Parity Party XIV, "Avec cages"
(Published on 4. January 2021, 13:13 by glum_hippo)
F-puzzles version with colors, and version without colors (recommended).
Normal Sudoku rules apply, that is: each row, 3x3 box, and column must include the digits 1-9.
The cages are 'Party Cages', not killer cages. Think of each cage as a 1-cell-wide snake. Digits need to be placed such that, when reading from the head of the snake AS WELL AS reading from the tail of the snake, the digits can fulfill the same Parity Party sum (explained below). Digits may not repeat within a cage. Not all sums are given! If a cage has no sum, it must nevertheless fulfill the same parity party clue from either end.
Parity Party rules: all the digits up to and including the first odd OR first even digit you encounter must add up to the given sum. Thus, you could be adding 0-4 evens and an odd OR 0-5 odds and then an even. Please keep in mind that a PP clue can be fulfilled by a single digit!
Note that the parity sequences in a cage could leave a gap, leave no gap, or overlap. So 15 in five digits could be — among others — 87 1 96 (gap) or 645 87 (no gap) or 62715 (overlapping). 8+7 = 6+9 = 6+4+5 = 7+8 = 6+2+7 = 5+1+7+2 = 15.
Colors serve no purpose other than to make cages easier to see.
Solution code: Column 5, then row 9 (18 digits total)
Comments
on 21. February 2022, 15:32 by Mody
Der vierte Anlauf ging bedeutend schneller ;)
Mir hat es viel Spaß gemacht.
on 18. March 2021, 20:59 by ffricke
Eine sehr schöne Variante. Ich finde die Bewertung mit 4 Sternen für angemessen. Ich mag diese Parity Bedingung sehr und das daraus erforderliche Finden von möglichen Kombinationen. Ich bin gespannt, was dir noch so an Varianten einfällt, ich bin ja bald durch mit der bisherigen Serie.
— Ja, danke, ffricke - die Ideen gehen noch nicht aus, aber ich muß auch die PP-Rätsel von Christoph wärmstens empfehlen: 000566 und 0005FJ. g_h
on 5. January 2021, 20:58 by Big Tiger
Right, so ... a theoretical 9 cage could read mentally as "We start on even digits, of which there are none, and 'switch' to one odd digit, which is a nine at the head of the cage."
——if it helps, put the word ‘switch’ out of your mind. We are adding everything up to and including the first odd or first even digit. Put another way: any string of digits with evens and odds in it could be represented by two different PP clues, but one of those is always just the value of the first digit. I should mention that Richard Stolk made a puzzle inspired by this rule set called 'Parity Parade' (see 0003LP) - but in his version the word 'switch' does come into play. "Basic inspiration for this type comes from glum_hippo but with a small change in the ruleset. A sum cannot consist of only one digit[,] to give more power to the parity swap."
on 5. January 2021, 15:14 by glum_hippo
add note about single-digit solution possibility.
on 5. January 2021, 06:38 by marcmees
very enjoyable. ... solves smoothly if one keeps in mind that single digits can make up a sum (with 0 of the other parity). Very nice but wouldn't rate it 5 stars though.
— Thank you marcmees; I’ll let it float at 5 stars until it gets an established rating. It’s always gratifying to see your name in the solvers list!
on 5. January 2021, 04:49 by glum_hippo
Hi Big Tiger. The parity party clue (8, in the case you cite), is the answer to the question “What’s the sum of all the odd numbers you see, plus the first even?” OR “What’s the sum of all the even digits you see, plus the first odd digit?” It’s possible you don’t encounter any odds before hitting an even, or vice versa. A "switch in parity" is not compulsory and in some cases not possible.
[Note that 8, of course, cannot be the sum of zero evens and an odd, but some other number like 5 can :)]
on 5. January 2021, 04:28 by Big Tiger
I'm confused: How can you add "Zero to Four" evens? (And "zero" to five odds?) In all of your examples, there is a definite switch in parity by one digit. For example:
5+1+7 (switch to evens) + 2
So I'm looking at the 8 cage and am completely stumped as to how that could work. From one end it could easily be two odds and then switch to the one even, but from the other end, I start with evens and switch to One Odd, which results in an odd sum every time.
So the question is: What am I not understanding about the rule set?
on 4. January 2021, 22:39 by SudokuExplorer
Very enjoyable parity logic! Its very important to remember that one could add zero terms of one parity.
on 4. January 2021, 18:12 by SudokuExplorer
Its an interesting idea. I used a similar concept but with skyscraper sums (id=0004MD) which might interest you. I'll probably try this later today or tomorrow.
on 4. January 2021, 16:52 by glum_hippo
Feedback is very welcome. This does not require guessing, but the logic is quite labyrinthine nonetheless. I am not sure if this one is too frustrating or whether the difficulty is "the nice kind".
Ich freue mich um alle Rückmeldungen, besonders bezüglich der Schwierigkeit und ob sie die Langwierigkeitsgrenze überschwappt.
