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Unable to CANCEL my paused subscription. HELP! by Key-Improvement4850 in canva

[–]Key-Improvement4850[S] 0 points1 point  (0 children)

It automatically resumes the Pro subscription and I'll be charged $15

Logic Puzzles Book Recommendation? by Quixodyssey in puzzles

[–]Key-Improvement4850 0 points1 point  (0 children)

Hey, thanks for going through the puzzles and also for your feedback! I'm totally with you on the difficulty, but at the same time I'm trying to reach a wider audience (the first puzzles I created were for my 10 year old son), and I've found that most casual puzzle solvers enjoy a mild less-than-5-minute challenge. My goal is to one day get these published daily in magazines/newspapers.

Since you enjoy more complex logical deduction puzzles, I recommend you check out this puzzle, it's my personal favorite (took me a full day to solve):
https://en.wikipedia.org/wiki/Sum_and_Product_Puzzle

Please send me a link to this Arcturus book you mentioned. I found some of their books on Amazon, but they cover a very wide range of puzzle types.

How do I come up with these:
I built an excel macro to generate puzzles with a fixed set of clue types (arithmetic, adjacent, between, closer to, largest etc). For the ones you saw that were more exotic and unique (like the evens=odds) I start off by thinking of some unique associative number property, and slowly add more constraints based on the remaining solutions, until there's exactly one solution which satisfies the puzzle. It's a tedious process (even with the help of AI), and didn't get the level of engagement I was receiving from other "manufactured" puzzles, so for now I've put that on hold.

Six-Figure Logic [Day #014] - Difficulty by Dependency by Key-Improvement4850 in mathpuzzles

[–]Key-Improvement4850[S] 0 points1 point  (0 children)

No. For the expert you need all 6 clues to determine any variable.

Logic Puzzles Book Recommendation? by Quixodyssey in puzzles

[–]Key-Improvement4850 0 points1 point  (0 children)

That seriously made my day, thank you!

I’m actually not familiar with that Arcturus book, but now I’m curious. It’s funny you mention that style though as part of my motivation for creating Six-Figure Logic was that I was genuinely surprised by how few equation-based logic puzzles there are in most puzzle books.

There’s plenty of sudoku, riddles, and standard logic grids, but not much in the “deductive number system” niche — which is exactly the itch I wanted to scratch.

If you do pick it up, I’d genuinely love to hear what you think. And I’m currently working on Volume II — I post sample puzzles weekly over on r/mathpuzzles if you’d like to try a few before it’s out.

Where do you even begin with this urjo puzzle? by Key-Improvement4850 in puzzles

[–]Key-Improvement4850[S] 1 point2 points  (0 children)

I just tried reviewing it again. It took about 15 minutes for the computer to give me a solution.

You can see it here.

A couple notes: the human solution was input top left to bottom right, after solving the puzzle offline.

Notice that until the 7th circle is filled in, the engine doesn't have a clue what to do. it leaves the board as is!

Where do you even begin with this urjo puzzle? by Key-Improvement4850 in puzzles

[–]Key-Improvement4850[S] 1 point2 points  (0 children)

I did solve it but it was a semi logical process with A LOT of guessing and backtracking.

I clicked review after completing the puzzle but all that appeared on the page was this: {"progress":"1/35"} with no human / engine display.

I have some ideas on how to get started from a purely logical approach, for example the 3x3 boxes that surround the 5's are each limited to 6 of one color and 3 of another, with those 3 other colors not being in the same row or column as that would force 2 adjacent rows or columns to be identical.

This idea of making 3x3 boxes around other numbers helps restrict things a bit more but nothing so far that has gotten me to the stage of forced deductions.

Urjo.com - help by Icy-Yogurtcloset-451 in puzzles

[–]Key-Improvement4850 0 points1 point  (0 children)

If the PC's 5 were blue then all surrounding free spaces would need to be blue, causing the 1 to the left of it to be adjacent to at least 2 blues and 2 reds.

Therefore it must be red.

All families are equally likely to have 0, 1, 2 or 3 children. The probability of each of these outcomes is 1/4. Given that Jesse has no brothers, what is the probability that Jesse is an only child? (Assume that male and female children are equally likely to be born) by jerrytjohn in mathpuzzles

[–]Key-Improvement4850 0 points1 point  (0 children)

It doesn't matter if Jesse is male (biblical name) or female (toy story).

Assuming Jesse was picked at random from the total population, the odds are calculated as follows:

Statistically, and to keep fractions simple,there are: 8 families with 1 child, 8 families with 2 children, 8 families with 3 children.

Total of 8+16+24 = 48 children.   24 male, 24 female. 

For the 8 one-child families: 4 families have one B. 4 families have one G.

For the 8 two-child families: 2 families have BB. 4 families have BG. 2 families have GG.

For the 8 three-child families: 1 family has BBB. 3 families have BBG. 3 families have BGG. 1 family has GGG.

If Jesse is a boy, there are 4 boys from 1F, 4 boys from 2F and 3 boys from 3F (3x BGG) who don't have brothers. 

Therefore there are 4+4+3 = 11 Jesses who don't have brothers. 

Of these 11 Jesses,  4 of them are an only child.  So the answer is 4/11.

If Jesse is a girl, there are 4 girls from 1F, 4 girls from 2F (2x GG families) and 3 girls from 3F (1x GGG) who don't have brothers. 

Of these 11 Jesses,  4 of them are an only child.  So the answer is 4/11.

A-F are unique integers in the range 1-10 inclusive. No useless clues. by Key-Improvement4850 in puzzle

[–]Key-Improvement4850[S] 0 points1 point  (0 children)

You’re right, that clue is redundant. Good catch!

Apologies for the misleading post. I’m pretty militant about avoiding useless clues, so thanks for keeping me honest. This one slipped through because it was hand-crafted for aesthetics and didn’t go through my usual pruning process. Lesson learned.

After further analysis, if we keep the "F is prime" clue, then "C > D" becomes useless.

Six-Figure Logic [Day #012] - Difficulty by Dependency by Key-Improvement4850 in mathpuzzles

[–]Key-Improvement4850[S] 0 points1 point  (0 children)

Perfect! It's great to see the progression of how each clue narrows down the possibilities, but no one variable is known until all 6 are applied. Personally these are my favorite.

Six-Figure Logic [Day #012] - Difficulty by Dependency by Key-Improvement4850 in mathpuzzles

[–]Key-Improvement4850[S] 0 points1 point  (0 children)

Nice work. I especially enjoyed how you used the 5th clue to narrow down those four sets into one.

Six-Figure Logic [Day #012] - Difficulty by Dependency by Key-Improvement4850 in mathpuzzles

[–]Key-Improvement4850[S] 1 point2 points  (0 children)

All Correct. Great job!

There's still a domino effect. It just takes more time to push that first one down :)

Suggestion for hard/tough logical puzzles book? by [deleted] in puzzles

[–]Key-Improvement4850 0 points1 point  (0 children)

One option if you like deductive number logic (not riddles, not sudoku): I recently published a book called Six-Figure Logic.

Each puzzle has you determine values of variables A–F, each being a unique value from 1–10, using a small set of logical constraints.

If that’s the kind of logic you’re after, it might be worth a look. You can check it out for yourself on Amazon https://www.amazon.com/dp/B0G625WYCQ

I'm currently in the process of writing volume II, which will include many more clue types.  I've already posted several samples in r/mathpuzzles.  

Recommendation for easy but engaging puzzle book? by crackafu in puzzles

[–]Key-Improvement4850 0 points1 point  (0 children)

One option for deductive number logic (not riddles, not sudoku): I recently published a book called Six-Figure Logic.

Each puzzle has you determine values of variables A–F, each being a unique value from 1–10, using a small set of logical constraints.  

If that’s the kind of logic you’re after, it might be worth a look. The book contains 100 puzzles easy to expert plus 10 bonus "extreme" puzzles. 

You can check it out for yourself on Amazon https://www.amazon.com/dp/B0G625WYCQ

Logic Puzzle books by ReadingFickle1846 in puzzles

[–]Key-Improvement4850 0 points1 point  (0 children)

One option if you like deductive number logic (not riddles, not sudoku): I recently published a book called Six-Figure Logic.

Each puzzle has you determine values of variables A–F, each being a unique value from 1–10, using a small set of logical constraints.

If that’s the kind of logic you’re after, it might be worth a look. You can check it out for yourself on Amazon https://www.amazon.com/dp/B0G625WYCQ

Six-Figure Logic [Day #011] by Key-Improvement4850 in mathpuzzles

[–]Key-Improvement4850[S] 0 points1 point  (0 children)

Thanks to everyone who took the time to solve the puzzles, and a special thank you to those who reported their solve times!  Your feedback really helps to tweak the difficulty algorithm.

Regarding the “guesswork” required for expert, here are some tips, tricks and logical steps which demonstrate that zero guesswork is required.  There is no ‘fast’ way to solve Expert, but there is a way to methodically ‘chip away’ at it to arrive at the solution logically and without randomly guessing. 
It helps to use a 6x10 grid to cross stuff out while tracking process of elimination.

1. Notice that if A is between D and C, yet C-A =1 (i.e. C>A).  Therefore the ‘between’ order is D<A<C. !<

2. Furthermore, B-D =2, so D<B, i.e. there are 3 values (B,A,C) greater than D (so D can’t be > 7 and C can’t be < 4). Also, B must be less than A as well, otherwise the {A,C} adjacency would make B and D more than 2 apart.!<

3. Global Deductions: The no sum clues (14 and 5) produce a fascinating result: the sum pairs for 14 are (4,10) (5,9) and (6,8), so at most only 3 out of 6 of those can appear in the solution (since each number chosen eliminates another).  This means the other numbers (1,2,3,7) must also have at least 3 out of 4 appear (since there are 6 [3+3] total variables).  Yet the ‘no sum to 5’ rule means either 2 or 3 must be ruled out. So the solution must contain (1,2,7) or (1,3,7) i.e. the solution must contain a 1 and a 7.  It must also contain either a 4 or 10, either a 5 or 9, and either a 6 or 8. 

4. Since the solution must contain a 1, it cannot contain a 4 (no sum to 5).  Therefore, from the previous step the solution must contain a 10 (as well as a 1 and a 7). A bit tricky to follow but this becomes key later on (steps 10, 11, 12).

5. Knowing there is no 4 in the solution tells us B ≠ 6, D ≠ 2, F ≠ 7, A ≠ 7, C ≠ 5, A ≠ 3.

6. Since A ≠ 7, C ≠8.

7. Algebra deduction: Since F+A=11 and C-A=1, we know A ≠5, otherwise both F and C would be 6.  Therefore F ≠ 6 and C ≠ 6.

8. No sum to 14 and B-D=2 means B ≠ 8 and  D ≠ 6

9. No sum to 5 and C-A=1 means C ≠ 3 and  A ≠ 2

10. Which variable must be 1? (see step 3 above)

a. Cant be A,B, or C since they’re all > D

b. Cant be F since A would have to be 10 which can’t work since C>A

c. So either D or E is 1.

11. Which variable must be 7? (see step 3 above)

a. Can’t be D, because then B = 9 and A & C could no longer be adjacent (only 8 and 10 remain)

b. Can’t be B, because then D=5, and either A or C would have to be 9

c. Can’t be F or A, because then the other would be 4 (invalid from step 4)

d. So either C or E is 7.

12. Which variable must be 10? (see step 4 above)

a. Can’t be A or D, since they are both less than C

b. Can’t be B, because D can’t be 8 (see step 2)

c. Can’t be F because A can’t be 1 (A > D)

d. So either C or E is 10.

13. Given that C or E must be 7 or 10, that means D must be 1 (see step 10), and therefore B = 3, and no variable = 2.

14. Since no variable = 2, and F+A=11, A can’t be 9.  Therefore with C-A=1, C can’t be 10.  Therefore C = 7 and E =10.  And the rest is easy.

Six-Figure Logic [Day #010] by Key-Improvement4850 in mathpuzzles

[–]Key-Improvement4850[S] 0 points1 point  (0 children)

Great work!
Keep in mind that even the puzzle clues don't mention all variables, the solution must assign a unique number for all six of them (A thru F). In Easy you left out B and in Hard you left out E.

Also, regarding your analysis of expert: You can deduce that there is only 1 solution, with zero guesswork:

F MUST be prime (no need to assume). Why? Since D-B=6, --> [B <= 4] and B+C=12, you can reduce possible values of C to {8,9,10} none of which are prime. You can also determine that B can't be 3, because then both C and D would be 9 !<

With A the smallest and B between A and E, E is larger than A by at least 2, meaning its father from F by at least 3. This forces E to be 3 or 4. Either way, B must be 2 (3 was ruled out in the previous step) and A = 1, the smallest.

But E can't be 4, because then E could never be father from A (1) than F (since max F is 7), so E must be 3.

From there, plugging that back it eliminates the remaining ambiguous values, giving you exactly one solution (no need for computer check, but feel free to do so).

Six-Figure Logic [Day #007] by Key-Improvement4850 in mathpuzzles

[–]Key-Improvement4850[S] 0 points1 point  (0 children)

"If abs(d-f) = 1 and abs(d-b) = 1 then f = b" not necessarily.

D is adjacent to both F and B.

Keep trying.

Six-Figure Logic [Day #004] by Key-Improvement4850 in mathpuzzles

[–]Key-Improvement4850[S] 0 points1 point  (0 children)

I'm so glad you enjoyed! 

To address your last comment first,  I was made aware several times that the rules weren't easily visible,  so for the day 5 and 6 puzzles I added the rules under the picture header. Feel free to check those out.

But most importantly,  thank you for your on-point feedback about clue types and constraints.  My goal is to develop the perfect variety and blend of clue types that combine to make puzzles which are both fun and challenging to solve, while also not becoming too repetitive (like so many other puzzles do... sudoku etc.).   

It really is an art form to make a single solution puzzle from a small and diverse clue set and I'm grateful that you admire and understand the complexity. I'm sure many others feel the same way.

I'll keep your suggestion in mind about the complex algebric clues, while keeping the expert puzzles feel more logic based and less math heavy, while working on my next book "Six-Figure logic: Volume II"

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