl testOddEven
100: @ standard end of the program
mov r0, #0 @ return code
mov r7, #EXIT @ request to exit program
svc #0 @ perform the system call
iAdrszCarriageReturn: .int szCarriageReturn
iAdrsMessResultOdd: .int sMessResultOdd
iAdrsMessResultEven: .int sMessResultEven
iAdrsZoneConv: .int sZoneConv
/***************************************************/
/* test if number is odd or even */
/***************************************************/
// r0 contains à number
testOddEven:
push {r2-r8,lr} @ save registers
tst r0,#1 @ test bit 0 to one
beq 1f @ if result are all zéro, go to even
ldr r1,iAdrsZoneConv @ else display odd message
bl conversion10 @ call decimal conversion
ldr r0,iAdrsMessResultOdd
ldr r1,iAdrsZoneConv @ insert value conversion in message
bl strInsertAtCharInc
bl affichageMess
b 100f
1:
ldr r1,iAdrsZoneConv
bl conversion10 @ call decimal conversion
ldr r0,iAdrsMessResultEven
ldr r1,iAdrsZoneConv @ insert conversion in message
bl strInsertAtCharInc
bl affichageMess
100:
pop {r2-r8,lr} @ restaur registers
bx lr @ return
/***************************************************/
/* ROUTINES INCLUDE */
/***************************************************/
.include "../affichage.inc"
|
http://rosettacode.org/wiki/Euler_method | Euler method | Euler's method numerically approximates solutions of first-order ordinary differential equations (ODEs) with a given initial value. It is an explicit method for solving initial value problems (IVPs), as described in the wikipedia page.
The ODE has to be provided in the following form:
d
y
(
t
)
d
t
=
f
(
t
,
y
(
t
)
)
{\displaystyle {\frac {dy(t)}{dt}}=f(t,y(t))}
with an initial value
y
(
t
0
)
=
y
0
{\displaystyle y(t_{0})=y_{0}}
To get a numeric solution, we replace the derivative on the LHS with a finite difference approximation:
d
y
(
t
)
d
t
≈
y
(
t
+
h
)
−
y
(
t
)
h
{\displaystyle {\frac {dy(t)}{dt}}\approx {\frac {y(t+h)-y(t)}{h}}}
then solve for
y
(
t
+
h
)
{\displaystyle y(t+h)}
:
y
(
t
+
h
)
≈
y
(
t
)
+
h
d
y
(
t
)
d
t
{\displaystyle y(t+h)\approx y(t)+h\,{\frac {dy(t)}{dt}}}
which is the same as
y
(
t
+
h
)
≈
y
(
t
)
+
h
f
(
t
,
y
(
t
)
)
{\displaystyle y(t+h)\approx y(t)+h\,f(t,y(t))}
The iterative solution rule is then:
y
n
+
1
=
y
n
+
h
f
(
t
n
,
y
n
)
{\displaystyle y_{n+1}=y_{n}+h\,f(t_{n},y_{n})}
where
h
{\displaystyle h}
is the step size, the most relevant parameter for accuracy of the solution. A smaller step size increases accuracy but also the computation cost, so it has always has to be hand-picked according to the problem at hand.
Example: Newton's Cooling Law
Newton's cooling law describes how an object of initial temperature
T
(
t
0
)
=
T
0
{\displaystyle T(t_{0})=T_{0}}
cools down in an environment of temperature
T
R
{\displaystyle T_{R}}
:
d
T
(
t
)
d
t
=
−
k
Δ
T
{\displaystyle {\frac {dT(t)}{dt}}=-k\,\Delta T}
or
d
T
(
t
)
d
t
=
−
k
(
T
(
t
)
−
T
R
)
{\displaystyle {\frac {dT(t)}{dt}}=-k\,(T(t)-T_{R})}
It says that the cooling rate
d
T
(
t
)
d
t
{\displaystyle {\frac {dT(t)}{dt}}}
of the object is proportional to the current temperature difference
Δ
T
=
(
T
(
t
)
−
T
R
)
{\displaystyle \Delta T=(T(t)-T_{R})}
to the surrounding environment.
The analytical solution, which we will compare to the numerical approximation, is
T
(
t
)
=
T
R
+
(
T
0
−
T
R
)
e
−
k
t
{\displaystyle T(t)=T_{R}+(T_{0}-T_{R})\;e^{-kt}}
Task
Implement a routine of Euler's method and then to use it to solve the given example of Newton's cooling law with it for three different step sizes of:
2 s
5 s and
10 s
and to compare with the analytical solution.
Initial values
initial temperature
T
0
{\displaystyle T_{0}}
shall be 100 °C
room temperature
T
R
{\displaystyle T_{R}}
shall be 20 °C
cooling constant
k
{\displaystyle k}
shall be 0.07
time interval to calculate shall be from 0 s ──► 100 s
A reference solution (Common Lisp) can be seen below. We see that bigger step sizes lead to reduced approximation accuracy.
| #Kotlin | Kotlin | // version 1.1.2
typealias Deriv = (Double) -> Double // only one parameter needed here
const val FMT = " %7.3f"
fun euler(f: Deriv, y: Double, step: Int, end: Int) {
var yy = y
print(" Step %2d: ".format(step))
for (t in 0..end step step) {
if (t % 10 == 0) print(FMT.format(yy))
yy += step * f(yy)
}
println()
}
fun analytic() {
print(" Time: ")
for (t in 0..100 step 10) print(" %7d".format(t))
print("\nAnalytic: ")
for (t in 0..100 step 10)
print(FMT.format(20.0 + 80.0 * Math.exp(-0.07 * t)))
println()
}
fun cooling(temp: Double) = -0.07 * (temp - 20.0)
fun main(args: Array<String>) {
analytic()
for (i in listOf(2, 5, 10))
euler(::cooling, 100.0, i, 100)
} |
http://rosettacode.org/wiki/Evaluate_binomial_coefficients | Evaluate binomial coefficients | This programming task, is to calculate ANY binomial coefficient.
However, it has to be able to output
(
5
3
)
{\displaystyle {\binom {5}{3}}}
, which is 10.
This formula is recommended:
(
n
k
)
=
n
!
(
n
−
k
)
!
k
!
=
n
(
n
−
1
)
(
n
−
2
)
…
(
n
−
k
+
1
)
k
(
k
−
1
)
(
k
−
2
)
…
1
{\displaystyle {\binom {n}{k}}={\frac {n!}{(n-k)!k!}}={\frac {n(n-1)(n-2)\ldots (n-k+1)}{k(k-1)(k-2)\ldots 1}}}
See Also:
Combinations and permutations
Pascal's triangle
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
| #Commodore_BASIC | Commodore BASIC |
10 REM BINOMIAL COEFFICIENTS
20 REM COMMODORE BASIC 2.0
30 REM 2021-08-24
40 REM BY ALVALONGO
100 Z=0:U=1
110 FOR N=U TO 10
120 PRINT N;
130 FOR K=Z TO N
140 GOSUB 900
150 PRINT C;
160 NEXT K
170 PRINT
180 NEXT N
190 END
900 REM BINOMIAL COEFFICIENT
910 IF K<Z OR K>N THEN C=Z:RETURN
920 IF K=Z OR K=N THEN C=U:RETURN
930 P=K:IF N-K<P THEN P=N-K
940 C=U
950 FOR I=Z TO P-U
960 C=C/(I+U)*(N-I)
980 NEXT I
990 RETURN
|
http://rosettacode.org/wiki/Fibonacci_sequence | Fibonacci sequence | The Fibonacci sequence is a sequence Fn of natural numbers defined recursively:
F0 = 0
F1 = 1
Fn = Fn-1 + Fn-2, if n>1
Task
Write a function to generate the nth Fibonacci number.
Solutions can be iterative or recursive (though recursive solutions are generally considered too slow and are mostly used as an exercise in recursion).
The sequence is sometimes extended into negative numbers by using a straightforward inverse of the positive definition:
Fn = Fn+2 - Fn+1, if n<0
support for negative n in the solution is optional.
Related tasks
Fibonacci n-step number sequences
Leonardo numbers
References
Wikipedia, Fibonacci number
Wikipedia, Lucas number
MathWorld, Fibonacci Number
Some identities for r-Fibonacci numbers
OEIS Fibonacci numbers
OEIS Lucas numbers
| #Vlang | Vlang | fn fib_iter(n int) int {
if n < 2 {
return n
}
mut prev, mut fib := 0, 1
for _ in 0..(n - 1){
prev, fib = fib, prev + fib
}
return fib
}
fn main() {
for val in 0..11 {
println('fibonacci(${val:2d}) = ${fib_iter(val):3d}')
}
} |
http://rosettacode.org/wiki/Entropy/Narcissist | Entropy/Narcissist |
Task
Write a computer program that computes and shows its own entropy.
Related Tasks
Fibonacci_word
Entropy
| #C.2B.2B | C++ | #include <iostream>
#include <fstream>
#include <cmath>
using namespace std;
string readFile (string path) {
string contents;
string line;
ifstream inFile(path);
while (getline (inFile, line)) {
contents.append(line);
contents.append("\n");
}
inFile.close();
return contents;
}
double entropy (string X) {
const int MAXCHAR = 127;
int N = X.length();
int count[MAXCHAR];
double count_i;
char ch;
double sum = 0.0;
for (int i = 0; i < MAXCHAR; i++) count[i] = 0;
for (int pos = 0; pos < N; pos++) {
ch = X[pos];
count[(int)ch]++;
}
for (int n_i = 0; n_i < MAXCHAR; n_i++) {
count_i = count[n_i];
if (count_i > 0) sum -= count_i / N * log2(count_i / N);
}
return sum;
}
int main () {
cout<<entropy(readFile("entropy.cpp"));
return 0;
} |
http://rosettacode.org/wiki/Entropy/Narcissist | Entropy/Narcissist |
Task
Write a computer program that computes and shows its own entropy.
Related Tasks
Fibonacci_word
Entropy
| #Crystal | Crystal | def entropy(s)
counts = s.chars.each_with_object(Hash(Char, Float64).new(0.0)) { |c, h| h[c] += 1 }
counts.values.sum do |count|
freq = count / s.size
-freq * Math.log2(freq)
end
end
puts entropy File.read(__FILE__) |
http://rosettacode.org/wiki/Enumerations | Enumerations | Task
Create an enumeration of constants with and without explicit values.
| #ACL2 | ACL2 | (defun symbol-to-constant (sym)
(intern (concatenate 'string "*" (symbol-name sym) "*")
"ACL2"))
(defmacro enum-with-vals (symbol value &rest args)
(if (endp args)
`(defconst ,(symbol-to-constant symbol) ,value)
`(progn (defconst ,(symbol-to-constant symbol) ,value)
(enum-with-vals ,@args))))
(defun interleave-with-nats-r (xs i)
(if (endp xs)
nil
(cons (first xs)
(cons i (interleave-with-nats-r (rest xs)
(1+ i))))))
(defun interleave-with-nats (xs)
(interleave-with-nats-r xs 0))
(defmacro enum (&rest symbols)
`(enum-with-vals ,@(interleave-with-nats symbols))) |
http://rosettacode.org/wiki/Enumerations | Enumerations | Task
Create an enumeration of constants with and without explicit values.
| #Ada | Ada | type Fruit is (apple, banana, cherry); -- No specification of the representation value;
for Fruit use (apple => 1, banana => 2, cherry => 4); -- specification of the representation values |
http://rosettacode.org/wiki/Ethiopian_multiplication | Ethiopian multiplication | Ethiopian multiplication is a method of multiplying integers using only addition, doubling, and halving.
Method:
Take two numbers to be multiplied and write them down at the top of two columns.
In the left-hand column repeatedly halve the last number, discarding any remainders, and write the result below the last in the same column, until you write a value of 1.
In the right-hand column repeatedly double the last number and write the result below. stop when you add a result in the same row as where the left hand column shows 1.
Examine the table produced and discard any row where the value in the left column is even.
Sum the values in the right-hand column that remain to produce the result of multiplying the original two numbers together
For example: 17 × 34
17 34
Halving the first column:
17 34
8
4
2
1
Doubling the second column:
17 34
8 68
4 136
2 272
1 544
Strike-out rows whose first cell is even:
17 34
8 68
4 136
2 272
1 544
Sum the remaining numbers in the right-hand column:
17 34
8 --
4 ---
2 ---
1 544
====
578
So 17 multiplied by 34, by the Ethiopian method is 578.
Task
The task is to define three named functions/methods/procedures/subroutines:
one to halve an integer,
one to double an integer, and
one to state if an integer is even.
Use these functions to create a function that does Ethiopian multiplication.
References
Ethiopian multiplication explained (BBC Video clip)
A Night Of Numbers - Go Forth And Multiply (Video)
Russian Peasant Multiplication
Programming Praxis: Russian Peasant Multiplication
| #ALGOL-M | ALGOL-M |
BEGIN
INTEGER FUNCTION HALF(I);
INTEGER I;
BEGIN
HALF := I / 2;
END;
INTEGER FUNCTION DOUBLE(I);
INTEGER I;
BEGIN
DOUBLE := I + I;
END;
% RETURN 1 IF EVEN, OTHERWISE 0 %
INTEGER FUNCTION EVEN(I);
INTEGER I;
BEGIN
EVEN := 1 - (I - 2 * (I / 2));
END;
% RETURN I * J AND OPTIONALLY SHOW COMPUTATIONAL STEPS %
INTEGER FUNCTION ETHIOPIAN(I, J, SHOW);
INTEGER I, J, SHOW;
BEGIN
INTEGER P, YES;
YES := 1;
P := 0;
WHILE I >= 1 DO
BEGIN
IF EVEN(I) = YES THEN
BEGIN
IF SHOW = YES THEN WRITE(I," ----", J);
END
ELSE
BEGIN
IF SHOW = YES THEN WRITE(I,J);
P := P + J;
END;
I := HALF(I);
J := DOUBLE(J);
END;
IF SHOW = YES THEN WRITE(" =");
ETHIOPIAN := P;
END;
% EXERCISE THE FUNCTION %
INTEGER YES;
YES := 1;
WRITE(ETHIOPIAN(17,34,YES));
END |
http://rosettacode.org/wiki/Equilibrium_index | Equilibrium index | An equilibrium index of a sequence is an index into the sequence such that the sum of elements at lower indices is equal to the sum of elements at higher indices.
For example, in a sequence
A
{\displaystyle A}
:
A
0
=
−
7
{\displaystyle A_{0}=-7}
A
1
=
1
{\displaystyle A_{1}=1}
A
2
=
5
{\displaystyle A_{2}=5}
A
3
=
2
{\displaystyle A_{3}=2}
A
4
=
−
4
{\displaystyle A_{4}=-4}
A
5
=
3
{\displaystyle A_{5}=3}
A
6
=
0
{\displaystyle A_{6}=0}
3 is an equilibrium index, because:
A
0
+
A
1
+
A
2
=
A
4
+
A
5
+
A
6
{\displaystyle A_{0}+A_{1}+A_{2}=A_{4}+A_{5}+A_{6}}
6 is also an equilibrium index, because:
A
0
+
A
1
+
A
2
+
A
3
+
A
4
+
A
5
=
0
{\displaystyle A_{0}+A_{1}+A_{2}+A_{3}+A_{4}+A_{5}=0}
(sum of zero elements is zero)
7 is not an equilibrium index, because it is not a valid index of sequence
A
{\displaystyle A}
.
Task;
Write a function that, given a sequence, returns its equilibrium indices (if any).
Assume that the sequence may be very long.
| #Arturo | Arturo | eqIndex: function [row][
suml: 0
delayed: 0
sumr: sum row
result: new []
loop.with:'i row 'r [
suml: suml + delayed
sumr: sumr - r
delayed: r
if suml = sumr -> 'result ++ i
]
return result
]
data: @[
@[neg 7, 1, 5, 2, neg 4, 3, 0]
@[2 4 6]
@[2 9 2]
@[1 neg 1 1 neg 1 1 neg 1 1]
]
loop data 'd ->
print [pad.right join.with:", " to [:string] d 25 "=> equilibrium index:" eqIndex d] |
http://rosettacode.org/wiki/Environment_variables | Environment variables | Task
Show how to get one of your process's environment variables.
The available variables vary by system; some of the common ones available on Unix include:
PATH
HOME
USER
| #C.23 | C# | using System;
namespace RosettaCode {
class Program {
static void Main() {
string temp = Environment.GetEnvironmentVariable("TEMP");
Console.WriteLine("TEMP is " + temp);
}
}
} |
http://rosettacode.org/wiki/Environment_variables | Environment variables | Task
Show how to get one of your process's environment variables.
The available variables vary by system; some of the common ones available on Unix include:
PATH
HOME
USER
| #C.2B.2B | C++ | #include <cstdlib>
#include <cstdio>
int main()
{
puts(getenv("HOME"));
return 0;
} |
http://rosettacode.org/wiki/Environment_variables | Environment variables | Task
Show how to get one of your process's environment variables.
The available variables vary by system; some of the common ones available on Unix include:
PATH
HOME
USER
| #Clojure | Clojure | (System/getenv "HOME") |
http://rosettacode.org/wiki/Esthetic_numbers | Esthetic numbers | An esthetic number is a positive integer where every adjacent digit differs from its neighbour by 1.
E.G.
12 is an esthetic number. One and two differ by 1.
5654 is an esthetic number. Each digit is exactly 1 away from its neighbour.
890 is not an esthetic number. Nine and zero differ by 9.
These examples are nominally in base 10 but the concept extends easily to numbers in other bases. Traditionally, single digit numbers are included in esthetic numbers; zero may or may not be. For our purposes, for this task, do not include zero (0) as an esthetic number. Do not include numbers with leading zeros.
Esthetic numbers are also sometimes referred to as stepping numbers.
Task
Write a routine (function, procedure, whatever) to find esthetic numbers in a given base.
Use that routine to find esthetic numbers in bases 2 through 16 and display, here on this page, the esthectic numbers from index (base × 4) through index (base × 6), inclusive. (E.G. for base 2: 8th through 12th, for base 6: 24th through 36th, etc.)
Find and display, here on this page, the base 10 esthetic numbers with a magnitude between 1000 and 9999.
Stretch: Find and display, here on this page, the base 10 esthetic numbers with a magnitude between 1.0e8 and 1.3e8.
Related task
numbers with equal rises and falls
See also
OEIS A033075 - Positive numbers n such that all pairs of consecutive decimal digits differ by 1
Numbers Aplenty - Esthetic numbers
Geeks for Geeks - Stepping numbers
| #Kotlin | Kotlin | import kotlin.math.abs
fun isEsthetic(n: Long, b: Long): Boolean {
if (n == 0L) {
return false
}
var i = n % b
var n2 = n / b
while (n2 > 0) {
val j = n2 % b
if (abs(i - j) != 1L) {
return false
}
n2 /= b
i = j
}
return true
}
fun listEsths(n: Long, n2: Long, m: Long, m2: Long, perLine: Int, all: Boolean) {
val esths = mutableListOf<Long>()
fun dfs(n: Long, m: Long, i: Long) {
if (i in n..m) {
esths.add(i)
}
if (i == 0L || i > m) {
return
}
val d = i % 10
val i1 = i * 10 + d - 1
val i2 = i1 + 2
when (d) {
0L -> {
dfs(n, m, i2)
}
9L -> {
dfs(n, m, i1)
}
else -> {
dfs(n, m, i1)
dfs(n, m, i2)
}
}
}
for (i in 0L until 10L) {
dfs(n2, m2, i)
}
val le = esths.size
println("Base 10: $le esthetic numbers between $n and $m:")
if (all) {
for (c_esth in esths.withIndex()) {
print("${c_esth.value} ")
if ((c_esth.index + 1) % perLine == 0) {
println()
}
}
println()
} else {
for (i in 0 until perLine) {
print("${esths[i]} ")
}
println()
println("............")
for (i in le - perLine until le) {
print("${esths[i]} ")
}
println()
}
println()
}
fun main() {
for (b in 2..16) {
println("Base $b: ${4 * b}th to ${6 * b}th esthetic numbers:")
var n = 1L
var c = 0L
while (c < 6 * b) {
if (isEsthetic(n, b.toLong())) {
c++
if (c >= 4 * b) {
print("${n.toString(b)} ")
}
}
n++
}
println()
}
println()
// the following all use the obvious range limitations for the numbers in question
listEsths(1000, 1010, 9999, 9898, 16, true);
listEsths(1e8.toLong(), 101_010_101, 13 * 1e7.toLong(), 123_456_789, 9, true);
listEsths(1e11.toLong(), 101_010_101_010, 13 * 1e10.toLong(), 123_456_789_898, 7, false);
listEsths(1e14.toLong(), 101_010_101_010_101, 13 * 1e13.toLong(), 123_456_789_898_989, 5, false);
listEsths(1e17.toLong(), 101_010_101_010_101_010, 13 * 1e16.toLong(), 123_456_789_898_989_898, 4, false);
} |
http://rosettacode.org/wiki/Euler%27s_sum_of_powers_conjecture | Euler's sum of powers conjecture | There is a conjecture in mathematics that held for over two hundred years before it was disproved by the finding of a counterexample in 1966 by Lander and Parkin.
Euler's (disproved) sum of powers conjecture
At least k positive kth powers are required to sum to a kth power,
except for the trivial case of one kth power: yk = yk
In 1966, Leon J. Lander and Thomas R. Parkin used a brute-force search on a CDC 6600 computer restricting numbers to those less than 250.
Task
Write a program to search for an integer solution for:
x05 + x15 + x25 + x35 == y5
Where all xi's and y are distinct integers between 0 and 250 (exclusive).
Show an answer here.
Related tasks
Pythagorean quadruples.
Pythagorean triples.
| #C.23 | C# | using System;
namespace EulerSumOfPowers {
class Program {
const int MAX_NUMBER = 250;
static void Main(string[] args) {
bool found = false;
long[] fifth = new long[MAX_NUMBER];
for (int i = 1; i <= MAX_NUMBER; i++) {
long i2 = i * i;
fifth[i - 1] = i2 * i2 * i;
}
for (int a = 0; a < MAX_NUMBER && !found; a++) {
for (int b = a; b < MAX_NUMBER && !found; b++) {
for (int c = b; c < MAX_NUMBER && !found; c++) {
for (int d = c; d < MAX_NUMBER && !found; d++) {
long sum = fifth[a] + fifth[b] + fifth[c] + fifth[d];
int e = Array.BinarySearch(fifth, sum);
found = e >= 0;
if (found) {
Console.WriteLine("{0}^5 + {1}^5 + {2}^5 + {3}^5 = {4}^5", a + 1, b + 1, c + 1, d + 1, e + 1);
}
}
}
}
}
}
}
} |
http://rosettacode.org/wiki/Factorial | Factorial | Definitions
The factorial of 0 (zero) is defined as being 1 (unity).
The Factorial Function of a positive integer, n, is defined as the product of the sequence:
n, n-1, n-2, ... 1
Task
Write a function to return the factorial of a number.
Solutions can be iterative or recursive.
Support for trapping negative n errors is optional.
Related task
Primorial numbers
| #MIPS_Assembly | MIPS Assembly |
##################################
# Factorial; iterative #
# By Keith Stellyes :) #
# Targets Mars implementation #
# August 24, 2016 #
##################################
# This example reads an integer from user, stores in register a1
# Then, it uses a0 as a multiplier and target, it is set to 1
# Pseudocode:
# a0 = 1
# a1 = read_int_from_user()
# while(a1 > 1)
# {
# a0 = a0*a1
# DECREMENT a1
# }
# print(a0)
.text ### PROGRAM BEGIN ###
### GET INTEGER FROM USER ###
li $v0, 5 #set syscall arg to READ_INTEGER
syscall #make the syscall
move $a1, $v0 #int from READ_INTEGER is returned in $v0, but we need $v0
#this will be used as a counter
### SET $a1 TO INITAL VALUE OF 1 AS MULTIPLIER ###
li $a0,1
### Multiply our multiplier, $a1 by our counter, $a0 then store in $a1 ###
loop: ble $a1,1,exit # If the counter is greater than 1, go back to start
mul $a0,$a0,$a1 #a1 = a1*a0
subi $a1,$a1,1 # Decrement counter
j loop # Go back to start
exit:
### PRINT RESULT ###
li $v0,1 #set syscall arg to PRINT_INTEGER
#NOTE: syscall 1 (PRINT_INTEGER) takes a0 as its argument. Conveniently, that
# is our result.
syscall #make the syscall
#exit
li $v0, 10 #set syscall arg to EXIT
syscall #make the syscall
|
http://rosettacode.org/wiki/Even_or_odd | Even or odd | Task
Test whether an integer is even or odd.
There is more than one way to solve this task:
Use the even and odd predicates, if the language provides them.
Check the least significant digit. With binary integers, i bitwise-and 1 equals 0 iff i is even, or equals 1 iff i is odd.
Divide i by 2. The remainder equals 0 iff i is even. The remainder equals +1 or -1 iff i is odd.
Use modular congruences:
i ≡ 0 (mod 2) iff i is even.
i ≡ 1 (mod 2) iff i is odd.
| #ArnoldC | ArnoldC | LISTEN TO ME VERY CAREFULLY isOdd
I NEED YOUR CLOTHES YOUR BOOTS AND YOUR MOTORCYCLE n
GIVE THESE PEOPLE AIR
HEY CHRISTMAS TREE result
YOU SET US UP @I LIED
GET TO THE CHOPPER result
HERE IS MY INVITATION n
I LET HIM GO 2
ENOUGH TALK
I'LL BE BACK result
HASTA LA VISTA, BABY
LISTEN TO ME VERY CAREFULLY showParity
I NEED YOUR CLOTHES YOUR BOOTS AND YOUR MOTORCYCLE n
TALK TO THE HAND n
HEY CHRISTMAS TREE parity
YOU SET US UP @I LIED
GET YOUR A** TO MARS parity
DO IT NOW isOdd n
BECAUSE I'M GOING TO SAY PLEASE parity
TALK TO THE HAND "odd"
BULLS***
TALK TO THE HAND "even"
YOU HAVE NO RESPECT FOR LOGIC
TALK TO THE HAND ""
HASTA LA VISTA, BABY
IT'S SHOWTIME
DO IT NOW showParity 5
DO IT NOW showParity 6
DO IT NOW showParity -11
YOU HAVE BEEN TERMINATED |
http://rosettacode.org/wiki/Euler_method | Euler method | Euler's method numerically approximates solutions of first-order ordinary differential equations (ODEs) with a given initial value. It is an explicit method for solving initial value problems (IVPs), as described in the wikipedia page.
The ODE has to be provided in the following form:
d
y
(
t
)
d
t
=
f
(
t
,
y
(
t
)
)
{\displaystyle {\frac {dy(t)}{dt}}=f(t,y(t))}
with an initial value
y
(
t
0
)
=
y
0
{\displaystyle y(t_{0})=y_{0}}
To get a numeric solution, we replace the derivative on the LHS with a finite difference approximation:
d
y
(
t
)
d
t
≈
y
(
t
+
h
)
−
y
(
t
)
h
{\displaystyle {\frac {dy(t)}{dt}}\approx {\frac {y(t+h)-y(t)}{h}}}
then solve for
y
(
t
+
h
)
{\displaystyle y(t+h)}
:
y
(
t
+
h
)
≈
y
(
t
)
+
h
d
y
(
t
)
d
t
{\displaystyle y(t+h)\approx y(t)+h\,{\frac {dy(t)}{dt}}}
which is the same as
y
(
t
+
h
)
≈
y
(
t
)
+
h
f
(
t
,
y
(
t
)
)
{\displaystyle y(t+h)\approx y(t)+h\,f(t,y(t))}
The iterative solution rule is then:
y
n
+
1
=
y
n
+
h
f
(
t
n
,
y
n
)
{\displaystyle y_{n+1}=y_{n}+h\,f(t_{n},y_{n})}
where
h
{\displaystyle h}
is the step size, the most relevant parameter for accuracy of the solution. A smaller step size increases accuracy but also the computation cost, so it has always has to be hand-picked according to the problem at hand.
Example: Newton's Cooling Law
Newton's cooling law describes how an object of initial temperature
T
(
t
0
)
=
T
0
{\displaystyle T(t_{0})=T_{0}}
cools down in an environment of temperature
T
R
{\displaystyle T_{R}}
:
d
T
(
t
)
d
t
=
−
k
Δ
T
{\displaystyle {\frac {dT(t)}{dt}}=-k\,\Delta T}
or
d
T
(
t
)
d
t
=
−
k
(
T
(
t
)
−
T
R
)
{\displaystyle {\frac {dT(t)}{dt}}=-k\,(T(t)-T_{R})}
It says that the cooling rate
d
T
(
t
)
d
t
{\displaystyle {\frac {dT(t)}{dt}}}
of the object is proportional to the current temperature difference
Δ
T
=
(
T
(
t
)
−
T
R
)
{\displaystyle \Delta T=(T(t)-T_{R})}
to the surrounding environment.
The analytical solution, which we will compare to the numerical approximation, is
T
(
t
)
=
T
R
+
(
T
0
−
T
R
)
e
−
k
t
{\displaystyle T(t)=T_{R}+(T_{0}-T_{R})\;e^{-kt}}
Task
Implement a routine of Euler's method and then to use it to solve the given example of Newton's cooling law with it for three different step sizes of:
2 s
5 s and
10 s
and to compare with the analytical solution.
Initial values
initial temperature
T
0
{\displaystyle T_{0}}
shall be 100 °C
room temperature
T
R
{\displaystyle T_{R}}
shall be 20 °C
cooling constant
k
{\displaystyle k}
shall be 0.07
time interval to calculate shall be from 0 s ──► 100 s
A reference solution (Common Lisp) can be seen below. We see that bigger step sizes lead to reduced approximation accuracy.
| #Lambdatalk | Lambdatalk |
{def eulersMethod
{def eulersMethod.r
{lambda {:f :b :h :t :y}
{if {<= :t :b}
then {tr {td :t} {td {/ {round {* :y 1000}} 1000}}}
{eulersMethod.r :f :b :h {+ :t :h} {+ :y {* :h {:f :t :y}}}}
else}}}
{lambda {:f :y0 :a :b :h}
{table {eulersMethod.r :f :b :h :a :y0}}}}
{def cooling
{lambda {:time :temp}
{* -0.07 {- :temp 20}}}}
{eulersMethod cooling 100 0 100 10}
->
0 100
10 44
20 27.2
30 22.16
40 20.648
50 20.194
60 20.058
70 20.017
80 20.005
90 20.002
100 20
|
http://rosettacode.org/wiki/Evaluate_binomial_coefficients | Evaluate binomial coefficients | This programming task, is to calculate ANY binomial coefficient.
However, it has to be able to output
(
5
3
)
{\displaystyle {\binom {5}{3}}}
, which is 10.
This formula is recommended:
(
n
k
)
=
n
!
(
n
−
k
)
!
k
!
=
n
(
n
−
1
)
(
n
−
2
)
…
(
n
−
k
+
1
)
k
(
k
−
1
)
(
k
−
2
)
…
1
{\displaystyle {\binom {n}{k}}={\frac {n!}{(n-k)!k!}}={\frac {n(n-1)(n-2)\ldots (n-k+1)}{k(k-1)(k-2)\ldots 1}}}
See Also:
Combinations and permutations
Pascal's triangle
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
| #Common_Lisp | Common Lisp |
(defun choose (n k)
(labels ((prod-enum (s e)
(do ((i s (1+ i)) (r 1 (* i r))) ((> i e) r)))
(fact (n) (prod-enum 1 n)))
(/ (prod-enum (- (1+ n) k) n) (fact k))))
|
http://rosettacode.org/wiki/Evaluate_binomial_coefficients | Evaluate binomial coefficients | This programming task, is to calculate ANY binomial coefficient.
However, it has to be able to output
(
5
3
)
{\displaystyle {\binom {5}{3}}}
, which is 10.
This formula is recommended:
(
n
k
)
=
n
!
(
n
−
k
)
!
k
!
=
n
(
n
−
1
)
(
n
−
2
)
…
(
n
−
k
+
1
)
k
(
k
−
1
)
(
k
−
2
)
…
1
{\displaystyle {\binom {n}{k}}={\frac {n!}{(n-k)!k!}}={\frac {n(n-1)(n-2)\ldots (n-k+1)}{k(k-1)(k-2)\ldots 1}}}
See Also:
Combinations and permutations
Pascal's triangle
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
| #D | D | T binomial(T)(in T n, T k) pure nothrow {
if (k > (n / 2))
k = n - k;
T bc = 1;
foreach (T i; T(2) .. k + 1)
bc = (bc * (n - k + i)) / i;
return bc;
}
void main() {
import std.stdio, std.bigint;
foreach (const d; [[5, 3], [100, 2], [100, 98]])
writefln("(%3d %3d) = %s", d[0], d[1], binomial(d[0], d[1]));
writeln("(100 50) = ", binomial(100.BigInt, 50.BigInt));
} |
http://rosettacode.org/wiki/Fibonacci_sequence | Fibonacci sequence | The Fibonacci sequence is a sequence Fn of natural numbers defined recursively:
F0 = 0
F1 = 1
Fn = Fn-1 + Fn-2, if n>1
Task
Write a function to generate the nth Fibonacci number.
Solutions can be iterative or recursive (though recursive solutions are generally considered too slow and are mostly used as an exercise in recursion).
The sequence is sometimes extended into negative numbers by using a straightforward inverse of the positive definition:
Fn = Fn+2 - Fn+1, if n<0
support for negative n in the solution is optional.
Related tasks
Fibonacci n-step number sequences
Leonardo numbers
References
Wikipedia, Fibonacci number
Wikipedia, Lucas number
MathWorld, Fibonacci Number
Some identities for r-Fibonacci numbers
OEIS Fibonacci numbers
OEIS Lucas numbers
| #Wart | Wart | def (fib n)
if (n < 2)
n
(+ (fib n-1) (fib n-2)) |
http://rosettacode.org/wiki/Entropy/Narcissist | Entropy/Narcissist |
Task
Write a computer program that computes and shows its own entropy.
Related Tasks
Fibonacci_word
Entropy
| #D | D | void main(in string[] args) {
import std.stdio, std.algorithm, std.math, std.file;
auto data = sort(cast(ubyte[])args[0].read);
return data
.group
.map!(g => g[1] / double(data.length))
.map!(p => -p * p.log2)
.sum
.writeln;
} |
http://rosettacode.org/wiki/Entropy/Narcissist | Entropy/Narcissist |
Task
Write a computer program that computes and shows its own entropy.
Related Tasks
Fibonacci_word
Entropy
| #Elixir | Elixir | File.open(__ENV__.file, [:read], fn(file) ->
text = IO.read(file, :all)
leng = String.length(text)
String.codepoints(text)
|> Enum.group_by(&(&1))
|> Enum.map(fn{_,value} -> length(value) end)
|> Enum.reduce(0, fn count, entropy ->
freq = count / leng
entropy - freq * :math.log2(freq)
end)
|> IO.puts
end) |
http://rosettacode.org/wiki/Entropy/Narcissist | Entropy/Narcissist |
Task
Write a computer program that computes and shows its own entropy.
Related Tasks
Fibonacci_word
Entropy
| #Emacs_Lisp | Emacs Lisp | (defun shannon-entropy (input)
(let ((freq-table (make-hash-table))
(entropy 0)
(length (+ (length input) 0.0)))
(mapcar (lambda (x)
(puthash x
(+ 1 (gethash x freq-table 0))
freq-table))
input)
(maphash (lambda (k v)
(set 'entropy (+ entropy
(* (/ v length)
(log (/ v length) 2)))))
freq-table)
(- entropy)))
(defun narcissist ()
(shannon-entropy (with-temp-buffer
(insert-file-contents "U:/rosetta/narcissist.el")
(buffer-string)))) |
http://rosettacode.org/wiki/Enumerations | Enumerations | Task
Create an enumeration of constants with and without explicit values.
| #ALGOL_68 | ALGOL 68 | BEGIN # example 1 #
MODE FRUIT = INT;
FRUIT apple = 1, banana = 2, cherry = 4;
FRUIT x := cherry;
CASE x IN
print(("It is an apple #",x, new line)),
print(("It is a banana #",x, new line)),
SKIP, # 3 not defined #
print(("It is a cherry #",x, new line))
OUT
SKIP # other values #
ESAC
END; |
http://rosettacode.org/wiki/Enumerations | Enumerations | Task
Create an enumeration of constants with and without explicit values.
| #AmigaE | AmigaE | ENUM APPLE, BANANA, CHERRY
PROC main()
DEF x
ForAll({x}, [APPLE, BANANA, CHERRY],
`WriteF('\d\n', x))
ENDPROC |
http://rosettacode.org/wiki/Enforced_immutability | Enforced immutability | Task
Demonstrate any means your language has to prevent the modification of values, or to create objects that cannot be modified after they have been created.
| #11l | 11l | -V min_size = 10 |
http://rosettacode.org/wiki/Ethiopian_multiplication | Ethiopian multiplication | Ethiopian multiplication is a method of multiplying integers using only addition, doubling, and halving.
Method:
Take two numbers to be multiplied and write them down at the top of two columns.
In the left-hand column repeatedly halve the last number, discarding any remainders, and write the result below the last in the same column, until you write a value of 1.
In the right-hand column repeatedly double the last number and write the result below. stop when you add a result in the same row as where the left hand column shows 1.
Examine the table produced and discard any row where the value in the left column is even.
Sum the values in the right-hand column that remain to produce the result of multiplying the original two numbers together
For example: 17 × 34
17 34
Halving the first column:
17 34
8
4
2
1
Doubling the second column:
17 34
8 68
4 136
2 272
1 544
Strike-out rows whose first cell is even:
17 34
8 68
4 136
2 272
1 544
Sum the remaining numbers in the right-hand column:
17 34
8 --
4 ---
2 ---
1 544
====
578
So 17 multiplied by 34, by the Ethiopian method is 578.
Task
The task is to define three named functions/methods/procedures/subroutines:
one to halve an integer,
one to double an integer, and
one to state if an integer is even.
Use these functions to create a function that does Ethiopian multiplication.
References
Ethiopian multiplication explained (BBC Video clip)
A Night Of Numbers - Go Forth And Multiply (Video)
Russian Peasant Multiplication
Programming Praxis: Russian Peasant Multiplication
| #ALGOL_W | ALGOL W | begin
% returns half of a %
integer procedure halve ( integer value a ) ; a div 2;
% returns a doubled %
integer procedure double ( integer value a ) ; a * 2;
% returns true if a is even, false otherwise %
logical procedure even ( integer value a ) ; not odd( a );
% returns the product of a and b using ethopian multiplication %
% rather than keep a table of the intermediate results, %
% we examine then as they are generated %
integer procedure ethopianMultiplication ( integer value a, b ) ;
begin
integer v, r, accumulator;
v := a;
r := b;
accumulator := 0;
i_w := 4; s_w := 0; % set output formatting %
while begin
write( v );
if even( v ) then writeon( " ---" )
else begin
accumulator := accumulator + r;
writeon( " ", r );
end;
v := halve( v );
r := double( r );
v > 0
end do begin end;
write( " =====" );
accumulator
end ethopianMultiplication ;
% task test case %
begin
integer m;
m := ethopianMultiplication( 17, 34 );
write( " ", m )
end
end. |
http://rosettacode.org/wiki/Equilibrium_index | Equilibrium index | An equilibrium index of a sequence is an index into the sequence such that the sum of elements at lower indices is equal to the sum of elements at higher indices.
For example, in a sequence
A
{\displaystyle A}
:
A
0
=
−
7
{\displaystyle A_{0}=-7}
A
1
=
1
{\displaystyle A_{1}=1}
A
2
=
5
{\displaystyle A_{2}=5}
A
3
=
2
{\displaystyle A_{3}=2}
A
4
=
−
4
{\displaystyle A_{4}=-4}
A
5
=
3
{\displaystyle A_{5}=3}
A
6
=
0
{\displaystyle A_{6}=0}
3 is an equilibrium index, because:
A
0
+
A
1
+
A
2
=
A
4
+
A
5
+
A
6
{\displaystyle A_{0}+A_{1}+A_{2}=A_{4}+A_{5}+A_{6}}
6 is also an equilibrium index, because:
A
0
+
A
1
+
A
2
+
A
3
+
A
4
+
A
5
=
0
{\displaystyle A_{0}+A_{1}+A_{2}+A_{3}+A_{4}+A_{5}=0}
(sum of zero elements is zero)
7 is not an equilibrium index, because it is not a valid index of sequence
A
{\displaystyle A}
.
Task;
Write a function that, given a sequence, returns its equilibrium indices (if any).
Assume that the sequence may be very long.
| #AutoHotkey | AutoHotkey | Equilibrium_index(list, BaseIndex=0){
StringSplit, A, list, `,
Loop % A0 {
i := A_Index , Pre := Post := 0
loop, % A0
if (A_Index < i)
Pre += A%A_Index%
else if (A_Index > i)
Post += A%A_Index%
if (Pre = Post)
Res .= (Res?", ":"") i - (BaseIndex?0:1)
}
return Res
} |
http://rosettacode.org/wiki/Equilibrium_index | Equilibrium index | An equilibrium index of a sequence is an index into the sequence such that the sum of elements at lower indices is equal to the sum of elements at higher indices.
For example, in a sequence
A
{\displaystyle A}
:
A
0
=
−
7
{\displaystyle A_{0}=-7}
A
1
=
1
{\displaystyle A_{1}=1}
A
2
=
5
{\displaystyle A_{2}=5}
A
3
=
2
{\displaystyle A_{3}=2}
A
4
=
−
4
{\displaystyle A_{4}=-4}
A
5
=
3
{\displaystyle A_{5}=3}
A
6
=
0
{\displaystyle A_{6}=0}
3 is an equilibrium index, because:
A
0
+
A
1
+
A
2
=
A
4
+
A
5
+
A
6
{\displaystyle A_{0}+A_{1}+A_{2}=A_{4}+A_{5}+A_{6}}
6 is also an equilibrium index, because:
A
0
+
A
1
+
A
2
+
A
3
+
A
4
+
A
5
=
0
{\displaystyle A_{0}+A_{1}+A_{2}+A_{3}+A_{4}+A_{5}=0}
(sum of zero elements is zero)
7 is not an equilibrium index, because it is not a valid index of sequence
A
{\displaystyle A}
.
Task;
Write a function that, given a sequence, returns its equilibrium indices (if any).
Assume that the sequence may be very long.
| #AWK | AWK |
# syntax: GAWK -f EQUILIBRIUM_INDEX.AWK
BEGIN {
main("-7 1 5 2 -4 3 0")
main("2 4 6")
main("2 9 2")
main("1 -1 1 -1 1 -1 1")
exit(0)
}
function main(numbers, x) {
x = equilibrium(numbers)
printf("numbers: %s\n",numbers)
printf("indices: %s\n\n",length(x)==0?"none":x)
}
function equilibrium(numbers, arr,i,leftsum,leng,str,sum) {
leng = split(numbers,arr," ")
for (i=1; i<=leng; i++) {
sum += arr[i]
}
for (i=1; i<=leng; i++) {
sum -= arr[i]
if (leftsum == sum) {
str = str i " "
}
leftsum += arr[i]
}
return(str)
}
|
http://rosettacode.org/wiki/Environment_variables | Environment variables | Task
Show how to get one of your process's environment variables.
The available variables vary by system; some of the common ones available on Unix include:
PATH
HOME
USER
| #COBOL | COBOL | IDENTIFICATION DIVISION.
PROGRAM-ID. Environment-Vars.
DATA DIVISION.
WORKING-STORAGE SECTION.
01 home PIC X(75).
PROCEDURE DIVISION.
* *> Method 1.
ACCEPT home FROM ENVIRONMENT "HOME"
DISPLAY home
* *> Method 2.
DISPLAY "HOME" UPON ENVIRONMENT-NAME
ACCEPT home FROM ENVIRONMENT-VALUE
GOBACK
. |
http://rosettacode.org/wiki/Environment_variables | Environment variables | Task
Show how to get one of your process's environment variables.
The available variables vary by system; some of the common ones available on Unix include:
PATH
HOME
USER
| #CoffeeScript | CoffeeScript | for var_name in ['PATH', 'HOME', 'LANG', 'USER']
console.log var_name, process.env[var_name] |
http://rosettacode.org/wiki/Environment_variables | Environment variables | Task
Show how to get one of your process's environment variables.
The available variables vary by system; some of the common ones available on Unix include:
PATH
HOME
USER
| #Common_Lisp | Common Lisp | (lispworks:environment-variable "USER") |
http://rosettacode.org/wiki/Esthetic_numbers | Esthetic numbers | An esthetic number is a positive integer where every adjacent digit differs from its neighbour by 1.
E.G.
12 is an esthetic number. One and two differ by 1.
5654 is an esthetic number. Each digit is exactly 1 away from its neighbour.
890 is not an esthetic number. Nine and zero differ by 9.
These examples are nominally in base 10 but the concept extends easily to numbers in other bases. Traditionally, single digit numbers are included in esthetic numbers; zero may or may not be. For our purposes, for this task, do not include zero (0) as an esthetic number. Do not include numbers with leading zeros.
Esthetic numbers are also sometimes referred to as stepping numbers.
Task
Write a routine (function, procedure, whatever) to find esthetic numbers in a given base.
Use that routine to find esthetic numbers in bases 2 through 16 and display, here on this page, the esthectic numbers from index (base × 4) through index (base × 6), inclusive. (E.G. for base 2: 8th through 12th, for base 6: 24th through 36th, etc.)
Find and display, here on this page, the base 10 esthetic numbers with a magnitude between 1000 and 9999.
Stretch: Find and display, here on this page, the base 10 esthetic numbers with a magnitude between 1.0e8 and 1.3e8.
Related task
numbers with equal rises and falls
See also
OEIS A033075 - Positive numbers n such that all pairs of consecutive decimal digits differ by 1
Numbers Aplenty - Esthetic numbers
Geeks for Geeks - Stepping numbers
| #Lua | Lua | function to(n, b)
local BASE = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"
if n == 0 then
return "0"
end
local ss = ""
while n > 0 do
local idx = (n % b) + 1
n = math.floor(n / b)
ss = ss .. BASE:sub(idx, idx)
end
return string.reverse(ss)
end
function isEsthetic(n, b)
function uabs(a, b)
if a < b then
return b - a
end
return a - b
end
if n == 0 then
return false
end
local i = n % b
n = math.floor(n / b)
while n > 0 do
local j = n % b
if uabs(i, j) ~= 1 then
return false
end
n = math.floor(n / b)
i = j
end
return true
end
function listEsths(n, n2, m, m2, perLine, all)
local esths = {}
function dfs(n, m, i)
if i >= n and i <= m then
table.insert(esths, i)
end
if i == 0 or i > m then
return
end
local d = i % 10
local i1 = 10 * i + d - 1
local i2 = i1 + 2
if d == 0 then
dfs(n, m, i2)
elseif d == 9 then
dfs(n, m, i1)
else
dfs(n, m, i1)
dfs(n, m, i2)
end
end
for i=0,9 do
dfs(n2, m2, i)
end
local le = #esths
print(string.format("Base 10: %s esthetic numbers between %s and %s:", le, math.floor(n), math.floor(m)))
if all then
for c,esth in pairs(esths) do
io.write(esth.." ")
if c % perLine == 0 then
print()
end
end
print()
else
for i=1,perLine do
io.write(esths[i] .. " ")
end
print("\n............")
for i = le - perLine + 1, le do
io.write(esths[i] .. " ")
end
print()
end
print()
end
for b=2,16 do
print(string.format("Base %d: %dth to %dth esthetic numbers:", b, 4 * b, 6 * b))
local n = 1
local c = 0
while c < 6 * b do
if isEsthetic(n, b) then
c = c + 1
if c >= 4 * b then
io.write(to(n, b).." ")
end
end
n = n + 1
end
print()
end
print()
-- the following all use the obvious range limitations for the numbers in question
listEsths(1000, 1010, 9999, 9898, 16, true)
listEsths(1e8, 101010101, 13 * 1e7, 123456789, 9, true)
listEsths(1e11, 101010101010, 13 * 1e10, 123456789898, 7, false)
listEsths(1e14, 101010101010101, 13 * 1e13, 123456789898989, 5, false)
listEsths(1e17, 101010101010101010, 13 * 1e16, 123456789898989898, 4, false) |
http://rosettacode.org/wiki/Euler%27s_sum_of_powers_conjecture | Euler's sum of powers conjecture | There is a conjecture in mathematics that held for over two hundred years before it was disproved by the finding of a counterexample in 1966 by Lander and Parkin.
Euler's (disproved) sum of powers conjecture
At least k positive kth powers are required to sum to a kth power,
except for the trivial case of one kth power: yk = yk
In 1966, Leon J. Lander and Thomas R. Parkin used a brute-force search on a CDC 6600 computer restricting numbers to those less than 250.
Task
Write a program to search for an integer solution for:
x05 + x15 + x25 + x35 == y5
Where all xi's and y are distinct integers between 0 and 250 (exclusive).
Show an answer here.
Related tasks
Pythagorean quadruples.
Pythagorean triples.
| #C.2B.2B | C++ | #include <algorithm>
#include <iostream>
#include <cmath>
#include <set>
#include <vector>
using namespace std;
bool find()
{
const auto MAX = 250;
vector<double> pow5(MAX);
for (auto i = 1; i < MAX; i++)
pow5[i] = (double)i * i * i * i * i;
for (auto x0 = 1; x0 < MAX; x0++) {
for (auto x1 = 1; x1 < x0; x1++) {
for (auto x2 = 1; x2 < x1; x2++) {
for (auto x3 = 1; x3 < x2; x3++) {
auto sum = pow5[x0] + pow5[x1] + pow5[x2] + pow5[x3];
if (binary_search(pow5.begin(), pow5.end(), sum))
{
cout << x0 << " " << x1 << " " << x2 << " " << x3 << " " << pow(sum, 1.0 / 5.0) << endl;
return true;
}
}
}
}
}
// not found
return false;
}
int main(void)
{
int tm = clock();
if (!find())
cout << "Nothing found!\n";
cout << "time=" << (int)((clock() - tm) * 1000 / CLOCKS_PER_SEC) << " milliseconds\r\n";
return 0;
} |
http://rosettacode.org/wiki/Factorial | Factorial | Definitions
The factorial of 0 (zero) is defined as being 1 (unity).
The Factorial Function of a positive integer, n, is defined as the product of the sequence:
n, n-1, n-2, ... 1
Task
Write a function to return the factorial of a number.
Solutions can be iterative or recursive.
Support for trapping negative n errors is optional.
Related task
Primorial numbers
| #Mirah | Mirah | def factorial_iterative(n:int)
2.upto(n-1) do |i|
n *= i
end
n
end
puts factorial_iterative 10 |
http://rosettacode.org/wiki/Even_or_odd | Even or odd | Task
Test whether an integer is even or odd.
There is more than one way to solve this task:
Use the even and odd predicates, if the language provides them.
Check the least significant digit. With binary integers, i bitwise-and 1 equals 0 iff i is even, or equals 1 iff i is odd.
Divide i by 2. The remainder equals 0 iff i is even. The remainder equals +1 or -1 iff i is odd.
Use modular congruences:
i ≡ 0 (mod 2) iff i is even.
i ≡ 1 (mod 2) iff i is odd.
| #Arturo | Arturo | loop (neg 5)..5 [x][
if? even? x -> print [pad to :string x 4 ": even"]
else -> print [pad to :string x 4 ": odd"]
] |
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