The worm was clearly safe in the fog, so it was your fault and not mine! But thank you for playing my puzzle, it was very fun watching you solve it! :)
25:59 R2C4 cannot be one higher than the original pencil mark (6) because that would force R1C5 1 higher as well! and thus move your overall total 2 higher to 26!
At 18:22, why couldn't the line go from R7C3 to R6C4 to R6C5? This one change would make the minimum sum exactly 16, and nothing is stopping such a construct. Edit: The original path ends up being correct, and you never had to rely on that line to progress through the puzzle, but you shouldn't have been able to assume the line looked like that at this point.
Yeah i think your right if R6C4 and then R6C5 were 1 and 3 you could still put a 2 in R6C2 and the 1 in R5C2 to maintain the 16. Looks like i didnt factor that in fully.
@@LogicLemurGaming I noticed this while solving but I found the right way to deduce it: the realization that r2c4 can't be 7 places 6 there and in r3c9, which forces 5 onto the 13 diagonal, maxing it out and forcing 3 into r3c7. Now the fog is cleared and the rest of the blue line must be minimized!
5:35 is this correct ? no rule says u cant have the same digits on a line. example: box 1 is a 2+3 0on the line, R4C3 a 1. R5C4 a 1 R6C5 a 2 and the bulb in R7c456. it breaks while i can't place a 9 on the bulb- u get it at 9:02 ;-)
I think i may have taken a bit of a stretch on that one but if following out the later logic of where can the other end be I think it ends up with the same result. I may be wrong there but looking back it doesn't seem like the end could ever be elsewhere than where shown due to the indexing math.
The worm was clearly safe in the fog, so it was your fault and not mine!
But thank you for playing my puzzle, it was very fun watching you solve it! :)
Haha....good point.
Oh no!!! Look at his poor face 🐛
Great logic, but I was able to solve it with some help of yours, so thank you for that! 🤗
25:59 R2C4 cannot be one higher than the original pencil mark (6) because that would force R1C5 1 higher as well! and thus move your overall total 2 higher to 26!
33:30 You got it!
Completed in 16m07s
Finished in 34:26. A lot of progress was made through lucky educated guesses, so I don't feel TOO good about it.
At 18:22, why couldn't the line go from R7C3 to R6C4 to R6C5? This one change would make the minimum sum exactly 16, and nothing is stopping such a construct.
Edit: The original path ends up being correct, and you never had to rely on that line to progress through the puzzle, but you shouldn't have been able to assume the line looked like that at this point.
Yeah i think your right if R6C4 and then R6C5 were 1 and 3 you could still put a 2 in R6C2 and the 1 in R5C2 to maintain the 16. Looks like i didnt factor that in fully.
@@LogicLemurGaming I noticed this while solving but I found the right way to deduce it: the realization that r2c4 can't be 7 places 6 there and in r3c9, which forces 5 onto the 13 diagonal, maxing it out and forcing 3 into r3c7. Now the fog is cleared and the rest of the blue line must be minimized!
5:35 is this correct ? no rule says u cant have the same digits on a line.
example: box 1 is a 2+3 0on the line, R4C3 a 1. R5C4 a 1 R6C5 a 2 and the bulb in R7c456. it breaks while i can't place a 9 on the bulb-
u get it at 9:02 ;-)
I think i may have taken a bit of a stretch on that one but if following out the later logic of where can the other end be I think it ends up with the same result. I may be wrong there but looking back it doesn't seem like the end could ever be elsewhere than where shown due to the indexing math.