Step by Step solution about How to resolve the algorithm Minimum positive multiple in base 10 using only 0 and 1 step by step in the Ruby programming language

The provided Ruby code calculates and prints values from the A004290 sequence for a given range of integers. Here's a detailed explanation of the code:

1. mod Function:

   • Defines a mod function that calculates the modulo of two numbers m and n.
   • It handles negative results by adding n to ensure a positive result.

2. getA004290 Function:

   • Takes an integer n as input.
   • If n is equal to 1, it returns 1.
   • Initializes a 2D array arr of size n x n and sets all elements to 0.
   • Sets the first two elements arr[0][0] and arr[0][1] to 1.
   • Enters a while loop that iterates until a specific condition is met.
   • Inside the loop:
     • Increments m by 1.
     • Checks if arr[m - 1][mod(-10 ** m, n)] is equal to 1. If it is, it breaks the loop.
     • Sets the first element arr[m][0] to 1.
     • Iterates through the columns (0 to n-1) and calculates the maximum value between arr[m - 1][k] and arr[m - 1][mod(k - 10 ** m, n)] for each column.
   • After the loop exits, it calculates r as 10 ** m.
   • Initializes k as mod(-r, n).
   • Iterates through rows j from m - 1 down to 1:
     • If arr[j - 1][k] is equal to 0, it increments r by 10 ** j and updates k to mod(k - 10 ** j, n).
   • Finally, it checks if k is equal to 1 and increments r by 1 if it is.
   • Returns the calculated value r.

3. testCases Definition:

   • Defines an array testCases containing a range of integers from 1 to 10, 95 to 105, and some specific values like 297, 576, etc.

4. Main Loop:

   • Iterates through the testCases array.
   • For each n in the array, it calls the getA004290 function to calculate the sequence value result.
   • Prints the result in the format: A004290(%d) = %d = %d * %d, where %d represents the corresponding values.

The code effectively calculates values from the A004290 sequence for a given range of integers and prints them in the specified format.

def mod(m, n)
    result = m % n
    if result < 0 then
        result = result + n
    end
    return result
end

def getA004290(n)
    if n == 1 then
        return 1
    end
    arr = Array.new(n) { Array.new(n, 0) }
    arr[0][0] = 1
    arr[0][1] = 1
    m = 0
    while true
        m = m + 1
        if arr[m - 1][mod(-10 ** m, n)] == 1 then
            break
        end
        arr[m][0] = 1
        for k in 1 .. n - 1
            arr[m][k] = [arr[m - 1][k], arr[m - 1][mod(k - 10 ** m, n)]].max
        end
    end
    r = 10 ** m
    k = mod(-r, n)
    (m - 1).downto(1) { |j|
        if arr[j - 1][k] == 0 then
            r = r + 10 ** j
            k = mod(k - 10 ** j, n)
        end
    }
    if k == 1 then
        r = r + 1
    end
    return r
end

testCases = Array(1 .. 10)
testCases.concat(Array(95 .. 105))
testCases.concat([297, 576, 594, 891, 909, 999, 1998, 2079, 2251, 2277, 2439, 2997, 4878])
for n in testCases
    result = getA004290(n)
    print "A004290(%d) = %d = %d * %d\n" % [n, result, n, result / n]
end