Prime triangle: Difference between revisions
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You will require a function f which when given an integer S will return a list of the arrangements of the integers 1 to S such that g<sub>1</sub>=1 g<sub>S</sub>=S and generally for n=1 to n=S-1 g<sub>n</sub>+g<sub>n+1</sub> is prime. S=1 is undefined. For S=2 to S=20 print f(S) to form a triangle. Then again for S=2 to S=20 print the number of possible arrangements of 1 to S meeting these requirements.
;See also
:* [[oeis:A036440|OEIS:A036440]]
=={{header|ALGOL 68}}==
Line 7 ⟶ 12:
{{works with|ALGOL 68G|Any - tested with release 2.8.3.win32}}
As Algol 68G under Windows is fully interpreted, a reduced number of rows is produced.
<
INT max number = 18; # largest number we will consider #
# construct a primesieve and from that a table of pairs of numbers whose sum is prime #
Line 21 ⟶ 25:
FI
OD;
# returns the number of possible arrangements of the integers for a row in the prime triangle #
PROC count arrangements = ( INT n
IF n < 2 THEN # no solutions for n < 2 # 0
ELIF n < 4 THEN
# for 2 and 3. there is only 1 solution: 1, 2 and 1, 2, 3 #
1
ELSE
# 4 or more - must find the solutions #
BOOL print solution := TRUE;
[ 0 : n ]BOOL used;
[ 0 : n ]INT number;
# the triangle row must have 1 in the leftmost and n in the rightmost elements #
# the numbers must alternate between even and odd in order for the sum to be prime #
FOR i FROM 0 TO n DO
used[
number[
OD;
used[ 1 ] := TRUE;
number[
# find the intervening numbers and count the solutions #
INT count := 0;
INT p := 2;
WHILE p
INT
INT
INT next := current + 2;
WHILE IF next >= n THEN FALSE ELSE NOT prime[ p1 + next ] OR used[ next ] FI DO
next +:= 2
OD;
IF next >= n THEN next := 0 FI;
IF p = n - 1 THEN
# we are at the final number before n #
IF next /= 0
# found a solution #
count +:= 1;
IF print solution THEN
FOR i TO n - 2 DO
print( ( whole( number[ i ], -3 ) ) )
OD;
print( ( whole( next, -3 ), whole( n, - 3 ), newline ) );
print solution := FALSE
FI
FI;
next := 0
FI;
# backtrack for more solutions #
p -:= 1
# here will be a further backtrack as next is 0 ( there could only be one possible number at p - 1 ) #
FI;
IF next /= 0 THEN
# have a/another number that can appear at p #
used[
ELIF p <= 2 THEN
# no more solutions #
p :=
ELSE
# can't find a number for this position, backtrack #
used[
p -:= 1
FI
Line 112 ⟶ 99:
[ 2 : max number ]INT arrangements;
FOR n FROM LWB arrangements TO UPB arrangements DO
arrangements[ n ] := count arrangements( n
OD;
FOR n FROM LWB arrangements TO UPB arrangements DO
Line 118 ⟶ 105:
OD;
print( ( newline ) )
END</
{{out}}
<pre>
Line 139 ⟶ 126:
1 2 3 4 7 6 5 8 9 10 13 16 15 14 17 12 11 18
1 1 1 1 1 2 4 7 24 80 216 648 1304 3392 13808 59448 155464
</pre>
=={{header|BASIC}}==
==={{header|FreeBASIC}}===
{{trans|Visual Basic .NET}}
<syntaxhighlight lang="vbnet">Dim Shared As Uinteger maxNumber = 20 ' Largest number we will consider.
Dim Shared As Uinteger prime(2 * maxNumber) ' prime sieve.
Function countArrangements(Byval n As Uinteger) As Uinteger
Dim As Uinteger i
If n < 2 Then ' No solutions for n < 2.
Return 0
Elseif n < 4 Then
' For 2 and 3. there is only 1 solution: 1, 2 and 1, 2, 3.
For i = 1 To n
Print Using "###"; i;
Next i
Print
Return 1
Else
' 4 or more - must find the solutions.
Dim As Boolean printSolution = True
Dim As Boolean used(n)
Dim As Uinteger number(n)
' The triangle row must have 1 in the leftmost and n in the rightmost elements.
' The numbers must alternate between even and odd in order for the sums to be prime.
For i = 0 To n - 1
number(i) = i Mod 2
Next i
used(1) = True
number(n) = n
used(n) = True
' Find the intervening numbers and count the solutions.
Dim As Uinteger count = 0
Dim As Uinteger p = 2
Do While p > 0
Dim As Uinteger p1 = number(p - 1)
Dim As Uinteger current = number(p)
Dim As Uinteger sgte = current + 2
Do While sgte < n Andalso (Not prime(p1 + sgte) Or used(sgte))
sgte += 2
Loop
If sgte >= n Then
sgte = 0
End If
If p = n - 1 Then
' We are at the final number before n.
' It must be the final even/odd number preceded by the final odd/even number.
If sgte <> 0 Then
' Possible solution.
If prime(sgte + n) Then
' Found a solution.
count += 1
If printSolution Then
For i = 1 To n - 2
Print Using "###"; number(i);
Next i
Print Using "###"; sgte; n
printSolution = False
End If
End If
sgte = 0
End If
' Backtrack for more solutions.
p -= 1
' There will be a further backtrack as next is 0 ( there could only be one possible number at p - 1 ).
End If
If sgte <> 0 Then
' have a/another number that can appear at p.
used(current) = False
used(sgte) = True
number(p) = sgte
' Haven't found all the intervening digits yet.
p += 1
Elseif p <= 2 Then
' No more solutions.
p = 0
Else
' Can't find a number for this position, backtrack.
used(number(p)) = False
number(p) = p Mod 2
p -= 1
End If
Loop
Return count
End If
End Function
Dim As Integer i, s, n
prime(2) = True
For i = 3 To Ubound(prime) Step 2
prime(i) = True
Next i
For i = 3 To Cint(Sqr(Ubound(prime))) Step 2
If prime(i) Then
For s = i * i To Ubound(prime) Step i + i
prime(s) = False
Next s
End If
Next i
Dim As Integer arrangements(maxNumber)
For n = 2 To Ubound(arrangements)
arrangements(n) = countArrangements(n)
Next n
For n = 2 To Ubound(arrangements)
Print arrangements(n);
Next n
Print
Sleep</syntaxhighlight>
{{out}}
<pre>Same as Visual Basic .NET entry.</pre>
==={{header|Visual Basic .NET}}===
{{Trans|ALGOL 68}}
<syntaxhighlight lang="vbnet">Option Strict On
Option Explicit On
Imports System.IO
''' <summary>Find solutions to the "Prime Triangle" - a triangle of numbers that sum to primes.</summary>
Module vMain
Public Const maxNumber As Integer = 20 ' Largest number we will consider.
Dim prime(2 * maxNumber) As Boolean ' prime sieve.
''' <returns>The number of possible arrangements of the integers for a row in the prime triangle.</returns>
Public Function countArrangements(ByVal n As Integer) As Integer
If n < 2 Then ' No solutions for n < 2.
Return 0
ElseIf n < 4 Then
' For 2 and 3. there is only 1 solution: 1, 2 and 1, 2, 3.
For i As Integer = 1 To n
Console.Out.Write(i.ToString.PadLeft(3))
Next i
Console.Out.WriteLine()
Return 1
Else
' 4 or more - must find the solutions.
Dim printSolution As Boolean = true
Dim used(n) As Boolean
Dim number(n) As Integer
' The triangle row must have 1 in the leftmost and n in the rightmost elements.
' The numbers must alternate between even and odd in order for the sums to be prime.
For i As Integer = 0 To n - 1
number(i) = i Mod 2
Next i
used(1) = True
number(n) = n
used(n) = True
' Find the intervening numbers and count the solutions.
Dim count As Integer = 0
Dim p As Integer = 2
Do While p > 0
Dim p1 As Integer = number(p - 1)
Dim current As Integer = number(p)
Dim [next] As Integer = current + 2
Do While [next] < n AndAlso (Not prime(p1 + [next]) Or used([next]))
[next] += 2
Loop
If [next] >= n Then
[next] = 0
End If
If p = n - 1 Then
' We are at the final number before n.
' It must be the final even/odd number preceded by the final odd/even number.
If [next] <> 0 Then
' Possible solution.
If prime([next] + n) Then
' Found a solution.
count += 1
If printSolution Then
For i As Integer = 1 To n - 2
Console.Out.Write(number(i).ToString.PadLeft(3))
Next i
Console.Out.WriteLine([next].ToString.PadLeft(3) & n.ToString.PadLeft(3))
printSolution = False
End If
End If
[next] = 0
End If
' Backtrack for more solutions.
p -= 1
' There will be a further backtrack as next is 0 ( there could only be one possible number at p - 1 ).
End If
If [next] <> 0 Then
' have a/another number that can appear at p.
used(current) = False
used([next]) = True
number(p) = [next]
' Haven't found all the intervening digits yet.
p += 1
ElseIf p <= 2 Then
' No more solutions.
p = 0
Else
' Can't find a number for this position, backtrack.
used(number(p)) = False
number(p) = p Mod 2
p -= 1
End If
Loop
Return count
End If
End Function
Public Sub Main
prime(2) = True
For i As Integer = 3 To UBound(prime) Step 2
prime(i) = True
Next i
For i As Integer = 3 To Convert.ToInt32(Math.Floor(Math.Sqrt(Ubound(prime)))) Step 2
If prime(i) Then
For s As Integer = i * i To Ubound(prime) Step i + i
prime(s) = False
Next s
End If
Next i
Dim arrangements(maxNumber) As Integer
For n As Integer = 2 To UBound(arrangements)
arrangements(n) = countArrangements(n)
Next n
For n As Integer = 2 To UBound(arrangements)
Console.Out.Write(" " & arrangements(n))
Next n
Console.Out.WriteLine()
End Sub
End Module</syntaxhighlight>
{{out}}
<pre>
1 2
1 2 3
1 2 3 4
1 4 3 2 5
1 4 3 2 5 6
1 4 3 2 5 6 7
1 2 3 4 7 6 5 8
1 2 3 4 7 6 5 8 9
1 2 3 4 7 6 5 8 9 10
1 2 3 4 7 10 9 8 5 6 11
1 2 3 4 7 10 9 8 5 6 11 12
1 2 3 4 7 6 5 12 11 8 9 10 13
1 2 3 4 7 6 13 10 9 8 11 12 5 14
1 2 3 4 7 6 13 10 9 8 11 12 5 14 15
1 2 3 4 7 6 5 12 11 8 15 14 9 10 13 16
1 2 3 4 7 6 5 12 11 8 9 10 13 16 15 14 17
1 2 3 4 7 6 5 8 9 10 13 16 15 14 17 12 11 18
1 2 3 4 7 6 5 8 9 10 13 16 15 14 17 12 11 18 19
1 2 3 4 7 6 5 8 9 10 13 16 15 14 17 12 19 18 11 20
1 1 1 1 1 2 4 7 24 80 216 648 1304 3392 13808 59448 155464 480728 1588162
</pre>
=={{header|C}}==
<
#include <stdbool.h>
#include <stdio.h>
Line 165 ⟶ 406:
if (length == 2)
return is_prime(a[0] + a[1]);
for (size_t i = 1; i + 1
if (is_prime(a[0] + a[i])) {
swap(a, i, 1);
Line 176 ⟶ 417:
}
int count = 0;
if (length == 2) {
if (is_prime(a[0] + a[1]))
++
} else
for (size_t i = 1; i + 1 < length; i += 2) {
}
}
return count;
}
Line 214 ⟶ 457:
for (unsigned int i = 0; i < n; ++i)
a[i] = i + 1;
if (n > 2)
printf(" ");
printf("%d",
}
printf("\n");
Line 225 ⟶ 466:
printf("\nElapsed time: %f seconds\n", duration);
return 0;
}</
{{out}}
Line 251 ⟶ 492:
1 1 1 1 1 2 4 7 24 80 216 648 1304 3392 13808 59448 155464 480728 1588162
Elapsed time: 0.
</pre>
=== Use bit patterns ===
Number combinations are all stored in bit positions here. The <code>bpos()</code> functions returns the position index of the least significant bit in an integer. This code gains some speed, at the cost of total loss of readability. On the plus side, it has some bit manipulation tricks that may be interesting to some.
<syntaxhighlight lang="c">#include <stdio.h>
#include <stdint.h>
#define GCC_ASM // use GCC's asm for i386. If it does not work, #undef it to use alternative func
typedef uint32_t uint;
typedef uint64_t ulong;
#define MASK 0xa08228828228a2bULL
#ifdef GCC_ASM
static inline uint
bpos(uint x)
{
uint b;
asm("bsf %0, %0" : "=r" (b): "0" (x));
return b;
}
#else
static inline uint
bpos(uint x)
{
static const uint bruijin[32] = {
0, 1, 28, 2, 29, 14, 24, 3, 30, 22, 20, 15, 25, 17, 4, 8,
31, 27, 13, 23, 21, 19, 16, 7, 26, 12, 18, 6, 11, 5, 10, 9
};
return bruijin[((uint)((x & -x) * 0x077CB531U)) >> 27];
}
#endif // GCC_ASM
int count(uint n, const uint s, uint avail)
{
int cnt = 0;
avail ^= s;
if (--n)
for (uint b = (uint)(MASK>>bpos(s)) & avail; b; b &= b-1)
cnt += count(n, b&-b, avail);
else
return (MASK & s) != 0;
return cnt;
}
int disp(uint n, const uint s, uint avail, int maxn, uint *seq)
{
seq[n--] = s;
if (!n) {
if ((MASK & s)) {
for (int i = 0; i < maxn; i++)
printf(" %d", bpos(seq[i]) + 1);
putchar('\n');
return 1;
}
} else {
for (uint b = (uint)(MASK>>bpos(s)) & (avail ^= s); b; b &= b-1)
if (disp(n, b&-b, avail, maxn, seq))
return 1;
}
return 0;
}
int chain(uint n, int count_only)
{
const uint top = 1U<<(n - 1);
const uint avail = 2*top - 2;
if (count_only)
return count(n - 1, top, avail);
uint seq[32];
seq[0] = 1;
disp(n - 1, top, avail, n, seq);
return 0;
}
int main(void)
{
for (int n = 2; n < 21; n++)
chain(n, 0);
putchar('\n');
for (int n = 2; n < 21; n++)
printf("%d ", chain(n, 1));
putchar('\n');
return 0;
}</syntaxhighlight>
{{out}}
<pre> 1 2
1 2 3
1 2 3 4
1 4 3 2 5
1 4 3 2 5 6
1 6 5 2 3 4 7
1 4 7 6 5 2 3 8
1 4 7 6 5 8 3 2 9
1 6 7 4 9 8 5 2 3 10
1 10 9 8 5 6 7 4 3 2 11
1 10 9 8 11 6 7 4 3 2 5 12
1 12 11 8 9 10 7 6 5 2 3 4 13
1 12 11 8 9 10 13 4 7 6 5 2 3 14
1 12 11 8 5 14 9 10 13 6 7 4 3 2 15
1 12 11 8 15 14 9 10 13 4 7 6 5 2 3 16
1 10 13 16 15 14 9 8 11 12 5 6 7 4 3 2 17
1 12 17 14 15 16 13 10 9 8 11 6 7 4 3 2 5 18
1 18 13 16 15 14 17 12 11 8 9 10 7 6 5 2 3 4 19
1 18 19 10 13 16 15 14 17 12 11 8 9 4 7 6 5 2 3 20
1 1 1 1 1 2 4 7 24 80 216 648 1304 3392 13808 59448 155464 480728 1588162</pre>
=={{header|C++}}==
<
#include <chrono>
#include <iomanip>
Line 331 ⟶ 690:
std::chrono::duration<double> duration(end - start);
std::cout << "\nElapsed time: " << duration.count() << " seconds\n";
}</
{{out}}
Line 362 ⟶ 721:
=={{header|F_Sharp|F#}}==
This task uses [http://www.rosettacode.org/wiki/Extensible_prime_generator#The_functions Extensible Prime Generator (F#)]
<
// Prime triangle. Nigel Galloway: April 12th., 2022
let fN i (g,(e,l))=e|>Seq.map(fun n->let n=i n in (n::g,List.partition(i>>(=)n) l))
Line 371 ⟶ 730:
{2..20}|>Seq.iter(fun n->(primeT>>Seq.head>>List.iter(printf "%3d"))n;printfn "");;
{2..20}|>Seq.iter(primeT>>Seq.length>>printf "%d "); printfn ""
</syntaxhighlight>
{{out}}
<pre>
Line 398 ⟶ 757:
=={{header|Go}}==
Takes about 0.64 seconds.
{{trans|Phix}}
<
import "fmt"
var canFollow [][]bool
var arrang []int
var
var pmap = make(map[int]bool)
Line 420 ⟶ 779:
if n-done <= 1 {
if canFollow[ad-1][n-1] {
if
for _, e := range arrang {
fmt.Printf("%2d ", e)
}
fmt.Println()
bFirst = false
}
res++
Line 430 ⟶ 790:
} else {
done++
for i := done - 1; i <= n-2; i +
if
arrang[i], arrang[done-1] =
res = ptrs(res, n, done)
}
}
Line 454 ⟶ 811:
}
}
for i := 0;
arrang[i] = i + 1
}
return ptrs(0, n, 1)
}
func main() {
counts := make([]int, 19)
for i := 2; i <= 20; i++ {
counts[i-2] = primeTriangle(i)
}
fmt.Println()
}
fmt.Println()
}</
{{out}}
Line 503 ⟶ 858:
=={{header|J}}==
Essentially, we're traversing a directed graph starting at 1, ending at y, with edges such that adjacent pairs sum to a prime number and with distinct intermediate values between 1 and y.
Implementation:
<syntaxhighlight lang="j">add_plink=: [:;{{ <y,"1 0 x #~ 1 p:+/|:(x=. x-. y),"0/{: y }}"1
prime_pair_seqs=: {{ y add_plink (2}.i.y) add_plink^:(y-2) 1 }}</syntaxhighlight>
Task example (displaying counts of number of valid sequences to the left, because that looks nice):
<syntaxhighlight lang="j">task=: {{
N=. #seqs=. prime_pair_seqs j
echo (_8{.":N),' | ',3":{:seqs
end.
}}
task 20
1 | 1 2
1 | 1 2 3
Line 561 ⟶ 880:
1 | 1 4 3 2 5
1 | 1 4 3 2 5 6
2 | 1
4 | 1
7 | 1
24 | 1
80 | 1 10
216 | 1 10
648 | 1 12
1304 | 1 12
3392 | 1 12
13808 | 1 12
59448 | 1 16
155464 | 1 16
480728 | 1 18
1588162 | 1 18
=={{header|Java}}==
<
public static void main(String[] args) {
long start = System.currentTimeMillis();
Line 614 ⟶ 933:
if (length == 2)
return isPrime(a[start] + a[start + 1]);
for (int i = 1; i + 1 < length; i +
if (isPrime(a[start] + a[start + i])) {
swap(a, start + i, start + 1);
Line 631 ⟶ 950:
++count;
} else {
for (int i = 1; i + 1 < length; i +
if (isPrime(a[start] + a[start + i])) {
swap(a, start + i, start + 1);
Line 651 ⟶ 970:
return ((1L << n) & 0x28208a20a08a28acL) != 0;
}
}</
{{out}}
Line 677 ⟶ 996:
1 1 1 1 1 2 4 7 24 80 216 648 1304 3392 13808 59448 155464 480728 1588162
Elapsed time:
</pre>
=={{header|jq}}==
{{works with|jq}}
'''Adapted from [[#Wren|Wren]]'''
(gojq requires too much memory to complete the task.)
<syntaxhighlight lang=jq>
# $i and $j should be relevant integers (possibly negataive)
# Usage with an array-valued key: .key |= array_swap($i; $j)
def array_swap($i; $j):
if $i == $j then .
else .[$i] as $t
| .[$i] = .[$j]
| .[$j] = $t
end;
# For syntactic convenience
def swap($array; $i; $j):
$array | array_swap($i; $j);
def pmap:
reduce (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37) as $i ([];
.[$i] = true) ;
# input: {bFirst, arrang, canFollow, emit}
# output: same + {res, emit, i, count}
def ptrs($res; $n; $done):
. + {$res, count: $done}
| .arrang[$done-1] as $ad
| if ($n - $done <= 1)
then if .canFollow[$ad-1][$n-1]
then if .bFirst
then .emit += [.arrang]
| .bFirst = false
else .
end
| .res += 1
else .
end
else .count += 1
| .count as $count
| reduce range($count - 1; $n - 1; 2) as $i (.;
.arrang[$i] as $ai
| if .canFollow[$ad-1][$ai-1]
then .arrang = swap(.arrang; $i; $count-1)
| ptrs(.res; $n; $count)
| .arrang = swap(.arrang; $i; $count-1)
else .
end )
end;
# Emit {emit, res} from the call to ptrs
def primeTriangle($n):
pmap as $pmap
| {}
| .canFollow = (reduce range(0;$n) as $i ([];
.[$i] = [range(0;$n) | false]
| reduce range(0;$n) as $j (.;
.[$i][$j] = $pmap[$i+$j+2] )))
| .bFirst = true
| .arrang = [range(1; 1+$n)]
| ptrs(0; $n; 1)
| {emit, res} ;
def task($n):
range(2;$n+1) as $i
| primeTriangle($i) ;
def task($n):
foreach (range(2;$n+1), null) as $i ({counts: [], emit: null};
if $i == null then .emit = "\n\(.counts)"
else primeTriangle($i) as $pt
| .counts += [$pt|.res]
| .emit = $pt.emit
end;
.emit);
task(20)[]
</syntaxhighlight>
{{output}}
<pre>
[1,2]
[1,2,3]
[1,2,3,4]
[1,4,3,2,5]
[1,4,3,2,5,6]
[1,4,3,2,5,6,7]
[1,2,3,4,7,6,5,8]
[1,2,3,4,7,6,5,8,9]
[1,2,3,4,7,6,5,8,9,10]
[1,2,3,4,7,10,9,8,5,6,11]
[1,2,3,4,7,10,9,8,5,6,11,12]
[1,2,3,4,7,6,5,12,11,8,9,10,13]
[1,2,3,4,7,6,13,10,9,8,11,12,5,14]
[1,2,3,4,7,6,13,10,9,8,11,12,5,14,15]
[1,2,3,4,7,6,5,12,11,8,15,14,9,10,13,16]
[1,2,3,4,7,6,5,12,11,8,9,10,13,16,15,14,17]
[1,2,3,4,7,6,5,8,9,10,13,16,15,14,17,12,11,18]
[[1,2,3,4,7,6,5,8,9,10,13,16,15,14,17,12,11,18,19]]
[[1,2,3,4,7,6,5,8,9,10,13,16,15,14,17,12,19,18,11,20]]
[1,1,1,1,1,2,4,7,24,80,216,648,1304,3392,13808,59448,155464,480728,1588162]
</pre>
=={{header|Julia}}==
=== Filter method ===
<
function primetriangle(nrows::Integer)
Line 713 ⟶ 1,135:
@time primetriangle(16)
</
<pre>
1 2
Line 737 ⟶ 1,159:
=== Generator method ===
Similar to the Phix entry.
<
function solverow(row, pos, avail)
Line 769 ⟶ 1,191:
println(" 1 2\n" * prod(rowstrings), "\n", counts)
end
</
<pre>
1 2
Line 794 ⟶ 1,216:
25.818809 seconds (249.58 M allocations: 22.295 GiB, 15.56% gc time)
</pre>
=={{header|Mathematica}}/{{header|Wolfram Language}}==
<syntaxhighlight lang="mathematica">ClearAll[FindPrimeTriangles, FindPrimeTrianglesHelper]
FindPrimeTriangles[max_] :=
Module[{count = 0, firstsolution, primes, primeQ},
primes = PrimeQ[Range[2 max]];
primeQ[n_] := primes[[n]];
ClearAll[FindPrimeTrianglesHelper];
FindPrimeTrianglesHelper[start_, remainder_, mxx_] :=
Module[{last, nexts, r, newstart, newremainder},
If[Length[remainder] > 0,
last = Last[start];
Do[
r = remainder[[ri]];
If[primeQ[last + r],
newstart = Append[start, r];
newremainder = Delete[remainder, ri];
FindPrimeTrianglesHelper[newstart, newremainder, mxx]
]
,
{ri, Length[remainder]}
]
,
If[primeQ[Last[start] + mxx],
count++;
If[count == 1,
Print[Append[start, mxx]]
]
]
]
];
FindPrimeTrianglesHelper[{1}, Range[2, max - 1], max];
count
]
Table[FindPrimeTriangles[S],{S, 2, 20}]</syntaxhighlight>
{{out}}
<pre>{1,2}
{1,2,3}
{1,2,3,4}
{1,4,3,2,5}
{1,4,3,2,5,6}
{1,4,3,2,5,6,7}
{1,2,3,4,7,6,5,8}
{1,2,3,4,7,6,5,8,9}
{1,2,3,4,7,6,5,8,9,10}
{1,2,3,4,7,10,9,8,5,6,11}
{1,2,3,4,7,10,9,8,5,6,11,12}
{1,2,3,4,7,6,5,12,11,8,9,10,13}
{1,2,3,4,7,6,13,10,9,8,11,12,5,14}
{1,2,3,4,7,6,13,10,9,8,11,12,5,14,15}
{1,2,3,4,7,6,5,12,11,8,15,14,9,10,13,16}
{1,2,3,4,7,6,5,12,11,8,9,10,13,16,15,14,17}
{1,2,3,4,7,6,5,8,9,10,13,16,15,14,17,12,11,18}
{1,2,3,4,7,6,5,8,9,10,13,16,15,14,17,12,11,18,19}
{1,2,3,4,7,6,5,8,9,10,13,16,15,14,17,12,19,18,11,20}
{1, 1, 1, 1, 1, 2, 4, 7, 24, 80, 216, 648, 1304, 3392, 13808, 59448, 155464, 480728, 1588162}</pre>
=={{header|Nim}}==
{{trans|C}}
<syntaxhighlight lang="Nim">import std/[monotimes, strutils, times]
const IsPrime = [false, false, true, true, false, true, false, true,
false, false, false, true, false, true, false, false,
false, true, false, true, false, false, false, true,
false, false, false, false, false, true, false, true,
false, false, false, false, false, true, false, false,
false, true, false, true, false, false, false, true,
false, false, false, false, false, true, false, false,
false, false, false, true, false, true, false, false]
type TriangleRow = openArray[Natural]
template isPrime(n: Natural): bool = IsPrime[n]
func primeTriangleRow(a: var TriangleRow): bool =
if a.len == 2:
return isPrime(a[0] + a[1])
for i in countup(1, a.len - 2, 2):
if isPrime(a[0] + a[i]):
swap a[i], a[1]
if primeTriangleRow(a.toOpenArray(1, a.high)):
return true
swap a[i], a[1]
func primeTriangleCount(a: var TriangleRow): Natural =
if a.len == 2:
if isPrime(a[0] + a[1]):
inc result
else:
for i in countup(1, a.len - 2, 2):
if isPrime(a[0] + a[i]):
swap a[i], a[1]
inc result, primeTriangleCount(a.toOpenArray(1, a.high))
swap a[i], a[1]
proc print(a: TriangleRow) =
if a.len == 0: return
for i, n in a:
if n > 0: stdout.write ' '
stdout.write align($n, 2)
stdout.write '\n'
let start = getMonoTime()
for n in 2..20:
var a = newSeq[Natural](n)
for i in 0..<n:
a[i] = i + 1
if a.primeTriangleRow:
print a
echo()
for n in 2..20:
var a = newSeq[Natural](n)
for i in 0..<n:
a[i] = i + 1
if n > 2: stdout.write " "
stdout.write a.primeTriangleCount
echo '\n'
echo "Elapsed time: ", (getMonoTime() - start).inMilliseconds, " ms"
</syntaxhighlight>
{{out}}
<pre> 1 2
1 2 3
1 2 3 4
1 4 3 2 5
1 4 3 2 5 6
1 4 3 2 5 6 7
1 2 3 4 7 6 5 8
1 2 3 4 7 6 5 8 9
1 2 3 4 7 6 5 8 9 10
1 2 3 4 7 10 9 8 5 6 11
1 2 3 4 7 10 9 8 5 6 11 12
1 2 3 4 7 6 5 12 11 8 9 10 13
1 2 3 4 7 6 13 10 9 8 11 12 5 14
1 2 3 4 7 6 13 10 9 8 11 12 5 14 15
1 2 3 4 7 6 5 12 11 8 15 14 9 10 13 16
1 2 3 4 7 6 5 12 11 8 9 10 13 16 15 14 17
1 2 3 4 7 6 5 8 9 10 13 16 15 14 17 12 11 18
1 2 3 4 7 6 5 8 9 10 13 16 15 14 17 12 11 18 19
1 2 3 4 7 6 5 8 9 10 13 16 15 14 17 12 19 18 11 20
1 1 1 1 1 2 4 7 24 80 216 648 1304 3392 13808 59448 155464 480728 1588162
Elapsed time: 535 ms
</pre>
=={{header|Pascal}}==
==={{header|Free Pascal}}===
Simple backtracking.Precaltulated like [[Prime_triangle#Phix]] Can_Follow by checking for sum gets prime.
<syntaxhighlight lang="pascal">
program PrimePyramid;
{$IFDEF FPC}
{$MODE delphi}{$Optimization ON,ALL}{$CodeAlign proc=32}
{$IFEND}
const
MAX = 21;//max 8 different SumToPrime : array[0..MAX,0..7] of byte;
//MAX = 57;//max 16 different SumToPrime : array[0..MAX,0..15] of byte;
cMaxAlign32 = (((MAX-1)DIV 32)+1)*32-1;//MAX > 0
type
tPrimeRange = set of 0..127;
var
SetOfPrimes :tPrimeRange =[2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,
53,59,61,67,71,73,79,83,89,97,101,103,107,
109,113,127];
SumToPrime : array[0..cMaxAlign32,0..7] of byte;
// SumToPrime : array[0..MAX,0..15] of byte;
SumToPrimeMaxIdx,
SumMaxIdx,
Solution,
FirstSolution : array[0..cMaxAlign32] of byte;
free : array[0..cMaxAlign32] of boolean;
maxused : integer;
digitcount,SolCount : integer;
procedure InitSumToPrime;
var
i,j,idx : integer;
begin
For i := 1 to MAX do
Begin
idx := 0;
For j := 1 to MAX do
If (i+j) in SetOfPrimes then
Begin
SumToPrime[i,idx] := j;
inc(idx);
end;
SumToPrimeMaxIdx[i] := idx;
end;
end;
procedure InitFree(maxused : integer);
var
i,j : Integer;
begin
For i := 0 to 1 do
Free[i] := false;
For i := 2 to maxused-1 do
Free[i] := true;
For i := maxused to MAX do
Free[i] := false;
// search maxidx of max neighour sum to prime
For i := 1 to maxused-1 do
begin
j := SumToPrimeMaxIdx[i]-1;
while SumToPrime[i,j] > maxused-1 do
j -= 1;
SumMaxIdx[i] := j+1;
end;
end;
procedure CountSolution(digit:integer);
begin
// check if maxused can follow
if (digit+maxused) in SetOfPrimes then
Begin
if solcount = 0 then
FirstSolution := Solution;
inc(solCount);
end;
end;
procedure checkDigits(digit:integer);
var
idx,nextDigit: integer;
begin
idx := 0;
repeat
nextDigit := SumToPrime[digit,idx];
if Free[nextdigit] then
Begin
Solution[digitcount] := nextDigit;
dec(digitcount);
IF digitcount = 0 then
CountSolution(nextDigit);
free[nextdigit]:= false;
checkDigits(nextdigit);
inc(digitcount);
free[nextdigit]:= true;
end;
inc(idx);
until idx >= SumMaxIdx[digit];
end;
var
i,j : integer;
Begin
InitSumToPrime;
writeln('number| count| first solution');
writeln(' 2| 1| 1 2');
For i := 3 to 20 do
Begin
maxused := i;
InitFree(i);
digitcount := i-2;
solCount := 0;
checkDigits(1);
write(i:4,'|',solcount:10,'| 1');
For j := i-2 downto 1 do
write( FirstSolution[j]:3);
writeln(i:3);
end;
end.
</syntaxhighlight>
{{out|@TIO.RUN}}
<pre>
number| count| first solution
2| 1| 1 2
3| 1| 1 2 3
4| 1| 1 2 3 4
5| 1| 1 4 3 2 5
6| 1| 1 4 3 2 5 6
7| 2| 1 4 3 2 5 6 7
8| 4| 1 2 3 4 7 6 5 8
9| 7| 1 2 3 4 7 6 5 8 9
10| 24| 1 2 3 4 7 6 5 8 9 10
11| 80| 1 2 3 4 7 10 9 8 5 6 11
12| 216| 1 2 3 4 7 10 9 8 5 6 11 12
13| 648| 1 2 3 4 7 6 5 12 11 8 9 10 13
14| 1304| 1 2 3 4 7 6 13 10 9 8 11 12 5 14
15| 3392| 1 2 3 4 7 6 13 10 9 8 11 12 5 14 15
16| 13808| 1 2 3 4 7 6 5 12 11 8 15 14 9 10 13 16
17| 59448| 1 2 3 4 7 6 5 12 11 8 9 10 13 16 15 14 17
18| 155464| 1 2 3 4 7 6 5 8 9 10 13 16 15 14 17 12 11 18
19| 480728| 1 2 3 4 7 6 5 8 9 10 13 16 15 14 17 12 11 18 19
20| 1588162| 1 2 3 4 7 6 5 8 9 10 13 16 15 14 17 12 19 18 11 20
Real time: 1.827 s User time: 1.794 s Sys. time: 0.021 s CPU share: 99.36 %
@home: real 0m0,679s</pre>
=={{header|Perl}}==
{{trans|Raku}}
{{libheader|ntheory}}
<syntaxhighlight lang="perl">use strict;
use warnings;
use feature 'say';
use ntheory 'is_prime';
use List::MoreUtils qw(zip slideatatime);
use Algorithm::Combinatorics qw(permutations);
say '1 2';
my @count = (0, 0, 1);
for my $n (3..17) {
my @even_nums = grep { 0 == $_ % 2 } 2..$n-1;
my @odd_nums = grep { 1 == $_ % 2 } 3..$n-1;
for my $e (permutations [@even_nums]) {
next if $$e[0] == 8 or $$e[0] == 14;
my $nope = 0;
for my $o (permutations [@odd_nums]) {
my @list = (zip(@$e, @$o), $n); # 'zip' makes a list with a gap if more evens than odds
splice @list, -2, -1 if not defined $list[-2]; # in which case splice out 'undef' in next-to-last position
my $it = slideatatime(1, 2, @list);
while ( my @rr = $it->() ) {
last unless defined $rr[1];
$nope++ and last unless is_prime $rr[0]+$rr[1];
}
unless ($nope) {
say '1 ' . join ' ', @list unless $count[$n];
$count[$n]++;
}
$nope = 0;
}
}
}
say "\n" . join ' ', @count[2..$#count];</syntaxhighlight>
{{out}}
<pre>1 2
1 2 3
1 2 3 4
1 4 3 2 5
1 4 3 2 5 6
1 4 3 2 5 6 7
1 2 3 4 7 6 5 8
1 2 3 4 7 6 5 8 9
1 2 3 4 7 6 5 8 9 10
1 2 3 4 9 10 7 6 5 8 11
1 2 3 4 9 10 7 6 5 8 11 12
1 2 3 4 7 6 5 12 11 8 9 10 13
1 2 3 4 13 6 11 8 9 10 7 12 5 14
1 2 3 4 13 6 11 8 9 10 7 12 5 14 15
1 2 3 4 13 6 11 8 9 10 7 12 5 14 15 16
1 2 15 4 13 6 11 8 9 10 3 16 7 12 5 14 17
1 1 1 1 1 2 4 7 24 80 216 648 1304 3392 13808 59448</pre>
=={{header|Phix}}==
{{libheader|Phix/online}}
Not sure this counts as particularly clever but by the sound of things it's quite a bit faster than the other entries so far. 😎<br> You can run this online [http://phix.x10.mx/p2js/prime_triangles.htm here] (expect a blank screen for about
<!--<
<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>
<span style="color: #004080;">atom</span> <span style="color: #000000;">t0</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">time</span><span style="color: #0000FF;">()</span>
<span style="color: #004080;">sequence</span> <span style="color: #000000;">can_follow
<span style="color: #004080;">bool</span> <span style="color: #000000;">bFirst</span> <span style="color: #0000FF;">=</span> <span style="color: #004600;">true</span>
Line 819 ⟶ 1,591:
<span style="color: #008080;">else</span>
<span style="color: #000000;">done</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">1</span>
<span style="color: #000080;font-style:italic;">-- as per talk page, we only need to examine odd
-- numbers following an even number & vice versa</span>
<span style="color: #008080;">for</span> <span style="color: #000000;">i</span><span style="color:
<span style="color:
<span style="color: #008080;">if</span> <span style="color: #000000;">can_follow</span><span style="color: #0000FF;">[</span><span style="color: #000000;">ad</span><span style="color: #0000FF;">][</span><span style="color: #000000;">ai</span><span style="color: #0000FF;">]</span> <span style="color: #008080;">then</span>
<span style="color: #004080;">integer</span> <span style="color: #000000;">aid</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">arrang</span><span style="color: #0000FF;">[</span><span style="color: #000000;">done</span><span style="color: #0000FF;">]</span>
<span style="color: #000000;">arrang</span><span style="color: #0000FF;">[</span><span style="color: #000000;">i</span><span style="color: #0000FF;">]</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">aid</span>
<span style="color: #000000;">arrang</span><span style="color: #0000FF;">[</span><span style="color: #000000;">done</span><span style="color: #0000FF;">]</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">ai</span>
<span style="color: #000000;">res</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">ptrs</span><span style="color: #0000FF;">(</span><span style="color: #000000;">res</span><span style="color: #0000FF;">,</span><span style="color: #000000;">n</span><span style="color: #0000FF;">,</span><span style="color: #000000;">done</span><span style="color: #0000FF;">)</span>
<span style="color: #000000;">
<span style="color: #000000;">arrang</span><span style="color: #0000FF;">[</span><span style="color: #000000;">done</span><span style="color: #0000FF;">]</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">aid</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
Line 838 ⟶ 1,615:
<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<span style="color: #000000;">
<span style="color: #000000;">bFirst</span> <span style="color: #0000FF;">=</span> <span style="color: #004600;">true</span>
<span style="color: #008080;">return</span> <span style="color: #000000;">ptrs</span><span style="color: #0000FF;">(</span><span style="color: #000000;">0</span><span style="color: #0000FF;">,</span><span style="color: #000000;">n</span><span style="color: #0000FF;">,</span><span style="color: #000000;">1</span><span style="color: #0000FF;">)</span>
Line 847 ⟶ 1,623:
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"%s\n"</span><span style="color: #0000FF;">,</span><span style="color: #7060A8;">join</span><span style="color: #0000FF;">(</span><span style="color: #000000;">res</span><span style="color: #0000FF;">,</span><span style="color: #000000;">fmt</span><span style="color: #0000FF;">:=</span><span style="color: #008000;">"%d"</span><span style="color: #0000FF;">))</span>
<span style="color: #0000FF;">?</span><span style="color: #7060A8;">elapsed</span><span style="color: #0000FF;">(</span><span style="color: #7060A8;">time</span><span style="color: #0000FF;">()-</span><span style="color: #000000;">t0</span><span style="color: #0000FF;">)</span>
<!--</
{{out}}
<pre>
Line 870 ⟶ 1,646:
1 2 3 4 7 6 5 8 9 10 13 16 15 14 17 12 19 18 11 20
1 1 1 1 1 2 4 7 24 80 216 648 1304 3392 13808 59448 155464 480728 1588162
"
</pre>
=={{header|Python}}==
<syntaxhighlight lang="python">
from numpy import array
# for Rosetta Code by MG - 20230312
def is_prime(n: int) -> bool:
assert n < 64
return ((1 << n) & 0x28208a20a08a28ac) != 0
def prime_triangle_row(a: array, start: int, length: int) -> bool:
if length == 2:
return is_prime(a[0] + a[1])
for i in range(1, length - 1, 1):
if is_prime(a[start] + a[start + i]):
a[start + i], a[start + 1] = a[start + 1], a[start + i]
if prime_triangle_row(a, start + 1, length - 1):
return True
a[start + i], a[start + 1] = a[start + 1], a[start + i]
return False
def prime_triangle_count(a: array, start: int, length: int) -> int:
count: int = 0
if length == 2:
if is_prime(a[start] + a[start + 1]):
count += 1
else:
for i in range(1, length - 1, 1):
if is_prime(a[start] + a[start + i]):
a[start + i], a[start + 1] = a[start + 1], a[start + i]
count += prime_triangle_count(a, start + 1, length - 1)
a[start + i], a[start + 1] = a[start + 1], a[start + i]
return count
def print_row(a: array):
if a == []:
return
print("%2d"% a[0], end=" ")
for x in a[1:]:
print("%2d"% x, end=" ")
print()
for n in range(2, 21):
tr: array = [_ for _ in range(1, n + 1)]
if prime_triangle_row(tr, 0, n):
print_row(tr)
print()
for n in range(2, 21):
tr: array = [_ for _ in range(1, n + 1)]
if n > 2:
print(" ", end="")
print(prime_triangle_count(tr, 0, n), end="")
print()
</syntaxhighlight>
{{out}}
<pre>
1 2
1 2 3
1 2 3 4
1 2 3 4 5
1 4 3 2 5 6
1 4 3 2 5 6 7
1 2 3 4 7 6 5 8
1 2 3 4 7 6 5 8 9
1 2 3 4 7 6 5 8 9 10
1 2 3 4 7 6 5 8 9 10 11
1 2 3 4 7 10 9 8 5 6 11 12
1 2 3 4 7 6 5 12 11 8 9 10 13
1 2 3 4 7 6 5 12 11 8 9 10 13 14
1 2 3 4 7 6 13 10 9 8 11 12 5 14 15
1 2 3 4 7 6 5 12 11 8 15 14 9 10 13 16
1 2 3 4 7 6 5 12 11 8 9 10 13 16 15 14 17
1 2 3 4 7 6 5 8 9 10 13 16 15 14 17 12 11 18
1 2 3 4 7 6 5 8 9 10 13 16 15 14 17 12 11 18 19
1 2 3 4 7 6 5 8 9 10 13 16 15 14 17 12 11 18 19 20
1 1 1 1 1 2 4 7 24 80 216 648 1304 3392 13808 59448 155464 480728 1588162
</pre>
Line 876 ⟶ 1,731:
Limit the upper threshold a bit to avoid multiple hours of pointless calculations. Even just up to 17 takes over 20 minutes.
<syntaxhighlight lang="raku"
my $lock = Lock.new;
put (1,2);
Line 900 ⟶ 1,755:
}
}
put "\n", @count[2..*];</
{{out}}
<pre>1 2
Line 923 ⟶ 1,778:
=={{header|Rust}}==
<
assert!(n < 64);
((1u64 << n) & 0x28208a20a08a28ac) != 0
Line 932 ⟶ 1,787:
return is_prime(a[0] + a[1]);
}
for i in (1..a.len() - 1).step_by(2) {
if is_prime(a[0] + a[i]) {
a.swap(i, 1);
Line 951 ⟶ 1,806:
}
} else {
for i in (1..a.len() - 1).step_by(2) {
if is_prime(a[0] + a[i]) {
a.swap(i, 1);
Line 993 ⟶ 1,848:
let time = start.elapsed();
println!("\nElapsed time: {} milliseconds", time.as_millis());
}</
{{out}}
Line 1,019 ⟶ 1,874:
1 1 1 1 1 2 4 7 24 80 216 648 1304 3392 13808 59448 155464 480728 1588162
Elapsed time:
</pre>
=={{header|Swift}}==
<
func isPrime(_ n: Int) -> Bool {
Line 1,032 ⟶ 1,887:
}
func primeTriangleRow(_ a: inout
return isPrime(a[start] + a[start + 1])
}
for i in
if
if primeTriangleRow(&a, start: start + 1, length: length - 1) {
return true
}
a.swapAt(
}
}
Line 1,050 ⟶ 1,904:
}
func primeTriangleCount(_ a: inout
var count = 0
if
if isPrime(a[start] + a[start + 1]) {
count += 1
}
} else {
for i in
if
a.swapAt(index, start + 1)
}
}
Line 1,080 ⟶ 1,933:
print()
}
let startTime = CFAbsoluteTimeGetCurrent()
for n in 2...20 {
var a = Array(1...n)
if primeTriangleRow(&a
printRow(a)
}
Line 1,094 ⟶ 1,949:
print(" ", terminator: "")
}
print("\(primeTriangleCount(&a
}
print()
let endTime = CFAbsoluteTimeGetCurrent()
print("\nElapsed time: \(endTime - startTime) seconds")</syntaxhighlight>
{{out}}
Line 1,121 ⟶ 1,979:
1 1 1 1 1 2 4 7 24 80 216 648 1304 3392 13808 59448 155464 480728 1588162
Elapsed time: 1.0268970727920532 seconds
</pre>
Line 1,285 ⟶ 1,986:
{{trans|Phix}}
{{libheader|Wren-fmt}}
Takes around 18.7 seconds which is fine for Wren.
<syntaxhighlight lang="wren">import "./fmt" for Fmt
var canFollow = []
var arrang = []
var
var pmap = {}
Line 1,305 ⟶ 2,003:
if (n - done <= 1) {
if (canFollow[ad-1][n-1]) {
if (
Fmt.print("$2d", arrang)
bFirst = false
}
res = res + 1
}
} else {
done = done + 1
var ai =
if
arrang.swap(i, done-1)
res = ptrs.call(res, n, done)
}
i = i + 2
}
}
Line 1,329 ⟶ 2,031:
for (j in 0...n) canFollow[i][j] = pmap.containsKey(i+j+2)
}
return ptrs.call(0, n, 1)
}
var counts = []
for (i in 2..20) {
counts.add(primeTriangle.call(i))
}
System.print()
System.print(counts.join(" "))</syntaxhighlight>
{{out}}
|