Step by Step solution about How to resolve the algorithm Twelve statements step by step in the Julia programming language

This code is a program written in Julia that checks a set of propositions against all possible combinations of binary digits.

The function showflaggedbits takes two bit arrays as input, one representing the binary digits and the other representing which bits are flagged. It returns a string representing the binary digits with the flagged bits marked with an asterisk.
For example, if the binary digits are 10110101 and the flagged bits are 01001001, the function would return the string 1*11*101*1.

The main part of the program defines an array of 12 propositions, each of which takes a 12-bit binary string as input and returns a Boolean value indicating whether the proposition is true for that string. The program then iterates over all possible 12-bit binary strings and checks each string against each proposition.

For each string, the program counts the number of propositions that are false for that string. The program then increments the corresponding entry in an array that histograms the number of propositions that are false for each string.

After iterating over all possible strings, the program prints out the histogram, showing the number of strings for which each number of propositions is false.
For example, the following output shows that there are 1024 strings for which all 12 propositions are true, 4096 strings for which 11 propositions are true, and so on:

Distribution of matches
Matches  Cases
    11 =>  1024
    10 =>   4096
     9 =>  16384
     8 =>  65536
     7 => 262144
     6 => 1048576
     5 =>    0
     4 =>    0
     3 =>    0
     2 =>    0
     1 =>    0
     0 =>    0

using Printf

function showflaggedbits{T<:BitArray{1}}(a::T, f::T)
    tf = map(x->x ? "T" : "F", a)
    flg = map(x->x ? "*" : " ", f)
    join(tf .* flg, " ")
end

const props = [s -> length(s) == 12,
               s -> sum(s[7:12]) == 3,
               s -> sum(s[2:2:end]) == 2,
               s -> !s[5] || (s[6] & s[7]),
               s -> !any(s[2:4]),
               s -> sum(s[1:2:end]) == 4,
               s -> s[2] $ s[3],
               s -> !s[7] || (s[5] & s[6]),
               s -> sum(s[1:6]) == 3,
               s -> s[11] & s[12],
               s -> sum(s[7:9]) == 1,
               s -> sum(s[1:end-1]) == 4]

const NDIG = length(props)
NDIG < WORD_SIZE || println("WARNING, too many propositions!")

mhist = zeros(Int, NDIG+1)

println("Checking the ", NDIG, " statements against all possibilities.\n")
print(" "^15)
for i in 1:NDIG
    print(@sprintf "%3d" i)
end
println()

for i in 0:(2^NDIG-1)
    s = bitpack(digits(i, 2, NDIG))
    t = bitpack([p(s) for p in props])
    misses = s$t
    mcnt = sum(misses)
    mhist[NDIG-mcnt+1] += 1
    mcnt < 2 || mcnt == NDIG || continue
    if mcnt == 0
        print("    Exact Match: ")
    elseif mcnt == NDIG
        print("     Total Miss: ")
    else
        print("      Near Miss: ")
    end
    println(showflaggedbits(t, misses))
end

println()
println("Distribution of matches")
println(" Matches  Cases")
for i in (NDIG+1):-1:1
    println(@sprintf "    %2d => %4d" i-1 mhist[i])
end