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11-07-2021 , 03:18 PM
Quote:
Originally Posted by g-bebe
Where's the best place to find solutions/help on CtC app puzzles?

Working on #26 in Arrow Sudoku, it's an arrow sandwich puzzle but I definitely don't get how they explain their hints to get started.
Try the CtC discord; maybe the puzzle-discussion forum. Or post something here and maybe we can help.
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11-08-2021 , 01:45 PM
If I can't find it online (outside of the app), where's a good resource to create it and post in here?
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11-08-2021 , 06:35 PM
What are you playing on? You should be able to take a screenshot. If you can't upload it to imgur, say, you could email it to me, then I could post it.
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11-08-2021 , 07:22 PM
Quote:
Originally Posted by g-bebe
If I can't find it online (outside of the app), where's a good resource to create it and post in here?
f-puzzles.com and you can screen shot it.
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11-08-2021 , 07:36 PM
Quote:
Originally Posted by Onuzq
f-puzzles.com and you can screen shot it.
Nice!
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11-08-2021 , 07:40 PM


This is probably an easy puzzle for most but the arrow ones really seem to trip me up.

So, there are some obvious restrictions on 9's in boxes 2/4/6/8 due to the arrow sums, but I get stuck on the logic to nail down how the arrow sums allow me to solve it.

The CtC hint #1:

Spoiler:
In each box with circles in, the circles must sum to 23 or 24 and the outlying arrow cell must be 1/3.

Don't understand how that is arrived at. The arrow sums are a 986 or a 987 triple, though I'm not sure how one would know that immediately. After that, why is the outlying arrow cell 1/3? Do I really have to go through every permutation of 986 and 987 triple arrow sums to eliminate every possible option from that outlying arrow cell?
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11-08-2021 , 07:43 PM
This an arrow/sandwich?
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11-08-2021 , 07:44 PM
Quote:
Originally Posted by Onuzq
This an arrow/sandwich?
yes
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11-08-2021 , 08:14 PM
Quote:
Originally Posted by g-bebe


This is probably an easy puzzle for most but the arrow ones really seem to trip me up.

So, there are some obvious restrictions on 9's in boxes 2/4/6/8 due to the arrow sums, but I get stuck on the logic to nail down how the arrow sums allow me to solve it.

The CtC hint #1:

Spoiler:
In each box with circles in, the circles must sum to 23 or 24 and the outlying arrow cell must be 1/3.

Don't understand how that is arrived at. The arrow sums are a 986 or a 987 triple, though I'm not sure how one would know that immediately. After that, why is the outlying arrow cell 1/3? Do I really have to go through every permutation of 986 and 987 triple arrow sums to eliminate every possible option from that outlying arrow cell?
Okay, I got it solved in 14:55. So I'll give you the general idea:

For their hint
Spoiler:
Notice the triples of arrows cover the whole box. A whole box sums to 45. Each arrow represents each number twice. So either you have 2*(6+8+9)=45+1 or 2*(7+8+9)=48. Any other attempt to create a sum would be too small, e.g. 5,8,9 gives 44 but you're covering a full 3x3 box + an extra cell which leads to a contradiction.

Furthermore,
Spoiler:
notice how box 4 and 6 vs. 2 and 8 are perpendicular to each other for the circles on the arrows. A first thing you should think about here is what happens when you have 4 values in a row? Of course you're going to get a quadruple. Since 8 and 9 have to be in each box by the previous statment that they're (6/7), 8, 9 triples then you're limited down to two cases in R4C9 and R1C6.

The way to break that parity is here if you want another hint:
Spoiler:
If you look at row 7, you have a 9 clue with a 1,3 pair already in the row. Consider if you put the pair as 1,3 vs. 3,1. From there, where do you put the 9 in the row? Can you find a spot where you put the 9 where the sandwich isn't too small? Too large?


I hope that helps get you started.

Last edited by Onuzq; 11-08-2021 at 08:32 PM.
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11-09-2021 , 12:30 PM
Here's another one, probably fairly easy, but I hope it is interesting/fun.

Mountain Streams

Standard sudoku rules apply.
Cells separated by a black dot are in a 2:1 ratio.
Cells separated by a white dot are consecutive.

Blocks 1, 3, 7, 9 have mountain ranges in them.
Blocks 2, 4, 6, 8 have streams in them
Mountain ranges and streams divide their blocks so that even cells are on one side and odd cells are on the other.

Mountain cells must be higher than orthogonally adjacent non-mountain cells. Streams have no other effect.

Block 5 has a mountain and a valley cell, which must be lower than all orthogonally adjacent non-valley cells.

https://f-puzzles.com/?id=yg5hv49o


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11-09-2021 , 05:28 PM
Done.. will send screenshot and comments

I did it in under 20 mins, so others will probably find this easy.

The ruleset seems hectic at first, but it provides a ton of cell exclusion info.

I was fun. Thx Eric...
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11-09-2021 , 08:06 PM
Quote:
Originally Posted by Onuzq
Okay, I got it solved in 14:55. So I'll give you the general idea:

For their hint
Spoiler:
Notice the triples of arrows cover the whole box. A whole box sums to 45. Each arrow represents each number twice. So either you have 2*(6+8+9)=45+1 or 2*(7+8+9)=48. Any other attempt to create a sum would be too small, e.g. 5,8,9 gives 44 but you're covering a full 3x3 box + an extra cell which leads to a contradiction.

Furthermore,
Spoiler:
notice how box 4 and 6 vs. 2 and 8 are perpendicular to each other for the circles on the arrows. A first thing you should think about here is what happens when you have 4 values in a row? Of course you're going to get a quadruple. Since 8 and 9 have to be in each box by the previous statment that they're (6/7), 8, 9 triples then you're limited down to two cases in R4C9 and R1C6.

The way to break that parity is here if you want another hint:
Spoiler:
If you look at row 7, you have a 9 clue with a 1,3 pair already in the row. Consider if you put the pair as 1,3 vs. 3,1. From there, where do you put the 9 in the row? Can you find a spot where you put the 9 where the sandwich isn't too small? Too large?


I hope that helps get you started.
Yes, it did, thank you very much. Something I recognize I am very bad at, which I've seen CtC use more and more is the idea of the numerical constraints of boxes/rows/columns. Bad considering we're doing Sudokus but those clues are exactly what I mean, realizing the patterns in a situation where the arrows/sums take up an entire box and what that means.
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11-09-2021 , 09:30 PM
13:21 on mountain/valley

The kropki is a very nice way to break every case down.

The colors on f-puzzles hides the rivers though, sadly.
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11-09-2021 , 09:41 PM
Quote:
Originally Posted by g-bebe
Yes, it did, thank you very much. Something I recognize I am very bad at, which I've seen CtC use more and more is the idea of the numerical constraints of boxes/rows/columns. Bad considering we're doing Sudokus but those clues are exactly what I mean, realizing the patterns in a situation where the arrows/sums take up an entire box and what that means.
Yeah, a lot of simon's puzzles are focusing more and more on unique examples of PIE/SET. But I doubt many of the puzzles they have on the app go as hard on it.

Tricks you'll want to get used to though are: numbers on black kropki dots always sum to a multiple of 3, white kropki dots always sum to an odd number, take any arrow and double it to get the value it covers (unless you have cases where arrows stack/split), among others.
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11-17-2021 , 01:06 PM
The Mirror of Eris

This one is possibly quite hard at the beginning; it may not be as elegant as I would like. After that, you will need to think carefully, but shouldn't be really hard.


Standard sudoku rules apply.

Cells in cages must sum to the little number at the top left corner of the cage.

Grey lines are palindromes: digits equidistant from either end must be the same.

Green lines are anti-palindromes: digits equidistant from either end must sum to 10.


https://f-puzzles.com/?id=yfgz9pe9


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11-18-2021 , 07:36 PM
Done.. screenshot sent.

That was hard. At least I found it hard.
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11-18-2021 , 07:53 PM
Quote:
Originally Posted by ArcticKnight
Done.. screenshot sent.

That was hard. At least I found it hard.
Thanks! I always worry a bit until someone besides me solves it. I know it's solvable, and not too easy.

So far, no one on CtC discord has tried it.
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11-19-2021 , 12:25 PM
My copies of Cracking the Cryptic Greatest Hits have shipped! One week after I receive them, I will draw to give one to a lucky POGGER. Remember, you get one entry for each solved puzzle that I have posted in this thread. Make sure you send me a solved grid for each one you have solved. I will update the current status when the books arrive.

By the way, zetamath has a couple of YouTube channels. On one of them, he creates a puzzle from scratch, which was later published on the discord as well as Logic Masters Germany.
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11-21-2021 , 12:29 AM
First puzzle I've tried, has a bit of a complex ruleset, ask me if you need more information:

Normal Sudoku rules apply

Each line is a juggling sequence. A juggling sequence is valid if no two throws land on the same spot in the cycle. This can be found by the following steps:
1. The number of balls in the pattern is in the circled cell, and is always the first toss in the pattern. Each following throw will be next along the line given. The value of the line must average to the number in the circled cell.
2. Take the value of the throw, and add the spot in the sequence it takes. (e.g. If x is the 3rd throw in the sequence you get x+3)
3. Take the remainder when dividing the new value by the length of the sequence.
4. If in this new sequence you get two (or more) numbers which are equal to each other after step, you have an invalid swap.

For example: 342 is a valid swap
For example: 435 is not a valid swap (all three new values map to 2)
For example: 334 is not a valid swap (they do not average to 3)

(You're going to be my first testers)

https://f-puzzles.com/?id=yej29wuw

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11-21-2021 , 12:33 AM
There's only 1 tricky spot in this if you understand the rules fully:

Spoiler:
I had to take partitions and found 3 cases that hit the average of 15
Spoiler:
Of that, 2 were the only ones that followed the rule of no two spots resulting in the same value
Spoiler:
After that, 1 broke the puzzle, while the other completed it

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11-22-2021 , 09:14 AM
If I understand the rules correctly, shouldn't the circle with the 5 have a line of length five coming from it? Similarly for the 3 on the left. I like the concept, however.
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11-22-2021 , 12:19 PM
Quote:
Originally Posted by Eric
My copies of Cracking the Cryptic Greatest Hits have shipped! One week after I receive them, I will draw to give one to a lucky POGGER. Remember, you get one entry for each solved puzzle that I have posted in this thread. Make sure you send me a solved grid for each one you have solved. I will update the current status when the books arrive.

By the way, zetamath has a couple of YouTube channels. On one of them, he creates a puzzle from scratch, which was later published on the discord as well as Logic Masters Germany.
You can put me back on the list for the draw. In the last while I have been exploring a variety of variants, so I would be happy to be the lucky one who go the book.. Thx Eric
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11-22-2021 , 12:26 PM
Quote:
Originally Posted by ArcticKnight
You can put me back on the list for the draw. In the last while I have been exploring a variety of variants, so I would be happy to be the lucky one who go the book.. Thx Eric
So let it be written.
So let it be done.
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11-22-2021 , 01:37 PM
Quote:
Originally Posted by Eric
If I understand the rules correctly, shouldn't the circle with the 5 have a line of length five coming from it? Similarly for the 3 on the left. I like the concept, however.
No. The average is equal to 5, so if you look at it like a little killer in a way you get 5*3 = 15 as the sum.

Similarly the average equal 3 then you get 3*5=15.

The 5 with a length 5 is hard to do. You'd need to have the sum of the line equal 25 using only 1-6 and limited repeats
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11-22-2021 , 02:39 PM
Quote:
Originally Posted by Onuzq

Each line is a juggling sequence. A juggling sequence is valid if no two throws land on the same spot in the cycle. This can be found by the following steps:
1. The number of balls in the pattern is in the circled cell, and is always the first toss in the pattern. Each following throw will be next along the line given. The value of the line must average to the number in the circled cell.
2. Take the value of the throw, and add the spot in the sequence it takes. (e.g. If x is the 3rd throw in the sequence you get x+3)
3. Take the remainder when dividing the new value by the length of the sequence.
4. If in this new sequence you get two (or more) numbers which are equal to each other after step, you have an invalid swap.

For example: 342 is a valid swap
For example: 435 is not a valid swap (all three new values map to 2)
For example: 334 is not a valid swap (they do not average to 3)

(You're going to be my first testers)

https://f-puzzles.com/?id=yej29wuw

Quote:
Originally Posted by Eric
If I understand the rules correctly, shouldn't the circle with the 5 have a line of length five coming from it? Similarly for the 3 on the left. I like the concept, however.
Quote:
Originally Posted by Onuzq
No. The average is equal to 5, so if you look at it like a little killer in a way you get 5*3 = 15 as the sum.

Similarly the average equal 3 then you get 3*5=15.

The 5 with a length 5 is hard to do. You'd need to have the sum of the line equal 25 using only 1-6 and limited repeats
So the bolded part above has no significance for the puzzle? It looked like it was saying that we would be placing four more numbers starting from the 5. The only significance, then, is that it gives the average of the numbers on the line.
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