A Collection of Code Snippets in as Many Programming Languages as Possible

This project is maintained by TheRenegadeCoder

Welcome to the Longest Common Subsequence in Commodore Basic page! Here, you'll find the source code for this program as well as a description of how the program works.

```
10 DIM A(31)
15 REM Storage for lists
20 DIM L1(31), L2(31)
25 REM Storage for subsequences:
26 REM - C%(i, j, 0) contains length of subsequence
27 REM - C%(i, j, 1:2) contains bitmap indicating which indices of first list
28 REM are in the subsequence
30 DIM C%(32, 32, 2)
35 REM Bit 0 to 15
40 DATA 1, 2, 4, 8, 16, 32, 64, 128
50 DATA 256, 512, 1024, 2048, 4096, 8192, 16384, -32768
60 DIM WM%(15)
70 FOR I = 0 TO 15
80 READ WM%(I)
90 NEXT I
95 REM Get list 1
100 GOSUB 2000
110 IF V = 0 OR C >= 0 THEN 300: REM invalid or not end of input/value
120 M = NA
130 FOR I = 0 TO NA - 1
140 L1(I) = A(I)
150 NEXT I
155 REM Get list 2
160 GOSUB 2000
170 IF V = 0 OR C >= 0 THEN 300: REM invalid or not end of input/value
180 N = NA
190 FOR I = 0 TO NA - 1
200 L2(I) = A(I)
210 NEXT I
215 REM Find longest common subsequence and display
220 GOSUB 3000
230 GOSUB 3500
240 END
300 Q$ = CHR$(34): REM quote
310 PRINT "Usage: please provide two lists in the format ";
320 PRINT Q$; "1, 2, 3, 4, 5"; Q$
330 END
1000 REM Read input value one character at a time since Commodore BASIC
1001 REM has trouble reading line from stdin properly
1002 REM NR = number
1003 REM V = 1 if valid number, 0 otherwise
1004 REM C = -2 if end of input, -1 if end of value,
1005 REM 32 if whitespace, ASCII code of last character otherwise
1006 REM Initialize
1010 NR = 0
1020 V = 0
1030 S = 1
1035 REM Loop while leading spaces
1040 GOSUB 1500
1050 IF C = 43 OR C = 45 THEN GOTO 1100: REM + or -
1060 IF C >= 48 AND C <= 57 THEN GOTO 1150: REM 0 to 9
1070 IF C = 32 THEN GOTO 1040: REM whitespace
1080 RETURN: REM other character
1085 REM Loop while sign
1090 GOSUB 1500
1100 IF C = 43 THEN GOTO 1090: REM +
1110 IF C >= 48 AND C <= 57 THEN GOTO 1150: REM 0 to 9
1120 IF C <> 45 THEN RETURN: REM not -
1130 S = -S
1140 GOTO 1090
1145 REM Set valid flag
1150 V = 1
1155 REM Loop while digits
1160 NR = (ABS(NR) * 10 + C - 48) * S
1170 GOSUB 1500
1180 IF C >= 48 AND C <= 57 THEN GOTO 1160: REM 0 to 9
1185 REM Loop while trailing spaces
1190 IF C < 0 OR C <> 32 THEN RETURN: REM end character or not whitespace
1200 GOSUB 1500
1210 GOTO 1180
1500 REM Get input character
1501 REM A$ = input character
1502 REM C = One of the following:
1502 REM - -1 if end of value
1503 REM - -2 if end of input
1504 REM - 32 if whitespace
1505 REM - ASCII code otherwise
1510 GET A$
1520 C = ASC(A$)
1530 IF C = 13 THEN C = -1
1540 IF C = 255 THEN C = -2
1550 IF C = 9 OR C = 10 THEN C = 32
1560 RETURN
2000 REM Read array value
2001 REM A contains array value
2002 REM NA contains length of array
2003 REM V = 1 if valid number, 0 otherwise
2004 REM C = -2 if end of input, -1 if end of value,
2005 REM 32 if whitespace, ASCII code of last character otherwise
2006 REM Initialize
2010 NA = 0
2020 GOSUB 1000: REM Read input value
2030 IF V = 0 THEN RETURN: REM invalid
2040 A(NA) = NR
2050 NA = NA + 1
2060 IF C < 0 THEN RETURN: REM end of input or value
2070 IF C = 44 THEN GOTO 2020: REM comma, get next value
2080 V = 0
2090 RETURN
3000 REM Longest common subsequence
3001 REM Source:
3002 REM https://en.wikipedia.org/wiki/Longest_common_subsequence#Example_in_C#
3003 REM Instead of storing just lengths, a bitmap indicating the indices of
3004 REM the first list is also stored
3005 REM Inputs:
3006 REM - L1 contains first list
3007 REM - M contains length of first list
3008 REM - L2 contains second list
3009 REM - N contains length of second list
3010 REM - WM% contains 16-bit words
3011 REM Outputs:
3012 REM - A contains subsequence
3013 REM - NA contains length of subsequence
3020 NA = 0
3030 IF M < 1 OR N < 1 THEN RETURN: REM exit if empty lists
3035 REM Initialize all subsequences to an empty sequence
3040 NW = INT((M + 15) / 16): REM Number of 16-bit words
3050 FOR I = 0 TO M
3060 FOR J = 0 TO N
3070 FOR K = 0 TO NW
3080 C%(I, J, K) = 0
3090 NEXT K
3100 NEXT J
3110 NEXT I
3115 REM Find the longest common subsequence using prior subsequences
3120 FOR I = 1 TO M
3130 FOR J = 1 TO N
3140 IF L1(I - 1) <> L2(J - 1) THEN GOTO 3230
3145 REM If common element found, create new subsequence based on
3146 REM prior subsequence with the common element appended
3150 C%(I, J, 0) = C%(I - 1, J - 1, 0) + 1
3160 FOR K = 1 TO NW
3170 C%(I, J, K) = C%(I - 1, J - 1, K)
3180 NEXT K
3190 WI = INT((I - 1) / 16): REM Word index
3200 BN = (I - 1) - 16 * WI: REM Bit number
3210 C%(I, J, WI + 1) = C%(I, J, WI + 1) OR WM%(BN)
3220 GOTO 3280
3230 REM Else, reuse the longer of the two prior subsequences
3240 II = I - 1: JJ = J
3240 IF C%(I, J - 1, 0) > C%(I - 1, J, 0) THEN II = I: JJ = J - 1
3250 FOR K = 0 TO NW
3260 C%(I, J, K) = C%(II, JJ, K)
3270 NEXT K
3280 NEXT J
3290 NEXT I
3295 REM Construct longest common subsequence from bitmap
3300 NA = 0
3310 NL = -1
3320 FOR WI = 0 TO NW - 1
3330 FOR BN = 0 TO 15
3340 NL = NL + 1
3350 IF (C%(M, N, WI + 1) AND WM%(BN)) = 0 THEN GOTO 3380
3360 A(NA) = L1(NL)
3370 NA = NA + 1
3380 NEXT BN
3390 NEXT WI
3400 RETURN
3500 REM Display array
3501 REM A contains array
3502 REM NA contains size of array
3510 IF NA < 1 THEN GOTO 3590
3520 FOR I = 0 TO NA - 1
3530 S$ = STR$(A(I))
3540 IF A(I) >= 0 THEN S$ = MID$(S$, 2): REM strip leading space
3550 PRINT S$;
3560 IF I < (NA - 1)THEN PRINT ", ";
3570 NEXT I
3580 PRINT
3590 RETURN
```

Longest Common Subsequence in Commodore Basic was written by:

- rzuckerm

If you see anything you'd like to change or update, please consider contributing.

No 'How to Implement the Solution' section available. Please consider contributing.

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