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Cloned SEACAS for EXODUS library with extra build files for internal package management.
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EXODUS/packages/seacas/applications/fastq/picktr.f

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C Copyright(C) 1999-2020 National Technology & Engineering Solutions
C of Sandia, LLC (NTESS). Under the terms of Contract DE-NA0003525 with
C NTESS, the U.S. Government retains certain rights in this software.
C
C See packages/seacas/LICENSE for details
SUBROUTINE PICKTR (NPER, X, Y, NID, ANGLE, HALFC, I1, I2, I3, I4,
& I5, I6, I7, I8)
C***********************************************************************
C SUBROUTINE PICKTR = DETERMINES A REASONABLE SHAPE FOR A BACK-TO-BACK
C SET OF TRIANGLES (TRANSITION REGION)
C***********************************************************************
PARAMETER (RLARGE = 1000000.)
DIMENSION X(NPER), Y(NPER), NID(NPER), ANGLE(NPER)
DIMENSION SMANG(7), INDEX(7)
DIMENSION ISORT(4)
LOGICAL HALFC
PI = ATAN2(0.0, -1.0)
PID2 = 0.5 * PI
TWOPI = 2.0 * PI
C FORM THE LIST OF SMALLEST ANGLES
NSA = 6
DO 100 I = 1, NSA
SMANG(I) = 10.
INDEX(I) = 0
100 CONTINUE
AGOLD = ATAN2 (Y (1) - Y (NPER), X (1) - X (NPER))
DO 130 J = 1, NPER
C GET THE ANGLE FORMED BY THIS SET OF POINTS
NEXT = J + 1
IF (NEXT .GT. NPER) NEXT = 1
AGNEW = ATAN2 (Y (NEXT) - Y (J), X (NEXT) - X (J))
DIFF = AGNEW - AGOLD
IF (DIFF .GT. PI) DIFF = DIFF - TWOPI
IF (DIFF .LT. - PI) DIFF = DIFF + TWOPI
ANGLE (J) = PI - DIFF
AGOLD = AGNEW
C SORT THIS ANGLE AGAINST PREVIOUS ANGLES TO SEE IF IT IS ONE OF
C THE SMALLEST
SMANG (NSA + 1) = ANGLE (J)
INDEX (NSA + 1) = J
DO 110 II = 1, NSA
I = NSA + 1 - II
IF (SMANG (I + 1) .GE. SMANG (I)) GO TO 120
TEMP = SMANG (I)
ITEMP = INDEX (I)
SMANG (I) = SMANG (I + 1)
INDEX (I) = INDEX (I + 1)
SMANG (I + 1) = TEMP
INDEX (I + 1) = ITEMP
110 CONTINUE
120 CONTINUE
130 CONTINUE
C DETERMINE TWO/FOUR BEST CORNER POINTS FOR SEMICIRCLE/TRANSITION REGION
ATOL = PI * 150. / 180.
C FIND SIDE DIVISION USING 4 SMALLEST ANGLES AND CHECK CONDITION
DO 140 I = 1, 4
ISORT (I) = INDEX (I)
140 CONTINUE
DO 160 I = 1, 3
DO 150 J = I + 1, 4
IF (ISORT (I) .GT. ISORT (J)) THEN
ITMP = ISORT (I)
ISORT (I) = ISORT (J)
ISORT (J) = ITMP
ENDIF
150 CONTINUE
160 CONTINUE
I1 = ISORT (1)
I2 = ISORT (2)
I3 = ISORT (3)
I4 = ISORT (4)
M1 = I2 - I1
IF (M1 .LT. 0) M1 = NPER + M1
M2 = I3 - I2
IF (M2 .LT. 0) M2 = NPER + M2
M3 = I4 - I3
IF (M3 .LT. 0) M3 = NPER + M3
M4 = NPER - M1 - M2 - M3
C USE THE LONGEST SIDE THAT DOES NOT HAVE OPPOSITE
C MATCHES AS THE CHOICE FOR THE BASE (OF TRANSITIONS)
C THE BASE MUST BE AT LEAST 4 INTERVALS LONG
IF ( (M1 .EQ. M3) .AND. (.NOT. HALFC)) THEN
MMAX = MAX0 (M2, M4)
IF (MMAX .GE. 4) THEN
IF (M2 .EQ. MMAX) THEN
IFIRST = I2
MBASE = M2
ELSE
IFIRST = I4
MBASE = M4
ENDIF
ENDIF
ELSEIF ( (M2 .EQ. M4) .AND. (.NOT. HALFC)) THEN
MMAX = MAX0 (M1, M3)
IF (MMAX .GE. 4) THEN
IF (M1 .EQ. MMAX) THEN
IFIRST = I1
MBASE = M1
ELSE
IFIRST = I3
MBASE = M3
ENDIF
ENDIF
ELSE
MMAX = MAX0 (M1, M2, M3, M4)
IF (MMAX .GE. 4) THEN
IF (M1 .EQ. MMAX) THEN
IFIRST = I1
MBASE = M1
ELSEIF (M2 .EQ. MMAX) THEN
IFIRST = I2
MBASE = M2
ELSEIF (M3 .EQ. MMAX) THEN
IFIRST = I3
MBASE = M3
ELSEIF (M4 .EQ. MMAX) THEN
IFIRST = I4
MBASE = M4
ENDIF
ENDIF
ENDIF
IF (MMAX .GE. 4) THEN
IF (HALFC) THEN
GBEST = ANGLE (I1) + ANGLE (I2) + ABS (PI - ANGLE (I3))
& + ABS (PI - ANGLE (I4))
ELSE
GBEST = ANGLE (I1) + ANGLE (I2) + ANGLE (I3) + ANGLE (I4)
ENDIF
ELSE
IFIRST = 1
GBEST = RLARGE
END IF
C GO AROUND THE PERIMETER USING THE 6 SMALLEST ANGLES AS POSSIBLE
C STARTING POINTS, AND THEN FIND THE BEST COMBINATION OF SIDE LENGTHS
DO 200 ISA = 1, NSA
IF (SMANG (ISA) .LE. ATOL) THEN
I1 = INDEX (ISA)
SUM1 = ANGLE (I1)
IF (HALFC) THEN
IF (SUM1 .GT. GBEST) GO TO 200
ELSE
IF (SUM1 .GE. GBEST) GO TO 200
ENDIF
C ASSIGN A TRIAL SECOND NODE
DO 190 N1 = 1, NPER - 4
I2 = I1 + N1
IF (I2 .GT. NPER) I2 = I2 - NPER
SUM2 = SUM1 + ANGLE (I2)
IF (HALFC) THEN
IF (SUM2 .GT. GBEST) GO TO 190
ELSE
IF (SUM2 .GE. GBEST) GO TO 190
ENDIF
C ASSIGN A TRIAL THIRD NODE
DO 180 N2 = 1, NPER - N1 - 3
I3 = I2 + N2
IF (I3 .GT. NPER) I3 = I3 - NPER
IF (HALFC) THEN
SUM3 = SUM2 + ABS (PI - ANGLE (I3))
ELSE
SUM3 = SUM2 + ANGLE (I3)
END IF
IF (HALFC) THEN
IF (SUM3 .GT. GBEST) GO TO 180
ELSE
IF (SUM3 .GE. GBEST) GO TO 180
ENDIF
C ASSIGN A TRIAL FOURTH NODE
DO 170 N3 = 1, NPER - N1 - N2 - 2
I4 = I3 + N3
IF (I4 .GT. NPER) I4 = I4 - NPER
IF (HALFC) THEN
GVAL = SUM3 + ABS (PI - ANGLE (I4))
ELSE
GVAL = SUM3 + ANGLE (I4)
END IF
IF (HALFC) THEN
IF (GVAL .GT. GBEST) GO TO 170
ELSE
IF (GVAL .GE. GBEST) GO TO 170
ENDIF
C FIND SIDE DIVISION AND CHECK CONDITION
M1 = I2 - I1
IF (M1 .LT. 0) M1 = NPER + M1
M2 = I3 - I2
IF (M2 .LT. 0) M2 = NPER + M2
M3 = I4 - I3
IF (M3 .LT. 0) M3 = NPER + M3
M4 = NPER - M1 - M2 - M3
C USE THE LONGEST SIDE THAT DOES NOT HAVE OPPOSITE
C MATCHES AS THE CHOICE FOR THE BASE (OF TRANSITIONS)
C THE BASE MUST BE AT LEAST 4 INTERVALS LONG
IF ( (M1 .EQ. M3) .AND. (.NOT. HALFC)) THEN
MMAX = MAX0 (M2, M4)
IF (MMAX .GE. 4) THEN
IF (M2 .EQ. MMAX) THEN
IFIRST = I2
MBASE = M2
ELSE
IFIRST = I4
MBASE = M4
ENDIF
ENDIF
ELSEIF ( (M2 .EQ. M4) .AND. (.NOT. HALFC)) THEN
MMAX = MAX0 (M1, M3)
IF (MMAX .GE. 4) THEN
IF (M1 .EQ. MMAX) THEN
IFIRST = I1
MBASE = M1
ELSE
IFIRST = I3
MBASE = M3
ENDIF
ENDIF
ELSE
MMAX = MAX0 (M1, M2, M3, M4)
IF (MMAX .GE. 4) THEN
IF (M1 .EQ. MMAX) THEN
IFIRST = I1
MBASE = M1
ELSEIF (M2 .EQ. MMAX) THEN
IFIRST = I2
MBASE = M2
ELSEIF (M3 .EQ. MMAX) THEN
IFIRST = I3
MBASE = M3
ELSEIF (M4 .EQ. MMAX) THEN
IFIRST = I4
MBASE = M4
ENDIF
ENDIF
IF (MMAX .GE. 4)GBEST = GVAL
ENDIF
170 CONTINUE
180 CONTINUE
190 CONTINUE
ENDIF
200 CONTINUE
C ROTATE THE PERIMETER AND THE ANGLES SO THE BASE LEADS THE LIST
IF (IFIRST .NE. 1) CALL FQ_ROTATE (NPER, X, Y, NID, IFIRST)
DO 220 I = 1, IFIRST - 1
AHOLD = ANGLE (1)
DO 210 J = 1, NPER - 1
ANGLE (J) = ANGLE (J + 1)
210 CONTINUE
ANGLE (NPER) = AHOLD
220 CONTINUE
C DECIDE THE TRIANGLE CORNERS
GBEST = RLARGE
C PICK AN ARBITRARY BASE CENTER (I3)
DO 250 I = 3, MBASE - 1
C FOR THIS BASE CENTER, PICK AN ARBITRARY I2 LOCATION
DO 240 J = 2, I - 1
C FOR THIS COMBINATION OF I3 AND I2, PICK AN ARBITRARY I4 LOCATION
DO 230 K = I + 1, MBASE
C CALCULATE I6 AND I8 AND ADD ANGLES TO FIND MINIMUM SUM
KN = MBASE + 1 - K
KK = I - J
KL = J - 1
KM = K - I
MLEFT = NPER - MBASE
KO = (MLEFT - KN + KL - KK - KM) / 2
KP = KN + KO - KL
C PROTECT AGAINST THE IMPOSSIBLE LENGTH PROBLEMS
C AND THE ODD NUMBER IN THE PERIMETER INPUT ERRORS
IF ( (KO .GT. 0) .AND. (KP .GT. 0)) THEN
IF (KP + KL .EQ. KN + KO) THEN
C NOW GET THE END POINTS GIVEN THESE SIDE LENGTHS
J6 = MBASE + 1 + KO
J7 = MBASE + 1 + KO + KM
J8 = MBASE + 1 + KO + KM + KK
C GET THE BASE ANGLE OF THE DIVIDER LINE
THETA1 = ATAN2 (Y (I + 1) - Y (I),
& X (I + 1) - X (I))
THETA2 = ATAN2 (Y (J7) - Y (I), X (J7) - X (I))
THETAB = ABS (THETA2 - THETA1)
IF (THETAB .GT. PI) THETAB = THETAB - PI
IF (THETAB .LT. PID2) THETAB = PI - THETAB
C GET THE TOP ANGLE OF THE DIVIDER LINE
THETA1 = ATAN2 (Y (J7 + 1) - Y (J7),
& X (J7 + 1) - X (J7))
THETAT = ABS (THETA2 - THETA1)
IF (THETAT .GT. PI) THETAT = THETAT - PI
IF (THETAT .LT. PID2) THETAT = PI - THETAT
C ADD THESE TO GET THE VALUE OF GVAL
IF (HALFC) THEN
GVAL = THETAB + THETAT + ABS (PI - ANGLE (J6))
& + ABS (PI - ANGLE (J8)) +
& (.1 * MAX0 (ABS (KK - KL),
& ABS (KM - KN)) / NPER)
ELSE
GVAL = ANGLE (J6) + ANGLE (J8) + THETAB + THETAT
ENDIF
IF (GVAL .LT. GBEST) THEN
GBEST = GVAL
I1 = 1
I2 = J
I3 = I
I4 = K
I5 = MBASE + 1
I6 = I5 + KO
I7 = I6 + KM
I8 = I7 + KK
ENDIF
ELSE
CALL MESSAGE('ODD PERIMETER PROBLEMS')
ENDIF
ENDIF
230 CONTINUE
240 CONTINUE
250 CONTINUE
RETURN
END