How to resolve the algorithm Pairs with common factors step by step in the Pascal programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Pairs with common factors step by step in the Pascal programming language
Table of Contents
Problem Statement
Generate the sequence where each term n is the count of the pairs (x,y) with 1 < x < y <= n, that have at least one common prime factor.
For instance, when n = 9, examine the pairs: Find all of the pairs that have at least one common prime factor: and count them:
Terms may be found directly using the formula: where 𝚽() is Phi; the Euler totient function.
For the term a(p), if p is prime, then a(p) is equal to the previous term.
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Pairs with common factors step by step in the Pascal programming language
Source code in the pascal programming language
program PairsWithCommonFactors;
{$IFdef FPC} {$MODE DELPHI} {$Optimization ON,ALL}{$IFEND}
{$IFdef Windows} {$APPTYPE CONSOLE}{$IFEND}
const
cLimit = 1000*1000*1000;
//global
type
TElem= Uint64;
tpElem = pUint64;
myString = String[31];
var
TotientList : array of TElem;
Sieve : Array of byte;
function Numb2USA(n:Uint64):myString;
const
//extend s by the count of comma to be inserted
deltaLength : array[0..24] of byte =
(0,0,0,0,1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,6,6,6,7,7,7);
var
pI :pChar;
i,j : NativeInt;
Begin
str(n,result);
i := length(result);
//extend s by the count of comma to be inserted
// j := i+ (i-1) div 3;
j := i+deltaLength[i];
if i<> j then
Begin
setlength(result,j);
pI := @result[1];
dec(pI);
while i > 3 do
Begin
//copy 3 digits
pI[j] := pI[i];
pI[j-1] := pI[i-1];
pI[j-2] := pI[i-2];
// insert comma
pI[j-3] := ',';
dec(i,3);
dec(j,4);
end;
end;
end;
procedure SieveInit(svLimit:NativeUint);
var
pSieve:pByte;
i,j,pr :NativeUint;
Begin
svlimit := (svLimit+1) DIV 2;
setlength(sieve,svlimit+1);
pSieve := @Sieve[0];
For i := 1 to svlimit do
Begin
IF pSieve[i]= 0 then
Begin
pr := 2*i+1;
j := (sqr(pr)-1) DIV 2;
IF j> svlimit then
BREAK;
repeat
pSieve[j]:= 1;
inc(j,pr);
until j> svlimit;
end;
end;
pr := 0;
j := 0;
For i := 1 to svlimit do
Begin
IF pSieve[i]= 0 then
Begin
pSieve[j] := i-pr;
inc(j);
pr := i;
end;
end;
setlength(sieve,j);
end;
procedure TotientInit(len: NativeUint);
var
pTotLst : tpElem;
pSieve : pByte;
i: NativeInt;
p,j,k,svLimit : NativeUint;
Begin
SieveInit(len);
setlength(TotientList,len+12);
pTotLst := @TotientList[0];
//Fill totient with simple start values for odd and even numbers
//and multiples of 3
j := 1;
k := 1;// k == j DIV 2
p := 1;// p == j div 3;
repeat
pTotLst[j] := j;//1
pTotLst[j+1] := k;//2 j DIV 2; //2
inc(k);
inc(j,2);
pTotLst[j] := j-p;//3
inc(p);
pTotLst[j+1] := k;//4 j div 2
inc(k);
inc(j,2);
pTotLst[j] := j;//5
pTotLst[j+1] := p;//6 j DIV 3 <= (div 2) * 2 DIV/3
inc(j,2);
inc(p);
inc(k);
until j>len+6;
//correct values of totient by prime factors
svLimit := High(sieve);
p := 3;// starting after 3
pSieve := @Sieve[svLimit+1];
i := -svlimit;
repeat
p := p+2*pSieve[i];
j := p;
while j <= cLimit do
Begin
k:= pTotLst[j];
pTotLst[j]:= k-(k DIV p);
inc(j,p);
end;
inc(i);
until i=0;
//primes not needed anymore
setlength(sieve,0);
end;
procedure CountOfPairs(len: NativeUint);
var
pTotLst : tpElem;
i,a_n,sum,Totient: tElem;
Begin
pTotLst := @TotientList[0];
sum := 1;
a_n := 2; // sums to i*(i-1)/2 +1
For i := 2 to len do
Begin
Totient := pTotLst[i];// relict for print data
sum += Totient;
pTotLst[i] := a_n-sum;
a_n += i;
end;
TotientList[1] := 0;
end;
var
i,k : NativeUint;
Begin
TotientInit(climit);
CountOfPairs(climit);
i := 1;
Repeat
For k := 9 downto 0 do
begin
write(TotientList[i]:6);
inc(i);
end;
writeln;
until i>99;
writeln;
writeln('Some values #');
i := 10;
repeat
writeln(Numb2USA(i):13,Numb2USA(TotientList[i]):25);
i *= 10;
until i > climit;
end.
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