Delphi Radix Sort supporting negative integers - sorting

Lately I used the great sorting algorythm made by Rebuilder from Habr.com. It served me well, but it sorts only positive integers, and recently I run into the need to sort negatives as well. For now I use QuickSort, but since the array is very large (10k+ elements), I wonder if one could modify RadixSort for this task.
There is the procedure as for now. Comment translation is mine, sorry if I get something wrong.
procedure RSort(var m: array of Longword);
//--------------------------------------------------
procedure Sort_step(var source, dest, offset: array of Longword; const num: Byte);
var i,temp : Longword;
k : Byte;
begin
for i := High(source) downto 0 do
begin
temp := source[i];
k := temp SHR num;
dec(offset[k]);
dest[offset[k]] := temp;
end;
end;
//--------------------------------------------------
// Объявляем массив корзин первым, для выравнивания на стеке
// Creating the bin array first for aligning at the stack
var s : array[0..3] of array[0..255] of Longword;
i,k : longword;
// Смещение байт внутри переменной k
// Byte offset inside of variable k
offset : array[0..3] of byte absolute k;
m_temp : array of Longword;
begin
SetLength(m_temp, Length(m));
// Быстрая очистка корзин
// Quick bin clear
FillChar(s[0], 256 * 4 * SizeOf(Longword), 0);
// Заполнение корзин
// Filling bins
for i := 0 to High(m) do
begin
k := m[i];
Inc(s[0,offset[0]]);
Inc(s[1,offset[1]]);
Inc(s[2,offset[2]]);
Inc(s[3,offset[3]]);
end;
// Пересчёт смещений для корзин
// Recalculating bin offsets
for i := 1 to 255 do
begin
Inc(s[0,i], s[0,i-1]);
Inc(s[1,i], s[1,i-1]);
Inc(s[2,i], s[2,i-1]);
Inc(s[3,i], s[3,i-1]);
end;
// Вызов сортировки по байтам от младших к старшим
// Sorting by byte, from least to most
Sort_step(m, m_temp, s[0], 0);
Sort_step(m_temp, m, s[1], 8);
Sort_step(m, m_temp, s[2], 16);
Sort_step(m_temp, m, s[3], 24);
SetLength(m_temp, 0);
end;
Link: https://habr.com/ru/post/484224/
I found some helpful advice on the Internet, including StackOverflow, but I met two problems:
There are too many different solutions and I can't choose the optimal one for Delphi.
I lack the knowledge and skill to implement them correctly. I've tried some and get wrong results.
So, could someone modify the given function and explain to me what they did and why?

A simple approach is to complement the sign bit somewhere during the process. Note this only affects the most significant "digit" (usually a byte for fast radix sort). The code to handle the most significant "digit" could be handled in a separate loop than the code that handles the other "digits".
The simplest approach would be to make an initial pass to complement the sign bit of every element in the array, do the radix sort, then a final pass complement the sign bit of every element again.

Related

Dynamic array in Turbo Pascal

I am working on my school project and I would like to use Dynamic (not static) array. I worked with ObjectPascal, so I am used to some syntax. But now I am programming in the old TurboPascal (I am using Turbo Pascal 7 for Windows).
It doesn't seem to know the ObjectPascal, so I thought, that you Turbo Pascal doesn't know dynamic arrays.
Could anyone tell me, if my theory is right or not? I tried to google, but I was not succesfull.
Basicly I am asking "how is it with dynamic arrays in Turbo Pascal 7" ?
Thank you for all reactions.
As MartynA says, there is no dynamic array type in Turbo Pascal. You need to manually allocate memory using pointers, and be careful if you use rangechecks.
Typically you define an array type
TYPE
TArrayT = array[0.. ((65535-spillbytes) div sizeof(T))-1] of T;
where spillbytes is a constant for a small deduction because you can't use the whole 64k, see what the compiler accepts. (Probably this deduction is for heapmanager structures inside the 64k block)
Then you define a pointer
PArrayT= ^TArrayT;
and a variable to it
var
P : PArrayT;
and you allocate nrelement elements using getmem;
getmem(P,SizeOf(T) * nrelements);
and optionally fill them with zero to initialize them:
fillchar(p^,SizeOf(T) * nrelements,#0);
You can access elements using
p^[index]
to free them, use freemem using the exact opposite of the getmem line.
freemem(P,Sizeof(T)*nrelements);
Which means you have to save the allocated number of elements somewhere. This was fixed/solved in Delphi and FPC.
Also keep in mind that you can't find bugs with rangechecking anymore.
If you want arrays larger than 64k, that is possible, but only with constraints, and it matters more which exact TP target (dos, dos-protected or Windows you use) I advise you to search for the online SWAG archive that has many examples. And of course I would recommend to go to FreePascal/Lazarus too where you can simply do:
var x : array of t;
begin
setlength(x,1000000);
and be done with it without additional lines and forget about all of this nonsense.
I'm using Turbo Pascal 5.5 and to create a dynamic array, perhaps the trick is to declare an array with zero dimension as follows:
dArray = array [0..0] of integer;
And then declare a pointer to that array:
pArray = ^dArray ;
And finally, create a pointer variable:
ArrayPtr : pArray;
You can now reference the pointer variable ArrayPtr as follows:
ArrayPtr^[i]; { The index 'i' is of type integer}
See the complete example below:
{
Title: dynarr.pas
A simple Pascal program demonstrating dynamic array.
Compiled and tested with Turbo Pascal 5.5.
}
program dynamic_array;
{Main Program starts here}
type
dArray = array [0..0] of integer;
pArray = ^dArray ;
var
i : integer;
ArrayPtr : pArray;
begin
for i := 0 to 9 do { In this case, array index starts at 0 instead of 1. }
ArrayPtr^[i] := i + 1;
writeln('The Dynamic Array now contains the following:');
writeln;
for i := 0 to 9 do
writeln(ArrayPtr^[i]);
end.
In this example, we have declared the array as:
array[0..0] of integer;
Therefore, the index starts at 0 and if we have n elements, the last element is at index n-1 which is similar to array indexing in C/C++.
Regular Pascal arrays start at 1 but for this case, it starts at 0.
unit Vector;
interface
const MaxVector = 8000;
// 64 k div SizeOf(float); number of float-values that fit in 64 K of stack
VectorError: boolean = False;
// toggle if error occurs. Calling routine can handle or abort
type
VectorStruc = record
Length: word;
Data: array [1..MaxVector] of float;
end;
VectorTyp = ^VectorStruc;
procedure CreateVector(var Vec: VectorTyp; Length: word; Value: float);
{ Generates a vector of length Length and sets all elements to Value }
procedure DestroyVector(var Vec: VectorTyp);
{ release memory occupied by vector }
procedure SetVectorElement(var Vec: VectorTyp; n: word; c: float);
function GetVectorElement(const Vec: VectorTyp; n: word): float;
implementation
var ch: char;
function WriteErrorMessage(Text: string): char;
begin
Write(Text);
Read(WriteErrorMessage);
VectorError := True; // toggle the error marker
end;
procedure CreateVector(var Vec: VectorTyp; Length: word; Value: float);
var
i: word;
begin
try
GetMem(Vec, Length * SizeOf(float) + SizeOf(word) + 6);
except
ch := WriteErrorMessage(' Not enough memory to create vector');
exit;
end;
Vec^.Length := Length;
for i := 1 to Length do
Vec^.Data[i] := Value;
end;
procedure DestroyVector(var Vec: VectorTyp);
var
x: word;
begin
x := Vec^.Length * SizeOf(float) + SizeOf(word) + 6;
FreeMem(Vec, x);
end;
function VectorLength(const Vec: VectorTyp): word;
begin
VectorLength := Vec^.Length;
end;
function GetVectorElement(const Vec: VectorTyp; n: word): float;
var
s1, s2: string;
begin
if (n <= VectorLength(Vec)) then
GetVectorElement := Vec^.Data[n]
else
begin
Str(n: 4, s1);
Str(VectorLength(Vec): 4, s2);
ch := WriteErrorMessage(' Attempt to read non-existent vector element No ' +
s1 + ' of ' + s2);
end;
end;
procedure SetVectorElement(var Vec: VectorTyp; n: word; C: float);
begin
if (n <= VectorLength(Vec)) then
Vec^.Data[n] := C
else
ch := WriteErrorMessage(' Attempt to write to non-existent vector element');
end;
end.
As long as your data fit on the stack, i.e., are smaller than 64 kB, the task is relatively simple. The only thing I don't know is where the 6 bit of extra size go, they are required, however.

Change array type: 8bit type to 6bit type

I have two types and two arrays of that types in file.ads
type Ebit is mod 2**8;
type Sbit is mod 2**6;
type Data_Type is array (Positive range <>) of Ebit;
type Changed_Data_Type is array (Positive range <>) of Sbit;
and function:
function ChangeDataType (D : in Data_Type) return Changed_Data_Type
with
Pre => D'Length rem 3 = 0 and D'Last < Positive'Last / 4,
Post => ChangeDataType'Result'Length = 4 * (D'Length / 3)
Ok i can understand all of this.
For example we have arrays of:
65, 66, 65, 65, 66, 65 in 8bit values function should give to us 16, 20, 9, 1, 16, 20, 9, 1 in 6bit values.
I dont know how i can build a 6bit table from 8 bit table.
My idea of sollutions is for example taking bit by bit from type:
fill all bites in 6bit type to 0 (propably default)
if first bit (2**1) is 1 set bit (2**1) in 6bit type to 1;
and do some iterations
But i dont know how to do this, always is a problem with types. Is this good idea or i can do this with easier way? I spend last nigt to try write this but without success.
Edit:
I wrote some code, its working but i have problem with array initialization.
function ChangeDataType (D: in Data_Type) return Changed_Data_Type
is
length: Natural := (4*(D'Length / 3));
ER: Changed_Data_type(length);
Temp: Ebit;
Temp1: Ebit;
Temp2: Ebit;
Actual: Ebit;
n: Natural;
k: Natural;
begin
n := 0;
k := 0;
Temp := 2#00000000#;
Temp1 := 2#00000000#;
Temp2 := 2#00000000#;
Array_loop:
for k in D'Range loop
case n is
when 0 =>
Actual := D(k);
Temp1 := Actual / 2**2;
ER(k) := Sbit(Temp1);
Temp := Actual * ( 2**4);
n := 2;
when 2 =>
Actual := D(k);
Temp1 := Actual / 2**4;
Temp2 := Temp1 or Temp;
ER(k) := Sbit(Temp2);
Temp := Actual * ( 2**2);
n := 4;
when 4 =>
Actual := D(k);
Temp1 := Actual / 2**6;
Temp2 := Temp1 or Temp;
ER(k) := Sbit(Temp2);
n := 6;
when 6 =>
Temp1 := Actual * ( 2**2);
Temp2 := Actual / 2**2;
ER(k) := Sbit(Temp2);
n := 0;
when others =>
n := 0;
end case;
end loop Array_Loop;
return ER;
end;
IF I understand what you're asking... it's that you want to re-pack the same 8-bit data into 6-bit values such that the "leftover" bits of the first EBit become the first bits (highest or lowest?) of the second Sbit.
One way you can do this - at least for fixed size arrays, e.g. your 6 words * 8 bits, 8 words * 6 bits example, is by specifying the exact layout in memory for each array type, using packing, and representation aspects (or pragmas, before Ada-2012) which are nicely described here.
I haven't tested the following, but it may serve as a starting point.
type Ebit is mod 2**8;
type Sbit is mod 2**6;
for Ebit'Size use 8;
for Sbit'Size use 6;
type Data_Type is array (1 .. 6) of Ebit
with Alignment => 0; -- this should pack tightly
type Changed_Data_Type is array (1 .. 8) of Sbit
with Alignment => 0;
Then you can instantiate the generic Unchecked_Conversion function with the two array types, and use that function to convert from one array to the other.
with Ada.Unchecked_Conversion;
function Change_Type is new Ada.Unchecked_Conversion(Data_Type, Changed_Data_Type);
declare
Packed_Bytes : Changed_Data_Type := Change_Type(Original_Bytes);
begin ...
In terms of code generated, it's not slow, because Unchecked_Conversion doesn't do anything, except tell the compile-time type checking to look the other way.
I view Unchecked_Conversion like the "I meant to do that" look my cat gives me after falling off the windowledge. Again...
Alternatively, if you wish to avoid copying, you can declare Original_Bytes as aliased, and use a similar trick with access types and Unchecked_Access to overlay both arrays on the same memory (like a Union in C). I think this is what DarkestKhan calls "array overlays" in a comment below. See also section 3 of this rather dated page which describes the technique further. It notes the overlaid variable must not only be declared aliased but also volatile so that accesses to one view aren't optimised into registers, but reflect any changes made via the other view. Another approach to overlays is in the Ada Wikibook here.
Now this may be vulnerable to endian-ness considerations, i.e. it may work on some platforms but not others. The second reference above gives an example of a record with exact bit-alignment of its members : we can at least take the Bit_Order aspect, as in with Alignment => 0, Bit_Order => Low_Order_First; for the arrays above...
-- code stolen from "Rationale" ... see link above p.11
type RR is record
Code: Opcode;
R1: Register;
R2: Register;
end record
with Alignment => 2, Bit_Order => High_Order_First;
for RR use record
Code at 0 range 0 .. 7;
R1 at 1 range 0 .. 3;
R2 at 1 range 4 .. 7;
end record;
One thing that's not clear to me is if there's a formulaic way to specify the exact layout of each element in an array, as is done in a record here - or even if there's a potential need to. If necessary, one workaround would be to replace the arrays above with records. But I'd love to see a better answer if there is one.

TPascal, Stack Overflow, Recursion

My program should inverse a string (ex. for Hello world returns dlrow olleH) and it works only for strings smaller than 20 characters. For 20 or more i get "Error 202 Stack overflow". Thank you :)
Program Inv;
var S, A: String;
n: integer;
Function I(X: String; z: integer):String;
begin
if z=1 then I:=X[z] else
I:=X[z]+I(X, z-1);
end;
begin
write ('Enter your text: ');
readln (S);
n:=length(S);
A:=I(S, n);
writeln (A);
readln;
end.
Unless you are required to show a recursive solution, you're usually better to sticking with iteration(a). Recursion uses an often-limited resource (the stack) to weave its magic and is often the cause of crashes when you exceed that limit.
Iterative solutions tend to be far less restrictive, such as the code below:
program PaxCode;
Function reverse(inp_str: string) : string;
var out_str : string = '';
var idx : integer = 1;
begin
while (idx <= length(inp_str)) do
begin
out_str := inp_str[idx] + out_str;
idx := idx + 1
end;
reverse := out_str
end;
var test_str: string = 'My hovercraft is full of eels and they will not let me drive it';
begin
writeln(test_str);
writeln(reverse(test_str))
end.
As you can see, the output is correct, and not limited to twenty (or twenty-nine) characters:
My hovercraft is full of eels and they will not let me drive it
ti evird em tel ton lliw yeht dna slee fo lluf si tfarcrevoh yM
(a) The best areas for recursion are those where each level removes a sizable proportion of the solution space. For example, binary searches remove fully 50% of the remaining solution space on every level so you could search through a structure holding four billion entries with just thirty-two levels, since 232 is a touch above 4.2 billion.
Something like reversing a 400-character string will take, ..., let me think, oh yes, 400 levels. That won't necessarily end well :-)

Most repeated character in a row

My task is to show most used letter in a row. For example if you put in aabbbbccbbb most repeated character is B and it is used 4 times. There was a very similar topic about the same task, but i didnt understand the code. Most repeating character in a string
Program Task;
var s:string;
i,k,g,count:integer;
c:char;
begin
Readln(s);
g:=0;
while Length(s) > 0 do
begin
c := s[1];
i:=1;
while i<= Length(s) do
begin
If (c=s[i]) then
delete(s,i,1)
else
Inc(i);
If (c=s[i]) then
Inc(g);
end;
end;
Writeln(g);
Readln;
end.
There are many problems i face. First is i dont know how to show which character is most used and second is i dont know how compare which one of repeating characters is most used.
For example if i write aaaabbbc it will give me answer of 7 because there is 4xa and 3xb.
All the help is most appreciated.
If it's just about english characters, you might just allocate an array to keep a count per character. In that case, the code could look like this.
I wrote this using Delphi. I hope it works as well in your flavour of Pascal.
program Task;
{$APPTYPE CONSOLE} // For Delphi
var
s: string[50];
i: Integer;
Counters: array[Char] of Integer;
Highest: Char;
begin
// Initialize counters.
for i := 0 to 255 do
Counters[Char(i)] := 0;
s := 'aabbbbccbbb';
// Count the characters.
for i := 1 to Length(s) do
Inc(Counters[s[i]]);
// Find out which one is highest.
Highest := #0;
for i := 0 to 255 do
if Counters[Char(i)] > Counters[Highest] then
Highest := Char(i);
// Output that character and its count.
WriteLn('The highest character is ', Highest, ' with ', Counters[Highest], ' occurrences.');
ReadLn;
end.
In less academic setups, using an array like this might not be the most efficient, because it contains a counter for every possible character, including those that don't occur in the string at all. That means, if you want to use this exact code for every possible character in the unicode table, your array would be a couple of megabytes large (still not really a problem on modern computers, but still).
You can improve this code by using a kind of dictionary or list to keep track of the items, so you need only to add those items you find, but if you have to write that yourself, it will make your program quite a bit larger.
EDIT:
As per request in comment: Counting the longest subsequent range of characters:
program Task;
{$APPTYPE CONSOLE} // For Delphi
var
s: String;
i: Integer;
Longest: Integer;
Current: Integer;
LongestChar: Char;
begin
s := 'aabbbbccbbb';
Longest := 0;
Current := 0;
// Count the characters.
for i := 1 to Length(s) do
begin
Inc(Current);
// If it's the last char or the next char is going to be different, restart the counting.
if (i = Length(s)) or (s[i] <> s[i+1]) then
begin
if Current > Longest then
begin
Longest := Current;
LongestChar := s[i];
end;
Current := 0;
end;
end;
// Output that character and its count.
WriteLn('The highest character is ', LongestChar, ' with ', Longest, ' occurrences.');
ReadLn;
end.
Current > Longest makes sure the first longest sequence is returned in case multiple character sequences have the same length. Change to Current >= Longest if you want the last sequence instead.

Generating permutations lazily

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

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