Step by Step solution about How to resolve the algorithm Fraction reduction step by step in the Ruby programming language

This Ruby code solves a mathematical problem that involves finding the number of fractions within a range where one digit can be omitted to make them equal. The code operates as follows:

1. indexOf: This is a helper function that returns the index of an element within an array. It iterates through the array, comparing each element to the input needle object. If a match is found, the corresponding index is returned.

2. getDigits: This function takes a number n, the expected length of the digit array le, and an array of digits digits. It extracts the individual digits from n and stores them in the digits array from right to left. It also checks for leading zeros and digits already present in the digits array, returning false if either condition is met.

3. POWS: This is an array of powers of 10 used for calculating the numeric value of the digits.

4. removeDigit: This function removes a digit at a specified index from the digits array and calculates the new numeric value of the resulting number. It iterates through the digits array, skipping the digit at the given index and multiplying the remaining digits by the appropriate power of 10.

5. main: This function serves as the entry point of the program. It iterates through four different ranges of numbers, defined by the lims array.

6. For each range, it initializes counters (count) and arrays (omitted) to keep track of the number of fractions and the number of times each digit is omitted. It then enters a loop to generate fractions within the specified range.

7. It uses getDigits to create an array of digits for each fraction. If getDigits returns false, the loop continues to the next fraction.

8. For each fraction, it generates a second fraction with a slightly higher value (d) and again uses getDigits to obtain its array of digits.

9. It then compares the digits of the two fractions to identify any digits that can be omitted to make them equal. If such a digit is found, it removes the digit from both fractions and recalculates their numeric values.

10. If the reduced fractions are indeed equal, it increments the corresponding counter (count) and records the omitted digit in the omitted array.

11. After processing all fractions within a range, it provides a summary of the results, including the number of equal fractions found and the frequency of each omitted digit.

12. Finally, it prints a comprehensive summary across all ranges, showing the total number of equal fractions and the specific digits that were omitted.

In summary, this code demonstrates a creative and efficient approach to solving a mathematical problem involving fractions and digit manipulation. It combines loop-based iteration, array manipulation, and conditional checks to find and count equal fractions within specified ranges, providing detailed insights into the patterns and relationships involved.

def indexOf(haystack, needle)
    idx = 0
    for straw in haystack
        if straw == needle then
            return idx
        else
            idx = idx + 1
        end
    end
    return -1
end

def getDigits(n, le, digits)
    while n > 0
        r = n % 10
        if r == 0 or indexOf(digits, r) >= 0 then
            return false
        end
        le = le - 1
        digits[le] = r
        n = (n / 10).floor
    end
    return true
end

POWS = [1, 10, 100, 1000, 10000]
def removeDigit(digits, le, idx)
    sum = 0
    pow = POWS[le - 2]
    i = 0
    while i < le
        if i == idx then
            i = i + 1
            next
        end
        sum = sum + digits[i] * pow
        pow = (pow / 10).floor
        i = i + 1
    end
    return sum
end

def main
    lims = [ [ 12, 97 ], [ 123, 986 ], [ 1234, 9875 ], [ 12345, 98764 ] ]
    count = Array.new(5, 0)
    omitted = Array.new(5) { Array.new(10, 0) }

    i = 0
    for lim in lims
        n = lim[0]
        while n < lim[1]
            nDigits = [0] * (i + 2)
            nOk = getDigits(n, i + 2, nDigits)
            if not nOk then
                n = n + 1
                next
            end
            d = n + 1
            while d <= lim[1] + 1
                dDigits = [0] * (i + 2)
                dOk = getDigits(d, i + 2, dDigits)
                if not dOk then
                    d = d + 1
                    next
                end
                nix = 0
                while nix < nDigits.length
                    digit = nDigits[nix]
                    dix = indexOf(dDigits, digit)
                    if dix >= 0 then
                        rn = removeDigit(nDigits, i + 2, nix)
                        rd = removeDigit(dDigits, i + 2, dix)
                        if (1.0 * n / d) == (1.0 * rn / rd) then
                            count[i] = count[i] + 1
                            omitted[i][digit] = omitted[i][digit] + 1
                            if count[i] <= 12 then
                                print "%d/%d = %d/%d by omitting %d's\n" % [n, d, rn, rd, digit]
                            end
                        end
                    end
                    nix = nix + 1
                end
                d = d + 1
            end
            n = n + 1
        end
        print "\n"
        i = i + 1
    end

    i = 2
    while i <= 5
        print "There are %d %d-digit fractions of which:\n" % [count[i - 2], i]
        j = 1
        while j <= 9
            if omitted[i - 2][j] == 0 then
                j = j + 1
                next
            end
            print "%6s have %d's omitted\n" % [omitted[i - 2][j], j]
            j = j + 1
        end
        print "\n"
        i = i + 1
    end
end

main()