# The following was written by Claude with the following prompt:
#
# I would like you to convert the Julia code for forward substitution given
# in the context to a near identical code for backward substitution.  Please
# preserve comments corrected as needed, indentation and style of the code.

function backward(A,b)
# Solves the upper triangular system of equations Ax = b
# A input argument, square upper triangular matrix
# b input argument
# x solution

    n,m=size(A) # finding the size of A
    if n != m
        error("input is not a square matrix")
    end
    if !(b isa Vector) || length(b) != n
        error("input dimensions do not match")
    end

    x = zeros(n) # initialise x to the same dimension
    if abs(A[n,n])>1e-12 # not comparing to zero because of possible
                         # rounding errors
        x[n] = b[n]/A[n,n] # solve for the last element of x
    else
        println("input singular") # A is singular if any of the diagonal
                                  # elements are zero
        return x
    end
    for k=n-1:-1:1 # the loop considers one row after the other
        if abs(A[k,k])>1e-12 # not comparing to zero because of possible
                             # rounding errors
            temp = 0
            for j=k+1:n
                temp = temp + A[k,j]*x[j] # Multiply the elements of
                                          # the k-th row of A after the
                                          # diagonal by the elements of x
                                          # already calculated
            end
            x[k] = (b[k]-temp)/A[k,k] # solve for the k-th element of x
        else
            error("input singular") # A is singular if any of the
                                    # diagonal elements are zero
        end
    end
    return x
end