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| subject: | Pythagorean Triples |
I changed my priorities. Now, I want a program that will go to the
highest numbers possible, regardless of speed. So I changed all the
integer operations ("\", MOD, Booleans) to floating-point ones, and
made all the numbers double-precision floating-point ones. The result
is this version. Previously, the program bombed at about A=46000,
because long-integers overflowed. This one will handle values of A of
more than 100 million!
Of course, it would take almost forever to print out all the triples up
to there. This version is *much* slower than the old one. But you can
start it wherever you want.
dow
----------------------------------------------------
' PYTRIP-L.BAS - Pythagorean Triples
' David O. Williams. 2004
' david.williams{at}ablelink.org
' Calculates and prints integer triples, A, B, C, such that
' A 1), and A^2 + B^2 = C^2.
' The list is printed in order of increasing A, then of B.
' A counter of triples is also shown.
' A more detailed explanation is at the end of the main module.
DECLARE FUNCTION NoComFacs# (X#, Y#)
DECLARE SUB PrintOut (A#, B#, C#)
DEFDBL A-Z
CLS
DO
PRINT "Both values must be >= 0, and Maximum >= Minimum"
PRINT
INPUT "Minimum value of smallest number in triple"; Mn
INPUT "Maximum value of smallest number in triple"; Mx
PRINT
Mx = INT(Mx)
Mn = INT(Mn)
IF Mn >= 0 AND Mx >= 0 AND Mx >= Mn THEN EXIT DO
BEEP
PRINT "Illegal value/s!"
PRINT
LOOP
T = Mx > Mn OR (Mn 4 AND Mn - 4 * INT(Mn / 4) 2)
IF Mx > 2 AND T THEN
PRINT "Count"; TAB(21); "A"; TAB(35); "B";
TAB(58); "C"
PRINT
ELSE
PRINT "There are no triples in this range!"
END
END IF
Z = SQR(2) + 1
H = SQR(Mn / Z)
T = 2 * INT(H / 2)
IF H - T > 1 THEN Q = T + 1 ELSE Q = T - 1
' Q is initial loop limit for odd cases
J = Q + 2
K = INT(J * J * Z)
R = INT(SQR(Mn / (Z + Z))) ' initial loop limit for even cases
L = R + 1
M = INT(L * L * Z)
' start of main loop
FOR A = Mn TO Mx
SELECT CASE A - 4 * INT(A / 4)
CASE 0
S = A / 2
IF S > M THEN
R = L
L = L + 1
M = INT(L * L * Z)
END IF
FOR F = R TO 1 STEP -1
G = S / F
IF G = INT(G) THEN
IF F + G - 2 * (INT(F / 2) + INT(G / 2)) = 1 THEN
IF NoComFacs(F, G) THEN
X = G * G
Y = F * F
B = X - Y
C = X + Y
PrintOut A, B, C
END IF
END IF
END IF
NEXT
CASE 1, 3
IF A > K THEN
Q = J
J = J + 2
K = INT(J * J * Z)
END IF
FOR D = Q TO 1 STEP -2
E = A / D
IF E = INT(E) THEN
IF NoComFacs(D, E) THEN
T = D * D
B = (E * E - T) / 2
C = B + T
PrintOut A, B, C
END IF
END IF
NEXT
END SELECT
NEXT
END
'----------------------------------------------------------
' Brief explanation:
' Pythagorean triples, if they are in lowest terms with no common
' factors, can be written as: Odd#^2 + Even#^2 = Big#^2. Big# is the
' largest integer (corresponding to the hypotenuse of the right-angled
' triangle). However, Odd# may be smaller or larger than Even#.
' If the three above numbers have no common factors, it is possible
' to define two odd positive integers, D and E, with E > D, such that:
' Odd# = D * E
' Even# = (E^2 - D^2) / 2
' Big# = (E^2 + D^2) / 2
' These definitions satisfy Pythagoras's Theorem. It is possible to
' prove that all valid triples, in lowest terms, can be written this
' way, with odd-integer values of D and E. (They must both be odd, so
' that their product is Odd#.)
' If a triple is written as A, B, C, with A < B < C, C must correspond
' to Big#, but A can be either Odd# or Even#, and B will be the other.
' The program treats these two possibilities separately. If A is odd,
' so it must be Odd#, the program simply searches for two odd integers
' whose product is A. After confirming that they have no common
' factors, which would mean that the triple is not in lowest terms,
' the program calculates B and C from them, and prints them out.
' If A is even, it is useful to define two further numbers, F and G,
' with G > F, such that:
' D = G - F
' E = F + G
' This means that:
' F = (E - D) / 2
' G = (D + E) / 2
' Since D and E are both odd, their sum and difference are both even,
' so F and G are integers. However, the sum and difference of F and G
' are E and D, which are odd, which implies that the parities of F
' and G must be opposites, so one is odd and the other even.
' Since, in this case, A is the same as Even#, it is given by:
' A = (E^2 - D^2) / 2
' Writing G - F for D and G + F for E, and simplifying, this gives:
' A / 2 = F * G
' Since F and G are of opposite parity, this means that A / 2 must be
' even, so A must be a multiple of 4. There are no valid triples when
' A MOD 4 = 2.
' When A is a multiple of 4, the program looks for factors of A / 2.
' It checks that they are of opposite parity, and have no common
' factors. It then uses these values of F and G to calculate B and
' C, using the easily-proved formulae:
' B = G^2 - F^2
' C = G^2 + F^2
' (Strictly, the parity check could be omitted. Since they are factors
' of an even number, at least one of the found values of F and G must
' be even. The possibility that they are both even would be detected
' by the common-factor test. However, the parity check is much simpler
' and faster than the common-factor routine, so having it in the
' program saves some time.)
' The FOR ... NEXT loops in the odd and even CASEs search for the
' factors D and E, or F and G, respectively, The loops run from
' high to low values of the counting variables, since this makes the
' triples appear in the desired order. Also, the ranges of the loops
' are limited so that only triples in which B > A appear. The
' variables Q and R govern these ranges. They are initialized near
' the start of the program, and are incremented in the main loop as
' the value of A increases. The numbers K and M are pre-calculated
' limits. When A (or S, which is just A / 2 when A is even) passes
' the relevant limit, Q or R is incremented appropriately, and a new
' value of K or M is calculated. In the vast majority of iterations
' around the loops, it is not necessary to increment Q or R, so all
' that has to be done is a simple comparison, e.g IF A > K THEN,
' which turns out not to be true. This saves a lot of time. Also,
' this method does not use the slow operation SQR inside any loops.
' A full mathematical treatment of the situation uses the number
' (SQR(2) + 1) several times. This number is therefore treated as a
' constant in the program, named Z.
' = end =
FUNCTION NoComFacs (X, Y) ' non-zero if X and Y have no common factors
U = Y
V = X
DO WHILE V > 1
W = U - V * INT(U / V)
U = V
V = W
LOOP
NoComFacs = V
END FUNCTION
SUB PrintOut (A, B, C)
STATIC N
N = N + 1
PRINT N; TAB(20); A; TAB(34); B; TAB(57); C
END SUB
----------------------------------------------------
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