aparler/RosettaCodeData
1
0
Fork 0
Code Issues 12 Pull requests 3 Projects Releases Packages Wiki Activity Actions

Data commit

Browse source
This commit is contained in:
Ingy döt Net 2023-07-01 11:58:00 -04:00
parent 7387c8f97b
commit cb5bb5e222
199093 changed files with 3378972 additions and 0 deletions
Download patch file Download diff file Expand all files Collapse all files

94
Task/Prime-triangle/ALGOL-68/prime-triangle.alg Normal file
View file

@ -0,0 +1,94 @@
BEGIN # find solutions to the "Prime Triangle" - a triangle of numbers that sum to primes #
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 #
[ 0 : 2 * max number ]BOOL prime;
prime[ 0 ] := prime[ 1 ] := FALSE;
prime[ 2 ] := TRUE;
FOR i FROM 3 BY 2 TO UPB prime DO prime[ i ] := TRUE OD;
FOR i FROM 4 BY 2 TO UPB prime DO prime[ i ] := FALSE OD;
FOR i FROM 3 BY 2 TO ENTIER sqrt( UPB prime ) DO
IF prime[ i ] THEN
FOR s FROM i * i BY i + i TO UPB prime DO prime[ s ] := FALSE OD
FI
OD;
# returns the number of possible arrangements of the integers for a row in the prime triangle #
PROC count arrangements = ( INT n )INT:
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 #
FOR i TO n DO print( ( whole( i, -3 ) ) ) OD; print( ( newline ) );
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[ i ] := FALSE;
number[ i ] := i MOD 2
OD;
used[ 1 ] := TRUE;
number[ n ] := n;
used[ n ] := TRUE;
# find the intervening numbers and count the solutions #
INT count := 0;
INT p := 2;
WHILE p > 0 DO
INT p1 = number[ p - 1 ];
INT current = number[ p ];
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 #
# 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 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[ current ] := FALSE;
used[ next ] := TRUE;
number[ p ] := next;
p +:= 1
ELIF 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
FI
OD;
count
FI # count arrangements # ;
[ 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
print( ( " ", whole( arrangements[ n ], 0 ) ) )
OD;
print( ( newline ) )
END