Symbolic Algebra Compressed Trees in ANF Sum-of-Products Form
$30-250 USD
Оплачується при отриманні
Develop three Linux command line functions CTPrint, CTXOR, and CTAND
based on the three C functions described in the attached file.
While implementation is preferred in normal ANSI C, it can also be implemented in Python.
The tools will be used for solution of large constraint sets of a hundred or so product terms so fast compact coding is critical. Recursion might simplify the coding. Be aware that worst case ANF can expand to 2^N terms. A lower state management scheme is likely to be more successful and faster, than calling frames and stack allocation.
It would be useful to use switch/case to parse the compressed tree stream, and avoid if/elseif/elseif/else
ANF is Algebraic Normal Form ... a special AND/XOR boolean algebra form. See the wikipedia description.
the anfct.c file has one omission, and that is if the first character of a CT stream is '~' the expression is in complement form, IE is "1 ^ ctexpression", such that the string "A\nB\nC\n~12132333\n" represents "1 ^ A&C ^ A&B ^ A&B&C" and the string "A[3]\n~12132333\n" represents "1 ^ A0&A2 ^ A0&A1 ^ A0&A1&A2". The CTAND example in anfct.c is wrong, the result should be 000 or <null>
CTPrint <filename>, will print the CT file in symbolic form. Examples would be to print fiile1 or file2 below from the CTAND example.
CTAND <input 1> <input 2> <output>, will Symbolic AND (boolean OR binary) the two input CT files and create the output CT file with the product.
It would probably be useful to include '(' and ')' as white space in the CT stream.
If file1 is the expression "B&C ^ A ^ A&C ^ A&B" IE (011 ^ 100 ^ 101 ^ 110) it would contain:
A
B
C
211322333033
#4
If file2 is the expression "C ^ B ^ A" IE (001 ^ 010 ^ 100) it would contain:
A
B
C
2213033003
#3
then CTAND file1 file2 output, will create the output file:
A
B
C
1223332333
#4
The intermediate expressions for the above CTAND example are:
(B&C ^ A ^ A&C ^ A&B) & A -> (A&B&C ^ A ^ A&C ^ A&B)
(B&C ^ A ^ A&C ^ A&B) & B -> (B&C ^ A&B ^ A&B&C ^ A&B) -> (B&C ^ A&B&C)
(B&C ^ A ^ A&C ^ A&B) & C -> (B&C ^ A&C ^ A&C ^ A&B&C)-> (B&C ^ A&B&C)
(B&C ^ A&B&C) ^ (B&C ^ A&B&C) ^ (A&B&C ^ A ^ A&C ^ A&B) -> ( A ^ A&C ^ A&B ^ A&B&C)
( A ^ A&C ^ A&B ^ A&B&C) is (100 ^ 101 ^ 110 ^ 111) which encodes as 1223332333
To help understand the BRANCH-END-END pairings I'll add some parens to emphasize
the pairing for each layer of branches in the three file streams:
file1: (2113(2(233)303)3)
file2: (2(21303)3003)
outfile: 1(2(233)3(233)3)
Note there are two 3's for every 2 in these groupings, one for the left/0 side, and one for the right/1 side.
One simple solution to CTAND is to extract every possible product term, do the symbolic AND function, and then merge the updated product into the output tree in memory. This is less desirable, as it limits constraint sets to what can fit entirely in memory.
A better solution is to deal with the two cases that require action to be taken. The first is simple, and that is where a 1 is OR'd into the stream where a 0 is, simply replace 0 with a 1. The second is a bit more difficult, and that is where a 1 is OR'd into the stream where a BRANCH is, which requires merging the left/0 side into the right/1 side, and deleting any duplicate pairs that might exist, effectively removing the BRANCH. If the left/0 and right/1 sides can fit in memory this can be done quickly, otherwise it will require left/right file descriptors/pointers or copying the left/0 and right/1 sides into temporary files, for the merge operation. This will have to be done at each point in the stream where a 1 is OR'd with a BRANCH, replacing a BRANCH with a 1. I have a 120 cores and a terabyte of SDRAM in my compute server, so in memory is possible for most modest to large projects.
In memory the CT stream can be compressed into 2bit tokens, with a 4X compression of the CT stream.
CTXOR <input 1> <input 2> <output2>, will merge the two input CT files and create the output CT file.
CTXOR fiile1 file2 output2 will produce:
A
B
C
2213233332333
#5
C ^ B ^ B&C ^ A&C ^ A&B -> 001 ^ 010 ^ 011 ^ 101 ^ 110
(2(213(233)3)3(233)3)
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