RogerBW's Blog

 The Weekly Challenge 224: Specially Additive Number Notes 09 July 2023 I’ve been doing the Weekly Challenges. The latest involved string construction and numerical sequence analysis. (Note that this ends today.) Task 1: Special Notes You are given two strings, \$source and \$target. Write a script to find out if using the characters (only once) from source, a target string can be created. Those who skip a week are apparently condemned to repeat it, since this is basically 221 part 1 with the order of the parameters reversed and only one input string rather than a whole list of them. (It's also quite like 216 part 2, from which I took `word2map` for the languages I didn't use in 221.1.) Perl: ``````sub word2map(\$word) { my %w; map {\$w{\$_}++} split '',lc(\$word); return \%w; } sub specialnotes(\$chars, \$word) { my \$cm = word2map(\$chars); my \$f = word2map(\$word); my \$valid = 1; foreach my \$c (keys %{\$f}) { if (!exists \$cm->{\$c} || \$f->{\$c} > \$cm->{\$c}) { \$valid = 0; last; } } return \$valid; } `````` Task 2: Additive Number You are given a string containing digits 0-9 only. Write a script to find out if the given string is additive number. An additive number is a string whose digits can form an additive sequence. A valid additive sequence should contain at least 3 numbers. Except the first 2 numbers, each subsequent number in the sequence must be the sum of the preceding two. e.g. "199100199" can be broken down as "1 + 99 = 100, 99 + 100 = 199". The best I came up with was a depth-first search scanning across the sequence. The three elements in the stack tuple are: start index of first value end index of first value end index of second value because the start of the second value is always one higher than the end of the first, so I can calculate that on the fly. The stack is pre-seeded with a start index of zero, and all possible end indices, for the first pair of numbers. For each pass through the stack, we check all possibilities for the end index of the third value, which starts immediately after the second. If the value defined by those indices matches the sum of the values from the first two, that's a valid continuation of the sequence - so what goes back on the stack is the indices for the second and third values, and we'll continue next pass. (Unless we've got to the end of the sequence, in which case, return `true` and all is well.) An optimisation I didn't bother with would be to reduce the scanning space: two values of `A` and `B` digits cannot add to one of more than `max(A, B) + 1` digits. However, since we allow zeroes, they may add to fewer than `max(A, B)` digits. PostScript, where I used stack marking to employ the actual language stack as my LIFO stack: ``````/exdigits { 3 dict begin /e exch def /s exch def /digits exch def 0 s 1 e { exch 10 mul exch digits exch get add } for end } bind def /additivenumber { 1 dict begin [ exch s2a { 48 sub } forall ] /digits exch def mark 0 1 digits length 3 sub { /i exch def i 1 add 1 digits length 2 sub { /j exch def [ 0 i j ] } for } for /valid false def { counttomark 0 eq { cleartomark exit } if aload pop /end_b exch def /end_a exch def /start_a exch def /start_b end_a 1 add def /val_ab digits start_a end_a exdigits digits start_b end_b exdigits add def /start_c end_b 1 add def start_c 1 digits length 1 sub { /end_c exch def val_ab digits start_c end_c exdigits eq { end_c digits length 1 sub eq { /valid true def exit } { [ start_b end_b end_c ] } ifelse } if } for valid { cleartomark exit } if } loop valid end } bind def `````` and for the same algorithm in a more conventional language, Python: ``````def exdigits(digits, start, end): x = 0 for i in range(start, end + 1): x *= 10 x += digits[i] return x def additivenumber(digitstring): digits = [int(x) for x in digitstring] stack = [] for i in range(len(digits) - 2): for j in range(i + 1, len(digits) - 1): stack.append((0, i, j)) while len(stack) > 0: start_a, end_a, end_b = stack.pop() start_b = end_a + 1 val_ab = exdigits(digits, start_a, end_a) + exdigits(digits, start_b, end_b) start_c = end_b + 1 for end_c in range(start_c, len(digits)): if val_ab == exdigits(digits, start_c, end_c): if end_c == len(digits) - 1: return True else: stack.append((start_b, end_b, end_c)) break return False `````` Full code on github. 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