sol1 Algorithm
The sol1 Algorithm, also known as "Squaring the Circle" algorithm, is a mathematical technique that aims to solve the ancient geometrical problem of constructing a square with the same area as a given circle using only compass and straightedge. This algorithm is based on the approximation of the value of Pi (π), which is the ratio of the circumference of a circle to its diameter. The main idea behind the sol1 Algorithm is to find the side length of a square that, when multiplied by itself, gives the same area as that of a circle with a given radius.
The sol1 Algorithm begins by drawing a circle with the desired radius, followed by constructing an inscribed square within the circle. The next step involves dividing the circle's circumference into a number of equal segments, which are then used to create a polygon that approximates the circle. The area of this polygon can be easily calculated using basic trigonometry, and as the number of segments increases, the approximation of the circle's area becomes more accurate. Finally, the side length of the square is determined by finding the square root of the approximated circle's area, and a square with this side length is constructed using a compass and straightedge. Although the sol1 Algorithm provides an approximation to the problem of squaring the circle, it has been proven mathematically impossible to achieve an exact solution using only compass and straightedge due to the transcendental nature of the number π.
""" Sum of digits sequence Problem 551 Let a(0), a(1),... be an integer sequence defined by: a(0) = 1 for n >= 1, a(n) is the sum of the digits of all preceding terms The sequence starts with 1, 1, 2, 4, 8, ... You are given a(10^6) = 31054319. Find a(10^15) """ ks = [k for k in range(2, 20 + 1)] base = [10 ** k for k in range(ks[-1] + 1)] memo = {} def next_term(a_i, k, i, n): """ Calculates and updates a_i in-place to either the n-th term or the smallest term for which c > 10^k when the terms are written in the form: a(i) = b * 10^k + c For any a(i), if digitsum(b) and c have the same value, the difference between subsequent terms will be the same until c >= 10^k. This difference is cached to greatly speed up the computation. Arguments: a_i -- array of digits starting from the one's place that represent the i-th term in the sequence k -- k when terms are written in the from a(i) = b*10^k + c. Term are calulcated until c > 10^k or the n-th term is reached. i -- position along the sequence n -- term to calculate up to if k is large enough Return: a tuple of difference between ending term and starting term, and the number of terms calculated. ex. if starting term is a_0=1, and ending term is a_10=62, then (61, 9) is returned. """ # ds_b - digitsum(b) ds_b = 0 for j in range(k, len(a_i)): ds_b += a_i[j] c = 0 for j in range(min(len(a_i), k)): c += a_i[j] * base[j] diff, dn = 0, 0 max_dn = n - i sub_memo = memo.get(ds_b) if sub_memo is not None: jumps = sub_memo.get(c) if jumps is not None and len(jumps) > 0: # find and make the largest jump without going over max_jump = -1 for _k in range(len(jumps) - 1, -1, -1): if jumps[_k][2] <= k and jumps[_k][1] <= max_dn: max_jump = _k break if max_jump >= 0: diff, dn, _kk = jumps[max_jump] # since the difference between jumps is cached, add c new_c = diff + c for j in range(min(k, len(a_i))): new_c, a_i[j] = divmod(new_c, 10) if new_c > 0: add(a_i, k, new_c) else: sub_memo[c] = [] else: sub_memo = {c: []} memo[ds_b] = sub_memo if dn >= max_dn or c + diff >= base[k]: return diff, dn if k > ks[0]: while True: # keep doing smaller jumps _diff, terms_jumped = next_term(a_i, k - 1, i + dn, n) diff += _diff dn += terms_jumped if dn >= max_dn or c + diff >= base[k]: break else: # would be too small a jump, just compute sequential terms instead _diff, terms_jumped = compute(a_i, k, i + dn, n) diff += _diff dn += terms_jumped jumps = sub_memo[c] # keep jumps sorted by # of terms skipped j = 0 while j < len(jumps): if jumps[j][1] > dn: break j += 1 # cache the jump for this value digitsum(b) and c sub_memo[c].insert(j, (diff, dn, k)) return (diff, dn) def compute(a_i, k, i, n): """ same as next_term(a_i, k, i, n) but computes terms without memoizing results. """ if i >= n: return 0, i if k > len(a_i): a_i.extend([0 for _ in range(k - len(a_i))]) # note: a_i -> b * 10^k + c # ds_b -> digitsum(b) # ds_c -> digitsum(c) start_i = i ds_b, ds_c, diff = 0, 0, 0 for j in range(len(a_i)): if j >= k: ds_b += a_i[j] else: ds_c += a_i[j] while i < n: i += 1 addend = ds_c + ds_b diff += addend ds_c = 0 for j in range(k): s = a_i[j] + addend addend, a_i[j] = divmod(s, 10) ds_c += a_i[j] if addend > 0: break if addend > 0: add(a_i, k, addend) return diff, i - start_i def add(digits, k, addend): """ adds addend to digit array given in digits starting at index k """ for j in range(k, len(digits)): s = digits[j] + addend if s >= 10: quotient, digits[j] = divmod(s, 10) addend = addend // 10 + quotient else: digits[j] = s addend = addend // 10 if addend == 0: break while addend > 0: addend, digit = divmod(addend, 10) digits.append(digit) def solution(n): """ returns n-th term of sequence >>> solution(10) 62 >>> solution(10**6) 31054319 >>> solution(10**15) 73597483551591773 """ digits = [1] i = 1 dn = 0 while True: diff, terms_jumped = next_term(digits, 20, i + dn, n) dn += terms_jumped if dn == n - i: break a_n = 0 for j in range(len(digits)): a_n += digits[j] * 10 ** j return a_n if __name__ == "__main__": print(solution(10 ** 15))
