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June 2018 Update

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Ingy döt Net 2018-06-22 20:57:24 +00:00
parent ba8067c3b7
commit 22f33d4004
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Task/Aliquot-sequence-classifications/ALGOL-68/aliquot-sequence-classifications.alg Normal file
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BEGIN
# aliquot sequence classification #
# maximum sequence length we consider #
INT max sequence length = 16;
# possible classifications #
STRING perfect classification = "perfect ";
STRING amicable classification = "amicable ";
STRING sociable classification = "sociable ";
STRING aspiring classification = "aspiring ";
STRING cyclic classification = "cyclic ";
STRING terminating classification = "terminating ";
STRING non terminating classification = "non terminating";
# structure to hold an aliquot sequence and its classification #
MODE ALIQUOT = STRUCT( STRING classification
, [ 1 : max sequence length ]LONG INT sequence
, INT length
);
# maximum value for sequence elements - if any element is more than this, #
# we assume it is non-teriminating #
LONG INT max element = 140 737 488 355 328;
# returns the sum of the proper divisors of n #
OP DIVISORSUM = ( LONG INT n )LONG INT:
BEGIN
LONG INT abs n = ABS n;
IF abs n < 2 THEN
0 # -1, 0 and 1 have no proper divisors #
ELSE
# have a number with possible divisors #
LONG INT result := 1; # 1 is always a divisor #
# a FOR loop counter can only be an INT, hence the WHILE loop #
LONG INT d := ENTIER long sqrt( abs n );
WHILE d > 1 DO
IF abs n MOD d = 0 THEN
# found another divisor #
result +:= d;
IF d * d /= abs n THEN
# add the other divisor #
result +:= abs n OVER d
FI
FI;
d -:= 1
OD;
result
FI
END # DIVISORSUM # ;
# generates the aliquot sequence of the number k and its classification #
# at most max elements of the sequence are considered #
OP CLASSIFY = ( LONG INT k )ALIQUOT :
BEGIN
ALIQUOT result;
classification OF result := "non-terminating";
INT lb = LWB sequence OF result;
INT ub = UPB sequence OF result;
( sequence OF result )[ lb ] := k; # the first element is always k #
length OF result := 1;
FOR i FROM lb + 1 TO ub DO
( sequence OF result )[ i ] := 0
OD;
BOOL classified := FALSE;
LONG INT prev k := k;
FOR i FROM lb + 1 TO ub WHILE NOT classified DO
length OF result +:= 1;
LONG INT next k := ( sequence OF result )[ i ] := DIVISORSUM prev k;
classified := TRUE;
IF next k = 0 THEN # the sequence terminates #
classification OF result := terminating classification
ELIF next k > max element THEN # the sequence gets too large #
classification OF result := non terminating classification
ELIF next k = k THEN # the sequence that returns to k #
classification OF result
:= IF i = lb + 1 THEN perfect classification
ELIF i = lb + 2 THEN amicable classification
ELSE sociable classification
FI
ELIF next k = prev k THEN # the sequence repeats with non-k #
classification OF result := aspiring classification
ELSE # check for repeating sequence with a period more than 1 #
classified := FALSE;
FOR prev pos FROM lb TO i - 2 WHILE NOT classified DO
IF classified := ( sequence OF result )[ prev pos ] = next k THEN
# found a repeatition #
classification OF result := cyclic classification
FI
OD
FI;
prev k := next k
OD;
result
END # CLASSIFY # ;
# test cases as per the task #
[]LONG INT test cases =
( 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
, 11, 12, 28, 496, 220, 1184, 12496, 1264460, 790, 909
, 562, 1064, 1488
, 15355717786080
);
FOR i FROM LWB test cases TO UPB test cases DO
LONG INT k := test cases[ i ];
ALIQUOT seq = CLASSIFY k;
print( ( whole( k, -14 ), ": ", classification OF seq, ":" ) );
FOR e FROM LWB sequence OF seq + 1 TO length OF seq DO
print( ( " ", whole( ( sequence OF seq )[ e ], 0 ) ) )
OD;
print( ( newline ) )
OD
END