Part of the program I have checks if an input number is a perfect number. We're supposed to find a solution that runs in O(sqrt(n)). The rest of my program runs in constant time, but this function is holding me back.
function Perfect(x: integer): boolean;
var
i: integer;
sum: integer=0;
begin
for i := 1 to x-1 do
if (x mod i = 0) then
sum := sum + i;
if sum = x then
exit(true)
else
exit(false);
end;
This runs in O(n) time, and I need to cut it down to O(sqrt(n)) time.
These are the options I've come up with:
(1) Find a way to make the for loop go from 1 to sqrt(x)...
(2) Find a way to check for a perfect number that doesn't use a for loop...
Any suggestions? I appreciate any hints, tips, instruction, etc. :)
You need to iterate the cycle not for i := 1 to x-1 but for i := 2 to trunc(sqrt(x)).
The highest integer divisor is x but we do not take it in into account when looking for perfect numbers. We increment sum by 1 instead (or initialize it with 1 - not 0).
The code if (x mod i = 0) then sum := sum + i; for this purpose can be converted to:
if (x mod i = 0) then
begin
sum := sum + i;
sum := sum + (x div i);
end;
And so we get the following code:
function Perfect(x: integer): boolean;
var
i: integer;
sum: integer = 1;
sqrtx: integer;
begin
sqrtx := trunc(sqrt(x));
i := 2;
while i <= sqrtx do
begin
if (x mod i = 0) then
begin
sum := sum + i;
sum := sum + (x div i) // you can also compare i and x div i
//to avoid adding the same number twice
//for example when x = 4 both 2 and 4 div 2 will be added
end;
inc(i);
end;
if sum = x then
exit(true)
else
exit(false);
end;
Related
I've beeng having some trouble with this code... I need to create an algorithm which makes the user input a number (X), and then the program calculates the sum of all the odd numbers below (x).
This what I've tried so far, but can't really wrap my head around the logic behind it:
Program odd_numbers;
Var
Num, Limite, Soma: integer;
Begin;
Soma := 0;
Writeln('Choose a limit:');
Readln(Limite);
While (Limite / 2 > 0) do
Begin;
Soma := ((Num < Limite) mod 2 > 0);
Writeln('The sum of odd numbers from 0 to ', Limite, ' é ', Soma);
End;
if (Limite mod 2 = 0) then
Begin;
Soma := ((Num < Limite) mod 2 = 0);
Writeln('The sum of odd numbers from 0 to ', Limite, ' é ', Soma);
End;
End.
*PS: Been writing the code with variables in Portuguese, so don't mind the variables appearing weird to understand. *
I see that everyone is happily looping, but this is not necessary. This is a simple arithmetic sequence, and the sum can be calculated without a loop.
Just think of the following:
1 + 3 = 2 * (1 + 3) / 2 = 2 * 2 = 4 ; limits 3 and 4
1 + 3 + 5 = 3 * (1 + 5) / 2 = 3 * 3 = 9 ; limits 5 and 6
1 + 3 + 5 + 7 = 4 * (1 + 7) / 2 = 4 * 4 = 16 ; limits 7 and 8
1 + 3 + 5 + 7 + 9 = 5 * (1 + 9) / 2 = 5 * 5 = 25 ; limits 9 and 10
1 + 3 + 5 + 7 + 9 + 11 = 6 * (1 + 11) / 2 = 6 * 6 = 36 ; limits 11 and 12
But not only that, you'll see that it is in fact always a perfect square: Sqr((n+1) div 2).
So just calculate:
program odd_numbers;
var
Num, Limite, Soma: Integer;
begin
Write('Choose a limit: ');
Readln(Limite);
Num := (Limite + 1) div 2;
Soma := Num * Num;
Writeln('The sum of odd numbers from 0 to ', Limite, ' is ', Soma);
end.
Looks a little simpler than what the others propose.
The loop While (Limite / 2 > 0) do ... uses real arithmetic and not integer arithmetic. I guess you mean While (Limite div 2 > 0) do ... And you should change Limite in the loop otherwise you get stuck because the exit condition can never be reached.
After you have asked the user to enter a number, Limite, you need to keep that unchanged, because you need it in the final message. You also need a loop where you go through all numbers from Limite towards 0.
You started with a while loop which is ok, you are just missing the loop control variable. That is a variable that eventually gets a terminating value which then stops the loop. Use for example the Num variable you already have declared. You can use the same variable to investigate the numbers between user input and 0, for being odd values.
num := limite-1; // give num a start value based on user input (-1 because of "... numbers below (x)")
while num > 0 do // stop the loop when 0 is reached
begin
// here you investigate if `num` is a odd number (e.g. using `mod` operator or
// possibly your pascal has a built in `function Odd(value: integer): boolean;`)
// and add it to `Soma` if it is
num := num - 1;// decrement num at every iteration
end;
Finally you need to consider changes to the above, to handle negative input from the user.
To test if an integer is an odd value, you could use following function:
function IsOdd( value : Integer) : Boolean;
begin
IsOdd := (value mod 2) <> 0;
end;
Many pascal compilers have a built-in function called Odd(), which you could use.
A while loop works well to solve this problem. If you start with lowest odd number above zero, i.e. one and continue upwards so long we do not exceed the limit value we have a simple start:
function GetOddSumBelowX( X : Integer) : Integer;
var
i,sum: Integer;
begin
i := 1; // Start with first odd number
sum := 0;
while (i < X) do begin // as long as i less than X, loop
if IsOdd(i) then begin
sum := sum + i; // add to sum
end;
i := i + 1; // Increment i
end;
GetOddSumBelowX := sum;
end;
Now, that was simple enough. Next step to simplify the loop is to increment the i variable by two instead, just to jump between all odd numbers:
function GetOddSumBelowX( X : Integer) : Integer;
var
i,sum: Integer;
begin
i := 1; // Start with first odd number
sum := 0;
while (i < X) do begin // as long as i less than X, loop
sum := sum + i; // add to sum
i := i + 2; // Increment to next odd number
end;
GetOddSumBelowX := sum;
end;
You play a game with 100 opponents. The game has k rounds. Every round you can eliminate some opponents (always atleast 1). You are rewarded for eliminating them.
The reward is: 100.000 * '# of eliminated opponents' / '# of opponents' <= in integers (rounded down)
I want to eliminate the opponents in a way, that gets me the largest amount of money possible.
Example game:
number of rounds = 3
first round we eliminate 50 opponents, so we get 100.000 * 50 / 100 = +50.000
second round we eliminate 30, so we get 100.000 * 30 / 50 = +60.000
last round we eliminate last 20 opponents, so we get 100.000 * 20 / 20 = +100.000
so the total winnings are: 210.000
I tried to write up something, but I don't think it's the most effective way to do it?
Program EliminationGame;
var
selectedHistory : array [1..10] of integer;
opponentCount,roundCount : integer;
maxOpponents,numberSelected : integer;
totalMoney : integer;
i : integer;
begin
totalMoney := 0;
maxOpponents := 100;
opponentCount := maxOpponents;
roundCount := 3; {test value}
for i:=1 to roundCount do begin
if (i = roundCount) then begin
numberSelected := opponentCount;
end else begin
numberSelected := floor(opponentCount / roundCount);
end;
selectedHistory[i] := numberSelected;
totalMoney := floor(totalMoney + (numberSelected / opponentCount * 100000));
opponentCount := opponentCount - numberSelected;
end;
writeln('Total money won:');
writeln(totalMoney);
writeln('Amount selected in rounds:');
for i:= 0 to Length(selectedHistory) do
write(selectedHistory[i],' ');
end.
Also it seems that floor function does not exist in pascal?
It seems the question has a maths answer that can be calculated in advance. As #Anton said it was obvious that the number of points given during the third round did not depend upon the number of eliminated enemies. So the third round should eliminate 1 enemy.
So We get the following function for a thre-round game.
f(x)=100000x/100+100000(99-x)/(100-x)+100000*1/1, where x- the number
of enemies eleminated at first round.
if we find the extrema (local maximum of the function) it appears equal to 90. That means the decision is the following: the first round eliminates 90 the second - 9, the third - 1 enemy.
Of course, for consideration: 90=100-sqrt(100).
In other words: the Pascal decision of the task is to loop a variable from 1 to 99 and see the maximum of this function. X-will be the answer.
program Project1;
var
x, xmax: byte;
MaxRes, tmp: real;
begin
xmax := 0;
MaxRes := 0;
for x := 1 to 99 do
begin
tmp := 100000 * x / 100 + 100000*(99 - x) / (100 - x) + 100000 * 1 / 1;
if tmp > MaxRes then
begin
MaxRes := tmp;
xmax := x;
end;
end;
writeln(xmax);
readln;
end.
The general decision for other number of enemies and rounds (using recursion) is the following (Delphi dialect):
program Project1;
{$APPTYPE CONSOLE}
{$R *.res}
Uses System.SysUtils;
var
s: string;
function Part(RemainingEnemies: byte; Depth: byte;
var OutputString: string): real;
var
i: byte;
tmp, MaxRes: real;
imax: byte;
DaughterString: string;
begin
OutputString := '';
if Depth = 0 then
exit(0);
imax := 0;
MaxRes := 0;
for i := 1 to RemainingEnemies - Depth + 1 do
begin
tmp := i / RemainingEnemies * 100000 + Part(RemainingEnemies - i, Depth - 1,
DaughterString);
if tmp > MaxRes then
begin
MaxRes := tmp;
imax := i;
OutputString := inttostr(imax) + ' ' + DaughterString;
end;
end;
result := MaxRes;
end;
begin
writeln(Part(100, 3, s):10:1);//first parameter-Enemies count,
//2-Number of rounds,
//3-output for eliminated enemies counter
writeln(s);
readln;
end.
This problem can be solved with a dynamic approach.
F(round,number_of_opponents_remained):
res = 0
opp // number_of_opponents_remained
for i in [1 opp]
res = max(res, opp/100 + F(round-1,opp - i) )
return res
I should say this not the complete solution and you add some details about it, and I am just giving you an idea. You should add some details such as base case and checking if opp>0 and some other details. The complexity of this algorithm is O(100*k).
I need an algorithm to print all possible sums of a number (partitions).
For example: for 5 I want to print:
1+1+1+1+1
1+1+1+2
1+1+3
1+2+2
1+4
2+3
5
I am writing my code in Pascal. So far I have this:
Program Partition;
Var
pole :Array [0..100] of integer;
n :integer;
{functions and procedures}
function Minimum(a, b :integer): integer;
Begin
if (a > b) then Minimum := b
else Minimum := a;
End;
procedure Rozloz(cislo, i :integer);
Var
j, soucet :integer;
Begin
soucet := 0;
if (cislo = 0) then
begin
for j := i - 1 downto 1 do
begin
soucet := soucet + pole[j];
if (soucet <> n) then
Write(pole[j], '+')
else Write(pole[j]);
end;
soucet := 0;
Writeln()
end
else
begin
for j := 1 to Minimum(cislo, pole[i - 1]) do
begin
pole[i] := j;
Rozloz(cislo - j, i + 1);
end;
end;
End;
{functions and procedures}
{Main program}
Begin
Read(n);
pole[0] := 101;
Rozloz(n, 1);
Readln;
End.
It works good but instead of output I want I get this:
1+1+1+1+1
2+1+1+1
2+2+1
3+1+1
3+2
4+1
5
I can't figure out how to print it in right way. Thank you for help
EDIT: changing for j:=i-1 downto 1 to for j:=1 to i-1 solves one problem. But my output is still this: (1+1+1+1+1) (2+1+1+1) (2+2+1) (3+1+1) (3+2) (4+1) (5) but it should be: (1+1+1+1+1) (1+1+1+2) (1+1+3) (1+2+2) (1+4) (2+3) (5) Main problem is with the 5th and the 6th element. They should be in the opposite order.
I won't attempt Pascal, but here is pseudocode for a solution that prints things in the order that you want.
procedure print_partition(partition);
print "("
print partition.join("+")
print ") "
procedure finish_and_print_all_partitions(partition, i, n):
for j in (i..(n/2)):
partition.append(j)
finish_and_print_all_partitions(partition, j, n-j)
partition.pop()
partition.append(n)
print_partition(partition)
partition.pop()
procedure print_all_partitions(n):
finish_and_print_all_partitions([], 1, n)
Before i must say this : Please, excuse me for my bad english...
I'm student.My teacher gave me problem in pascal for my course work...
I must write program that calculates 2^n for big values of n...I've wrote but there is a problem...My program returns 0 for values of n that bigger than 30...My code is below...Please help me:::Thanks beforehand...
function control(a: integer): boolean;
var
b: boolean;
begin
if (a >= 10) then b := true
else b := false;
control := b;
end;
const
n = 200000000;
var
a: array[1..n] of integer;
i, j, c, t, rsayi: longint; k: string;
begin
writeln('2^n');
write('n=');
read(k);
a[1] := 1;
rsayi := 1;
val(k, t, c);
for i := 1 to t do
for j := 1 to t div 2 do
begin
a[j] := a[j] * 2;
end;
for i := 1 to t div 2 do
begin
if control(a[j]) = true then
begin
a[j + 1] := a[j + 1] + (a[j] div 10);
a[j] := a[j] mod 10;
rsayi := rsayi + 1;
end;
end;
for j := rsayi downto 1 do write(a[j]);
end.
The first (nested) loop boils down to "t" multiplications by 2 on every single element of a.
30 multiplications by two is as far as you can go with a 32-bit integer (2^31-1 of positive values, so 2^31 is out of reach)
So the first loop doesn't work, and you probably have to rethink your strategy.
Here is a quick and dirty program to compute all 2^n up to some given, possibly large, n. The program repeatedly doubles the number in array a, which is stored in base 10; with lower digit in a[1]. Notice it's not particularly fast, so it would not be wise to use it for n = 200000000.
program powers;
const
n = 2000; { largest power to compute }
m = 700; { length of array, should be at least log(2)*n }
var
a: array[1 .. m] of integer;
carry, s, p, i, j: integer;
begin
p := 1;
a[1] := 1;
for i := 1 to n do
begin
carry := 0;
for j := 1 to p do
begin
s := 2*a[j] + carry;
if s >= 10 then
begin
carry := 1;
a[j] := s - 10
end
else
begin
carry := 0;
a[j] := s
end
end;
if carry > 0 then
begin
p := p + 1;
a[p] := 1
end;
write(i, ': ');
for j := p downto 1 do
write(a[j]);
writeln
end
end.
I'm looking for an algorithm to generate permutations of a set in such a way that I could make a lazy list of them in Clojure. i.e. I'd like to iterate over a list of permutations where each permutation is not calculated until I request it, and all of the permutations don't have to be stored in memory at once.
Alternatively I'm looking for an algorithm where given a certain set, it will return the "next" permutation of that set, in such a way that repeatedly calling the function on its own output will cycle through all permutations of the original set, in some order (what the order is doesn't matter).
Is there such an algorithm? Most of the permutation-generating algorithms I've seen tend to generate them all at once (usually recursively), which doesn't scale to very large sets. An implementation in Clojure (or another functional language) would be helpful but I can figure it out from pseudocode.
Yes, there is a "next permutation" algorithm, and it's quite simple too. The C++ standard template library (STL) even has a function called next_permutation.
The algorithm actually finds the next permutation -- the lexicographically next one. The idea is this: suppose you are given a sequence, say "32541". What is the next permutation?
If you think about it, you'll see that it is "34125". And your thoughts were probably something this: In "32541",
there is no way to keep the "32" fixed and find a later permutation in the "541" part, because that permutation is already the last one for 5,4, and 1 -- it is sorted in decreasing order.
So you'll have to change the "2" to something bigger -- in fact, to the smallest number bigger than it in the "541" part, namely 4.
Now, once you've decided that the permutation will start as "34", the rest of the numbers should be in increasing order, so the answer is "34125".
The algorithm is to implement precisely that line of reasoning:
Find the longest "tail" that is ordered in decreasing order. (The "541" part.)
Change the number just before the tail (the "2") to the smallest number bigger than it in the tail (the 4).
Sort the tail in increasing order.
You can do (1.) efficiently by starting at the end and going backwards as long as the previous element is not smaller than the current element. You can do (2.) by just swapping the "4" with the '2", so you'll have "34521". Once you do this, you can avoid using a sorting algorithm for (3.), because the tail was, and is still (think about this), sorted in decreasing order, so it only needs to be reversed.
The C++ code does precisely this (look at the source in /usr/include/c++/4.0.0/bits/stl_algo.h on your system, or see this article); it should be simple to translate it to your language: [Read "BidirectionalIterator" as "pointer", if you're unfamiliar with C++ iterators. The code returns false if there is no next permutation, i.e. we are already in decreasing order.]
template <class BidirectionalIterator>
bool next_permutation(BidirectionalIterator first,
BidirectionalIterator last) {
if (first == last) return false;
BidirectionalIterator i = first;
++i;
if (i == last) return false;
i = last;
--i;
for(;;) {
BidirectionalIterator ii = i--;
if (*i <*ii) {
BidirectionalIterator j = last;
while (!(*i <*--j));
iter_swap(i, j);
reverse(ii, last);
return true;
}
if (i == first) {
reverse(first, last);
return false;
}
}
}
It might seem that it can take O(n) time per permutation, but if you think about it more carefully, you can prove that it takes O(n!) time for all permutations in total, so only O(1) -- constant time -- per permutation.
The good thing is that the algorithm works even when you have a sequence with repeated elements: with, say, "232254421", it would find the tail as "54421", swap the "2" and "4" (so "232454221"), reverse the rest, giving "232412245", which is the next permutation.
Assuming that we're talking about lexicographic order over the values being permuted, there are two general approaches that you can use:
transform one permutation of the elements to the next permutation (as ShreevatsaR posted), or
directly compute the nth permutation, while counting n from 0 upward.
For those (like me ;-) who don't speak c++ as natives, approach 1 can be implemented from the following pseudo-code, assuming zero-based indexing of an array with index zero on the "left" (substituting some other structure, such as a list, is "left as an exercise" ;-):
1. scan the array from right-to-left (indices descending from N-1 to 0)
1.1. if the current element is less than its right-hand neighbor,
call the current element the pivot,
and stop scanning
1.2. if the left end is reached without finding a pivot,
reverse the array and return
(the permutation was the lexicographically last, so its time to start over)
2. scan the array from right-to-left again,
to find the rightmost element larger than the pivot
(call that one the successor)
3. swap the pivot and the successor
4. reverse the portion of the array to the right of where the pivot was found
5. return
Here's an example starting with a current permutation of CADB:
1. scanning from the right finds A as the pivot in position 1
2. scanning again finds B as the successor in position 3
3. swapping pivot and successor gives CBDA
4. reversing everything following position 1 (i.e. positions 2..3) gives CBAD
5. CBAD is the next permutation after CADB
For the second approach (direct computation of the nth permutation), remember that there are N! permutations of N elements. Therefore, if you are permuting N elements, the first (N-1)! permutations must begin with the smallest element, the next (N-1)! permutations must begin with the second smallest, and so on. This leads to the following recursive approach (again in pseudo-code, numbering the permutations and positions from 0):
To find permutation x of array A, where A has N elements:
0. if A has one element, return it
1. set p to ( x / (N-1)! ) mod N
2. the desired permutation will be A[p] followed by
permutation ( x mod (N-1)! )
of the elements remaining in A after position p is removed
So, for example, the 13th permutation of ABCD is found as follows:
perm 13 of ABCD: {p = (13 / 3!) mod 4 = (13 / 6) mod 4 = 2; ABCD[2] = C}
C followed by perm 1 of ABD {because 13 mod 3! = 13 mod 6 = 1}
perm 1 of ABD: {p = (1 / 2!) mod 3 = (1 / 2) mod 2 = 0; ABD[0] = A}
A followed by perm 1 of BD {because 1 mod 2! = 1 mod 2 = 1}
perm 1 of BD: {p = (1 / 1!) mod 2 = (1 / 1) mod 2 = 1; BD[1] = D}
D followed by perm 0 of B {because 1 mod 1! = 1 mod 1 = 0}
B (because there's only one element)
DB
ADB
CADB
Incidentally, the "removal" of elements can be represented by a parallel array of booleans which indicates which elements are still available, so it is not necessary to create a new array on each recursive call.
So, to iterate across the permutations of ABCD, just count from 0 to 23 (4!-1) and directly compute the corresponding permutation.
You should check the Permutations article on wikipeda. Also, there is the concept of Factoradic numbers.
Anyway, the mathematical problem is quite hard.
In C# you can use an iterator, and stop the permutation algorithm using yield. The problem with this is that you cannot go back and forth, or use an index.
More examples of permutation algorithms to generate them.
Source: http://www.ddj.com/architect/201200326
Uses the Fike's Algorithm, that is the one of fastest known.
Uses the Algo to the Lexographic order.
Uses the nonlexographic, but runs faster than item 2.
1.
PROGRAM TestFikePerm;
CONST marksize = 5;
VAR
marks : ARRAY [1..marksize] OF INTEGER;
ii : INTEGER;
permcount : INTEGER;
PROCEDURE WriteArray;
VAR i : INTEGER;
BEGIN
FOR i := 1 TO marksize
DO Write ;
WriteLn;
permcount := permcount + 1;
END;
PROCEDURE FikePerm ;
{Outputs permutations in nonlexicographic order. This is Fike.s algorithm}
{ with tuning by J.S. Rohl. The array marks[1..marksizn] is global. The }
{ procedure WriteArray is global and displays the results. This must be}
{ evoked with FikePerm(2) in the calling procedure.}
VAR
dn, dk, temp : INTEGER;
BEGIN
IF
THEN BEGIN { swap the pair }
WriteArray;
temp :=marks[marksize];
FOR dn := DOWNTO 1
DO BEGIN
marks[marksize] := marks[dn];
marks [dn] := temp;
WriteArray;
marks[dn] := marks[marksize]
END;
marks[marksize] := temp;
END {of bottom level sequence }
ELSE BEGIN
FikePerm;
temp := marks[k];
FOR dk := DOWNTO 1
DO BEGIN
marks[k] := marks[dk];
marks[dk][ := temp;
FikePerm;
marks[dk] := marks[k];
END; { of loop on dk }
marks[k] := temp;l
END { of sequence for other levels }
END; { of FikePerm procedure }
BEGIN { Main }
FOR ii := 1 TO marksize
DO marks[ii] := ii;
permcount := 0;
WriteLn ;
WrieLn;
FikePerm ; { It always starts with 2 }
WriteLn ;
ReadLn;
END.
2.
PROGRAM TestLexPerms;
CONST marksize = 5;
VAR
marks : ARRAY [1..marksize] OF INTEGER;
ii : INTEGER;
permcount : INTEGER;
PROCEDURE WriteArray;
VAR i : INTEGER;
BEGIN
FOR i := 1 TO marksize
DO Write ;
permcount := permcount + 1;
WriteLn;
END;
PROCEDURE LexPerm ;
{ Outputs permutations in lexicographic order. The array marks is global }
{ and has n or fewer marks. The procedure WriteArray () is global and }
{ displays the results. }
VAR
work : INTEGER:
mp, hlen, i : INTEGER;
BEGIN
IF
THEN BEGIN { Swap the pair }
work := marks[1];
marks[1] := marks[2];
marks[2] := work;
WriteArray ;
END
ELSE BEGIN
FOR mp := DOWNTO 1
DO BEGIN
LexPerm<>;
hlen := DIV 2;
FOR i := 1 TO hlen
DO BEGIN { Another swap }
work := marks[i];
marks[i] := marks[n - i];
marks[n - i] := work
END;
work := marks[n]; { More swapping }
marks[n[ := marks[mp];
marks[mp] := work;
WriteArray;
END;
LexPerm<>
END;
END;
BEGIN { Main }
FOR ii := 1 TO marksize
DO marks[ii] := ii;
permcount := 1; { The starting position is permutation }
WriteLn < Starting position: >;
WriteLn
LexPerm ;
WriteLn < PermCount is , permcount>;
ReadLn;
END.
3.
PROGRAM TestAllPerms;
CONST marksize = 5;
VAR
marks : ARRAY [1..marksize] of INTEGER;
ii : INTEGER;
permcount : INTEGER;
PROCEDURE WriteArray;
VAR i : INTEGER;
BEGIN
FOR i := 1 TO marksize
DO Write ;
WriteLn;
permcount := permcount + 1;
END;
PROCEDURE AllPerm (n : INTEGER);
{ Outputs permutations in nonlexicographic order. The array marks is }
{ global and has n or few marks. The procedure WriteArray is global and }
{ displays the results. }
VAR
work : INTEGER;
mp, swaptemp : INTEGER;
BEGIN
IF
THEN BEGIN { Swap the pair }
work := marks[1];
marks[1] := marks[2];
marks[2] := work;
WriteArray;
END
ELSE BEGIN
FOR mp := DOWNTO 1
DO BEGIN
ALLPerm<< n - 1>>;
IF >
THEN swaptemp := 1
ELSE swaptemp := mp;
work := marks[n];
marks[n] := marks[swaptemp};
marks[swaptemp} := work;
WriteArray;
AllPerm< n-1 >;
END;
END;
BEGIN { Main }
FOR ii := 1 TO marksize
DO marks[ii] := ii
permcount :=1;
WriteLn < Starting position; >;
WriteLn;
Allperm < marksize>;
WriteLn < Perm count is , permcount>;
ReadLn;
END.
the permutations function in clojure.contrib.lazy_seqs already claims to do just this.
It looks necromantic in 2022 but I'm sharing it anyway
Here an implementation of C++ next_permutation in Java can be found. The idea of using it in Clojure might be something like
(println (lazy-seq (iterator-seq (NextPermutationIterator. (list 'a 'b 'c)))))
disclaimer: I'm the author and maintainer of the project