CTAX7 00050 C DONT BE FRIGHTENED, ITS MOSTLY COMMENT CARDS 00070 0DIMENSION B(40,2),IC(79,70),ID(10,10,70),IE(10,10,70),IG(9),IH(79) 00080 1,IJ(79),IK(79),IR(79),IL(79),ALPH(100),CAL(10) 00090 CAL(1)=1H0 00100 CAL(2)=1H1 00110 CAL(3)=1H2 00120 CAL(4)=1H3 00130 CAL(5)=1H4 00140 CAL(6)=1H5 00150 CAL(7)=1H6 00160 CAL(8)=1H7 00170 CAL(9)=1H8 00180 C READ CARD CONTAINING NUMBER OF OTUS AND CHARACTERS 00190 1 READ INPUT TAPE 5,2,NS,NC,INDIC 00200 2 FORMAT(3I4) 00210 WRITE OUTPUT TAPE 6,830 00220 830 FORMAT(1H1,21HCHARACTER STATE TREES//1X,4HCHAR,2X,17H0 1 2 3 4 5 6 00230 1 7 8,2X,6HDELETE/2X) 00240 C NM IS THE NUMBER OF OTUS PLUS THE NUMBER OF HYPOTHETICAL ANCESTORS 00250 NM=2*NS-1 00260 C FOR EACH CHARACTER READ IN CODED CHARACTER STATE TREE, SET UP THE 00270 C CORRESPONDING PORTIONS OF THE ARRAYS ID AND IE 00280 450 DO 23 JA=1,NC 00290 READ INPUT TAPE 5,5,(IG(I),I=1,9),JB 00300 5 FORMAT( 10I2) 00310 DO 837 I=1,9 00320 IF(IG(I)-10) 834,832,834 00330 832 ALPH(I)=1H* 00340 GO TO 837 00350 834 IF(IG(I)) 835,836,836 00360 835 ALPH(I)=1H 00370 GO TO 837 00380 836 J=IG(I)&1 00390 ALPH(I)=CAL(J) 00400 837 CONTINUE 00410 WRITE OUTPUT TAPE 6,838,JA,(ALPH(I),I=1,9),JB 00420 838 FORMAT(1X,I3,2X,9(1X,A1),5X,I1) 00430 C IF CHARACTER IS TO BE IGNORED THEN JB=0, SO SET ID AND IE ENTRIES ZER 00440 IF(JB) 6,8,8 00450 6 DO 7 JC=1,10 00460 DO 7 JD=1,10 00470 ID(JC,JD,JA)=0 00480 7 IE(JC,JD,JA)=0 00490 GO TO 23 00500 8 IF(JB-1) 85,6,6 00510 85 DO 9 JC=1,9 00520 DO 9 JD=1,9 00530 9 ID(JC,JD,JA)=-1 00540 C THE STATE ANCESTRAL TO STATES 9 AND JC-1 IS STATE JC-1, THE NUMBER OF 00550 C TO EVOLVE STATE JC-1 FROM STATE 9 IS ZERO 00560 DO 10 JC=1,10 00570 IE(10,JC,JA)=JC-1 00580 IE(JC,10,JA)=JC-1 00590 ID(JC,10,JA)=0 00600 10 ID(10,JC,JA)=0 00610 C FROM EACH STATE WHICH IS PART OF THE CHARACTER STATE TREE, GO DOWN TH 00620 C TREE STEP BY STEP, SETTING UP THE CORRESPONDING ENTRIES OF ID AS YOU 00630 DO 14 JC=1,9 00640 IF(IG(JC)) 14,11,11 00650 11 ID(JC,JC,JA)=0 00660 JD=0 00670 JE=JC 00680 12 IF(IG(JE)-10) 13,14,14 00690 13 JD=JD&1 00700 JE=IG(JE)&1 00710 ID(JC,JE,JA)=JD 00720 GO TO 12 00730 14 CONTINUE 00740 C FOR EACH PAIR OF STATES JC-1 AND JD-1 WHICH ARE IN THE CHARACTER STAT 00750 C TREE EXAMINE ALL OTHER STATES JF-1 WHICH ARE IN THE TREE AND PICK AS 00760 C COMMON ANCESTOR OF JC-1 AND JD-1 THE STATE WHICH HAS THE SMALLEST 00770 C SUM OF THE DISTANCE DOWN THE TREE FROM JC-1 AND FROM JD-1. THIS IS TH 00780 C ENTRY USED IN IE 00790 DO 22 JC=1,9 00800 IF(IG(JC)) 22,15,15 00810 15 DO 21 JD=1,9 00820 IF(IG(JD)) 21,16,16 00830 16 JE=10 00840 DO 20 JF=1,9 00850 IF(ID(JC,JF,JA)) 20,17,17 00860 17 IF(ID(JD,JF,JA)) 20,18,18 00870 18 IF(ID(JC,JF,JA)&ID(JD,JF,JA)-JE) 19,20,20 00880 19 JE=ID(JC,JF,JA)&ID(JD,JF,JA) 00890 IE(JC,JD,JA)=JF-1 00900 20 CONTINUE 00910 21 CONTINUE 00920 22 CONTINUE 00930 C PREPARE ARRAY GIVING PRIMITIVE STATE FOR EACH CHARACTER 00940 DO 256 I=1,9 00950 IF(IG(I)-10) 256,255,256 00960 255 IR(JA)=I-1 00970 256 CONTINUE 00980 C NOW GO BACK AND TAKE THE NEXT CHARACTER 00990 23 CONTINUE 01000 WRITE OUTPUT TAPE 6,799 01010 C FOR EACH OTU READ NAME AND CHARACTERS 01020 799 FORMAT(1H1,5X,4HNAME,20X,10HCHARACTERS/2X) 01030 DO 800 I=1,40 01040 DO 800 J=1,70 01050 800 IC(I,J)=9 01060 DO 813 I=1,NS 01070 READ INPUT TAPE 5,801,AAN,BBN,(IK(J),J=1,NC) 01080 801 FORMAT(2A5,70I1) 01090 IF(AAN-5H*****) 810,802,810 01100 802 READ INPUT TAPE 5,803,ISP,AAN,BBN 01110 803 FORMAT(I4,2A5) 01120 WRITE OUTPUT TAPE 6,804,AAN,BBN,ISP 01130 804 FORMAT(2X/1X,8HGROUP - ,5X,2A5,5X,4HNEXT,I4/2X) 01140 B(I,1)=AAN 01150 B(I,2)=BBN 01160 KBB=1 01170 LUP=ISP 01180 855 IF(KBB-LUP) 805,805,856 01190 805 READ INPUT TAPE 5,801,AAN,BBN,(IK(JA),JA=1,NC) 01200 IF(AAN-5H*****) 854,852,854 01210 852 READ INPUT TAPE 5,803,NGG,AAN,BBN 01220 WRITE OUTPUT TAPE 6,853,AAN,BBN,NGG 01230 853 FORMAT(1X,8HGROUP - ,5X,2A5,5X,4HNEXT,I4) 01240 LUP=LUP-1&NGG 01250 GO TO 805 01260 854 DO 806 K=1,NC 01270 KKK=IK(K)&1 01280 KKL=IC(I,K)&1 01290 806 IC(I,K)=IE(KKK,KKL,K) 01300 WRITE OUTPUT TAPE 6,807,AAN,BBN,(IK(JA),JA=1,NC) 01310 807 FORMAT(1X,2A5,5X,14(1X,5I1)) 01320 KBB=KBB&1 01330 GO TO 855 01340 856 WRITE OUTPUT TAPE 6,809 01350 809 FORMAT(120X/2X) 01360 GO TO 813 01370 810 DO 811 J=1,NC 01380 811 IC(I,J)=IK(J) 01390 B(I,1)=AAN 01400 B(I,2)=BBN 01410 WRITE OUTPUT TAPE 6,807,AAN,BBN,(IK(JA),JA=1,NC) 01420 813 CONTINUE 01430 C IF INDIC IS 1,ORDER OTU@S IN ORDER OF INCREASING PRIMITIVENESS 01440 IF(INDIC) 826,826,814 01450 814 DO 816 I=1,NS 01460 IH(I)=0 01470 DO 816 J=1,NC 01480 IF(IC(I,J)-IR(J)) 816,815,816 01490 815 IH(I)=IH(I)&1 01500 816 CONTINUE 01510 821 I=1 01520 823 IF(I-NS&1) 817,817,824 01530 817 IF(I) 821,821,818 01540 818 IF(IH(I)-IH(I&1)) 822,822,819 01550 819 DO 820 J=1,NC 01560 III=IC(I,J) 01570 IC(I,J)=IC(I&1,J) 01580 820 IC(I&1,J)=III 01590 AAN=B(I,1) 01600 B(I,1)=B(I&1,1) 01610 B(I&1,1)=AAN 01620 AAN=B(I,2) 01630 B(I,2)=B(I&1,2) 01640 B(I&1,2)=AAN 01650 III=IH(I) 01660 IH(I)=IH(I&1) 01670 IH(I&1)=III 01680 I=I-1 01690 GO TO 817 01700 822 I=I&1 01710 GO TO 823 01720 824 WRITE OUTPUT TAPE 6,825 01730 825 FORMAT(2X//1X,20HOTU ORDER REARRANGED) 01740 C INITIALIZE THE ARRAYS IH AND IR. THE INITIAL TREE HAS OTU 1 WITH ITS 01750 C BEING NM&1, WHICH IS A NONEXISTENT COMMON ANCESTOR OF THE ENTIRE GROU 01760 C NOTE THAT ONE OF THE ANCESTORS NUMBERED NS&1 TO NM WILL ALSO BE ANCES 01770 C THE ENTIRE GROUP 01780 826 IH(1)=NM&1 01790 IR(1)=0 01800 IJ(1)=0 01810 DO 100 JA=2,NM 01820 IR(JA)=-1 01830 100 IH(JA)=0 01840 C WE WILL GO THROUGH THE FOLLOWING LOOP EACH TIME WE ADD ANOTHER OTU TO 01850 C TREE. EACH OTU IS ADDED TO WHATEVER POSITION MAKES THE SMALLEST EXTRA 01860 C NUMBER OF STEPS NECESSARY. THAT IS, WE TRY TO MINIMIZE THE ADDITIONAL 01870 C WHICH MUST BE ADDED TO THE PREVIOUS TREE TO ACCOUNT FOR EVOLUTION OF 01880 C NEWLY ADDED OTU 01890 DO 120 JA=2,NS 01900 C IBEST IS THE NUMBER OF EXTRA STEPS REQUIRED AT THE BEST POSITION 01910 C FOUND SO FAR. IT IS INITIIALIZED AT 1000 01920 IBEST=1000 01930 C IBN TELLS US THE NUMBER OF THE OTU OR HYPOTHETICAL ANCESTOR IMMEDIATE 01940 C BELOW WHICH IS THE BEST POSITION FOUND SO FAR 01950 IBN=1 01960 C DO OVER ALL OTUS AND ANCESTORS WHICH ARE ALREADY IN THE TREE 01970 DO 112 JB=1,NM 01980 IF(IR(JB)) 112,101,101 01990 101 ITOT=0 02000 C CALCULATE THE STEPS FROM THE OTU BEING ADDED TO ITS COMMON ANCESTOR W 02010 C OTU JB, AND FROM JB TO THE COMMON ANCESTOR 02020 DO 102 JC=1,NC 02030 KA=IC(JA,JC)&1 02040 KB=IC(JB,JC)&1 02050 IK(JC)=IE(KA,KB,JC) 02060 KC=IK(JC)&1 02070 102 ITOT=ITOT&ID(KA,KC,JC)&ID(KB,KC,JC) 02080 LA=JB 02090 C NOW CONTINUE DOWN THE TREE CALCULATING FOR EACH ANCESTOR LB THE EXTRA 02100 C STEPS NEEDED BETWEEN IT AND ITS IMMEDIATE ANCESTOR LA, 02110 103 LB=LA 02120 LA=IH(LA) 02130 IF(LA-NM-1) 104,110,110 02140 104 DO 105 JC=1,NC 02150 KA=IK(JC)&1 02160 KB=IC(LA,JC)&1 02170 IK(JC)=IE(KA,KB,JC) 02180 KC=IK(JC)&1 02190 105 ITOT=ITOT&ID(KA,KC,JC) 02200 C AND THE EXTRA STEPS NEEDED ON THE OTHER BRANCH WHICH CONVERGES ON LA 02210 DO 106 JC=1,NM 02220 IF(IH(JC)-LA) 106,107,106 02230 107 IF(JC-LB) 108,106,108 02240 106 CONTINUE 02250 108 DO 109 JD=1,NC 02260 KA=IC(JC,JD)&1 02270 KB=IK(JD)&1 02280 109 ITOT=ITOT&ID(KA,KB,JD) 02290 C ACTUALLY WE HAVE BEEN CALCULATING THE TOTAL STEPS NEEDED, NOW WE SUBT 02300 C PREVIOUS STEPS TO GET THE NUMBERS AS EXTRA STEPS 02310 ITOT=ITOT-IJ(JC) 02320 ITOT=ITOT-IJ(LB) 02330 GO TO 103 02340 C IT MAY SEEM THAT AT THE VERY BEGINNING OF THIS LAST PROCEDURE, AROUND 02350 C STATEMENT 102 A MISTAKE WAS MADE. BUT THE PROCEDURE IS CORRECT. 02360 C E PUR SI MUOVE 02370 110 IF(ITOT-IBEST) 111,112,112 02380 111 IBEST=ITOT 02390 IBN=JB 02400 112 CONTINUE 02410 C NOW WE WANT TO ADD THE NEW OTU TO ITS BEST POSITION. AGAIN WE MUST GO 02420 C DOWN THE TREE A STEP AT A TIME, MAKING MANY OF THE SAME CALCULATIONS, 02430 C ONLY THIS TIME RECORDING THE STEPS NEEDED AND THE PHENOTYPES OF THE A 02440 C FOR POSTERITY 02450 LA=NS&JA-1 02460 IH(LA)=IH(IBN) 02470 IH(JA)=LA 02480 IH(IBN)=LA 02490 IR(JA)=0 02500 IR(LA)=0 02510 DO 121 JB=1,NC 02520 KA=IC(JA,JB)&1 02530 KB=IC(IBN,JB)&1 02540 121 IC(LA,JB) = IE(KA,KB,JB) 02550 LA=JA 02560 113 LB=LA 02570 LA=IH(LA) 02580 IF(LA-NM-1) 114,120,120 02590 114 ISUM=0 02600 DO 115 JC=1,NC 02610 KA=IC(LB,JC)&1 02620 KB=IC(LA,JC)&1 02630 IC(LA,JC)=IE(KA,KB,JC) 02640 KC=IC(LA,JC)&1 02650 115 ISUM=ISUM&ID(KA,KC,JC) 02660 IJ(LB)=ISUM 02670 DO 116 JB=1,NM 02680 IF(IH(JB)-LA) 116,117,116 02690 117 IF(JB-LB) 118,116,118 02700 116 CONTINUE 02710 118 ISUM=0 02720 DO 119 JC=1,NC 02730 KA=IC(JB,JC)&1 02740 KB=IC(LA,JC)&1 02750 119 ISUM=ISUM&ID(KA,KB,JC) 02760 IJ(JB)=ISUM 02770 GO TO 113 02780 120 IJ(LB)=0 02790 C NOW WE HAVE CONSTRUCTED THE TREE WHICH WE WILL REARRANGE. IF ONE IS A 02800 C TO TERMINOLOGY ONE CAN CALL IT THE PROCLADOGRAM 02810 C WE WILL DO THE FOLLOWING OVER EACH ANCESTOR IN THE RANGE NS&1 TO NM, 02820 C THE ONE AT THE BOTTOM OF THE TREE WHOSE IMMEDIATE ANCESTOR IS THE NON 02830 C FORM NM&1 02840 33 JA=NS&1 02850 34 IF(JA-NM) 345,345,49 02860 345 IF(IH(JA)-NM-1) 35,482,482 02870 C WE FIND THE TWO FORMS WHICH HAVE JA AS THEIR ANCESTOR, AND THE FORM 02880 C WHICH IS ANCESTRAL TO JA, AND THE FORM WHICH HAS THE SAME ANCESTOR AS 02890 C THESE ARE RESPECTIVELY JB, JD, JF, AND JE 02900 35 DO 37 JB=1,NM 02910 IF(IH(JB)-JA) 37,38,37 02920 37 CONTINUE 02930 38 JC=JB&1 02940 DO 39 JD=JC,NM 02950 IF(IH(JD)-JA) 39,40,39 02960 39 CONTINUE 02970 40 DO 42 JE=1,NM 02980 IF(JE-JA) 41,42,41 02990 41 IF(IH(JE)-IH(JA)) 42,43,42 03000 42 CONTINUE 03010 43 JF=IH(JA) 03020 C NOW WE TRY THE ARRANGEMENT IN WHICH JB BRANCHES OFF FIRST, FOLLOWED 03030 C BY JD AND JE 03040 C WE CALCULATE THE STEPS REQUIRED IN THIS ARRANGEMENT 03050 JG=1 03060 435 DO 44 KA=1,NC 03070 KB=IC(JD,KA)&1 03080 KC=IC(JE,KA)&1 03090 IK(KA)=IE(KB,KC,KA) 03100 KB=IC(JB,KA)&1 03110 KC=IK(KA)&1 03120 44 IL(KA)=IE(KB,KC,KA) 03130 KA=0 03140 KB=0 03150 KC=0 03160 KE=0 03170 DO 45 KD=1,NC 03180 LA=IK(KD)&1 03190 LB=IL(KD)&1 03200 LC=IC(JD,KD)&1 03210 LD=IC(JE,KD)&1 03220 LE=IC(JB,KD)&1 03230 KA=KA&ID(LC,LA,KD) 03240 KB=KB&ID(LD,LA,KD) 03250 KC=KC&ID(LA,LB,KD) 03260 45 KE=KE&ID(LE,LB,KD) 03270 C NOW WE TEST TO SEE WHETHER THIS IS AN IMPROVEMENT 03280 IF(IJ(JD)&IJ(JE)&IJ(JA)&IJ(JB)-KA-KB-KC-KE) 48,48,46 03290 C IF IT IS, RETAIN IT AND GO BACK AND START REARRANGING BRANCHES AGAIN 03300 46 IH(JE)=JA 03310 IH(JB)=JF 03320 IJ(JD)=KA 03330 IJ(JE)=KB 03340 IJ(JB)=KE 03350 IJ(JA)=KC 03360 DO 47 KD=1,NC 03370 47 IC(JA,KD)=IK(KD) 03380 GO TO 33 03390 C IF IT IS NOT AN IMPROVEMENT,AND IF WE HAVE NOT YET TRIED THE ALTERNAT 03400 C REARRANGEMENT, TRY IT BY SWITCHING JB AND JD AND GOING BACK 03410 48 IF(JG-1) 481,481,482 03420 481 JG=2 03430 KA=JB 03440 JB=JD 03450 JD=KA 03460 GO TO 435 03470 C IF IT IS NOT AN IMPROVEMENT AND WE HAVE TRIED BOTH REARRANGEMENTS, GO 03480 C TO THE NEXT ANCESTOR 03490 482 JA=JA&1 03500 GO TO 34 03510 C NOW WE HAVE OUR TREE. THE REST IS OUTPUT. 03520 49 WRITE OUTPUT TAPE 6,50 03530 50 FORMAT(1H1,3X,4HNAME,3X,8HANCESTOR,2X,5HSTEPS,20X,10HCHARACTERS/2X 03540 1) 03550 DO 52 JA=1,NS 03560 0WRITE OUTPUT TAPE 6,51,B(JA,1),B(JA,2),IH(JA),IJ(JA),(IC(JA,N), 03570 1N=1,NC) 03580 51 FORMAT(1X,2A5,5X,I2,2X,I3,5X,14(5I1,1X)) 03590 52 CONTINUE 03600 WRITE OUTPUT TAPE 6,53 03610 53 FORMAT(1X/20X,22HHYPOTHETICAL ANCESTORS/2X) 03620 JA=NS&1 03630 DO 55 JB=JA,NM 03640 WRITE OUTPUT TAPE 6,54,JB,IH(JB),IJ(JB),(IC(JB,N),N=1,NC) 03650 54 FORMAT(1X,5X,I2,8X,I2,2X,I3,5X,14(5I1,1X)) 03660 55 CONTINUE 03670 IS=0 03680 DO 57 IT=1,NM 03690 57 IS=IS&IJ(IT) 03700 WRITE OUTPUT TAPE 6,58,IS 03710 58 FORMAT(1X/1X,13X,6HTOTAL ,I3) 03720 C WRITE NUMBER OF STEPS IN EACH CHARACTER FROM THE IMMEDIATE ANCESTOR OF 03730 C SPECIES OR HYPOTHETICAL ANCESTOR 03740 WRITE OUTPUT TAPE 6,350 03750 350 FORMAT(1H1,3X,4HNAME,3X,8HANCESTOR,2X,5HSTEPS,20X,47HSTEPS FROM IM 03760 1MEDIATE ANCESTOR IN EACH CHARACTER/2X) 03770 DO 352 IA=1,NS 03780 JB=IH(IA) 03790 DO 360 IB=1,NC 03800 JA=IC(IA,IB)&1 03810 JC=IC(JB,IB)&1 03820 IK(IB)=ID(JA,JC,IB) 03830 360 CONTINUE 03840 WRITE OUTPUT TAPE 6,351,B(IA,1),B(IA,2),IH(IA),IJ(IA),(IK(I),I=1, 03850 1NC) 03860 351 FORMAT(1X,2A5,5X,I2,2X,I3,5X,14(5I1,1X)) 03870 352 CONTINUE 03880 WRITE OUTPUT TAPE 6,355 03890 355 FORMAT(1X,120X) 03900 NN=NS&1 03910 DO 359 IA=NN,NM 03920 JB=IH(IA) 03930 DO 357 IB=1,NC 03940 IF(JB-NM-1) 356,354,354 03950 354 IK(IB)=0 03960 GO TO 357 03970 356 JA=IC(IA,IB)&1 03980 JC=IC(JB,IB)&1 03990 IK(IB)=ID(JA,JC,IB) 04000 357 CONTINUE 04010 WRITE OUTPUT TAPE 6,358,IA,JB,IJ(IA),(IK(I),I=1,NC) 04020 358 FORMAT(1X,I5,10X,I2,2X,I3,5X,14(5I1,1X)) 04030 359 CONTINUE 04040 C PREPARE TREE SHOWING FINAL RESULT 04050 DO 850 I=1,NM 04060 IJ(I)=I 04070 IK(I)=1 04080 850 IL(I)=0 04090 DO 302 I=1,NS 04100 KA=NS&1-I 04110 K=KA 04120 300 IF(K-NM-1) 301,302,302 04130 301 IL(K)=KA 04140 K=IH(K) 04150 GO TO 300 04160 302 CONTINUE 04170 DO 304 I=1,NM 04180 DO 304 J=1,NM 04190 IF(IH(J)-NM-1) 303,304,304 04200 303 K=IH(J) 04210 IK(K)=XMAX0F(IK(J)&1,IK(K)) 04220 304 CONTINUE 04230 DO 338 I=2,NS 04240 KA=NS&2-I 04250 K=KA 04260 335 IF(IL(K)-KA) 337,336,336 04270 336 K=IH(K) 04280 GO TO 335 04290 337 IL(KA)=IL(K) 04300 338 IK(KA)=IK(K) 04310 KA=1 04320 305 KB=2 04330 306 IF(KB-NS&1) 307,307,313 04340 307 IF(IL(KB&1)-KA) 312,308,312 04350 308 IF(IL(KB&1)-IJ(KB)) 309,312,309 04360 309 IF(IL(KB&1)-IL(KB)) 311,310,311 04370 310 IF(IK(KB&1)-IK(KB)) 311,312,312 04380 311 KC=IJ(KB) 04390 IJ(KB)=IJ(KB&1) 04400 IJ(KB&1)=KC 04410 KC=IK(KB) 04420 IK(KB)=IK(KB&1) 04430 IK(KB&1)=KC 04440 KC=IL(KB) 04450 IL(KB)=IL(KB&1) 04460 IL(KB&1)=KC 04470 GO TO 305 04480 312 KB=KB&1 04490 GO TO 306 04500 313 KA=KA&1 04510 IF(KA-NS) 305,314,314 04520 314 WRITE OUTPUT TAPE 6,315,B(1,1),B(1,2) 04530 315 FORMAT(1H1/1X,100(1H-),2X,2A5) 04540 KC=1 04550 DO 317 KA=2,NS 04560 IF(IK(KA)-KC) 317,317,316 04570 316 KC=IK(KA) 04580 317 CONTINUE 04590 DO 325 KA=2,NS 04600 LL=IK(KA) 04610 KD=(100*(KC-LL&1))/KC 04620 KE=KD-1 04630 KF=KD&1 04640 DO 318 KG=1,KE 04650 318 ALPH(KG)=1H 04660 ALPH(KD)=1HL 04670 DO 319 KG=KF,100 04680 319 ALPH(KG)=1H- 04690 IF(KA-NS) 320,323,323 04700 320 KB=KA&1 04710 KD=IK(KA) 04720 DO 322 KE=KB,NS 04730 IF(IK(KE)-KD) 322,322,321 04740 321 LL=IK(KE) 04750 KF=(100*(KC-LL&1))/KC 04760 ALPH(KF)=1HI 04770 KD=LL 04780 322 CONTINUE 04790 323 KD=IJ(KA) 04800 WRITE OUTPUT TAPE 6,324,(ALPH(I),I=1,100),B(KD,1),B(KD,2) 04810 324 FORMAT(1X,100A1,2X,2A5) 04820 325 CONTINUE 04830 WRITE OUTPUT TAPE 6,56 04840 C INSTRUCT PRINTER TO SKIP TO TOP OF NEXT PAGE. READ IN ANOTHER SET OF 04850 56 FORMAT(1H1) 04860 GO TO 1 04870 END 04880