Great solve, thanks very much for the feature! You were very on the ball with looking out for the 4th cage. Simon got totally caught out by that bit haha. You explained through everything really well and it was a lot of fun to watch
29:08 .... BRAVO to the setter. What a beautiful masterpiece. Started to suspect the location of the 4th killer in the last 10 minutes. Love it. Thanks for sharing, Bremster.
What a lovely puzzle. If I laugh when finding the final deduction, it's been an enjoyable solve, and I really enjoyed solving this one. very nice break-in, fun middle, and the ending was the 'Ah-ha' cherry on top. Also very linear - my favourite.
I stared at the grid for 10 minutes. Went and made coffee, and then watched you solve this fantastic puzzle. One day I hope to be able to see these kinds of things. It was great watching you work through the logic. Also, this is the first time I recall trying to verbally talk to you through the screen. Sadly it didn't work, but you fixed it in the end! Really fun video.
I actually remembered solving this before, and was able to find my solve: 29:28, not too bad. I do recall it stumping me a few times, though. 🙂 Props to Marty Sears for a great puzzle and to BremSter for a great solve!
Nice solve. I have to admit that I couldn't figure out how to get around guessing my first entry in the second cage, with the 432s. I knew it was 4 or 3, but I didn't know which. Good job getting around that. Sorry, I couldn't help laughing about the mistake. Every ten or fifteen seconds, you paused, face frozen in a sober expression, and I thought you'd caught the mistake, but then you continued onto another trajectory.
Well, indeed at 4:44 Bremster says the cage can only be in 2 boxes, but marks one cell of box6, too. So cells of 3 boxes. You're right anyway. Two of the threee 3s have to be the single cells in box 2 and box 6, a third 3 is in box 2, but the only further cell you could add is a 1 cell. 2 cells can't connect r2c6 and r4c9, but also the total would need to be in r2c6, or you'd need a cell in box 3, row 1, pushing the necessary number of cells even above 6, which is the maximum for 123. It should be mentioned to complete the logic, but I think this can be deduced in several ways.
That's where I got stuck. You can certainly put a 3 into the available cells in boxes 2, 3, and 6. But to connect them, you need 4 cells, which requires a 4 in the cage. That would require using 4 boxes, but we don't have access to a 4th box. The conclusion was correct, but the reasoning behind it got cut a bit short.
@@oldguydoesstuff120 no, a 4 would not be required, if a cage has the digits 123 it has 6 cells, enough to connect directly. One problem with that is the two 2s would both be in box 2. The other problem is, the total would need to be seen in the already visible leftmost cell of the cage in box 2, but it isn't. So there needs to be one cell in row 1 in box 3. And that in turn makes it impossible to reach box 6, even with a total of 6 cells each being 1,2, or 3.
@@OlafDoschke I think a 4 would be required. Connecting the known cage cell in box 2 to the potential cage cell in box 6 requires using 4 cells in box 3. The smallest those digits could be is 1234. Remember, cages are connected orthogonally - that is, edge to edge. Touching just by a corner isn't enough. But I now see your observation as well. Ignoring the connection problem, you also need two 2s, and there's no place for a second 2.
@@oldguydoesstuff120 Of course that thought is valid, too. When only concentrating on the cell count first, 6 is enough, but since 2s can only go in box 3, that's making 4 necessary, true. It depends on what you think of first, how you get to the conclusion, but you always come out at this not working.
Great solve, thanks very much for the feature! You were very on the ball with looking out for the 4th cage. Simon got totally caught out by that bit haha. You explained through everything really well and it was a lot of fun to watch
I spent all the end of my solve wondering where that last cage would be. What a great puzzle !
29:08 .... BRAVO to the setter. What a beautiful masterpiece. Started to suspect the location of the 4th killer in the last 10 minutes. Love it. Thanks for sharing, Bremster.
What a lovely puzzle. If I laugh when finding the final deduction, it's been an enjoyable solve, and I really enjoyed solving this one. very nice break-in, fun middle, and the ending was the 'Ah-ha' cherry on top. Also very linear - my favourite.
I stared at the grid for 10 minutes. Went and made coffee, and then watched you solve this fantastic puzzle. One day I hope to be able to see these kinds of things. It was great watching you work through the logic. Also, this is the first time I recall trying to verbally talk to you through the screen. Sadly it didn't work, but you fixed it in the end! Really fun video.
What a great puzzle. The 4th killer being the end key was amazing! Thanks for the puzzle 😊
I actually remembered solving this before, and was able to find my solve: 29:28, not too bad. I do recall it stumping me a few times, though. 🙂 Props to Marty Sears for a great puzzle and to BremSter for a great solve!
Took me 75 mins to solve.
What an awesome puzzle, with a fun finale.
Got some help from Pythagoras along the way.
I liked your solve more than Simon ‘s
Nice solve. I have to admit that I couldn't figure out how to get around guessing my first entry in the second cage, with the 432s. I knew it was 4 or 3, but I didn't know which. Good job getting around that.
Sorry, I couldn't help laughing about the mistake. Every ten or fifteen seconds, you paused, face frozen in a sober expression, and I thought you'd caught the mistake, but then you continued onto another trajectory.
Oops 😅. The sudoku went smoothly until I got stuck with deadly pattern. Came back to the video only to realise I forgot about the fourth cage
Neat puzzle , nice solve.
Great fun puzzle! I spent a while stuck at the end because I hadn't properly taken in that there were *exactly* four cages in the puzzle.
Nice puzzle.. took me 17 mins after understanding rules
That was fun!
Great solve - ref the minor error but you did indeed not use the 2 for anything, no problem!
Nice solve, you did not use the bad 2, I was yelling for you to fix it, guess you didn't hear me. 😂
5:13
could still be a 3 at that stage ... being in boxes 2,3 and 6
it does get rather hard to CONNECT the 3s
😜
Well, indeed at 4:44 Bremster says the cage can only be in 2 boxes, but marks one cell of box6, too. So cells of 3 boxes.
You're right anyway. Two of the threee 3s have to be the single cells in box 2 and box 6, a third 3 is in box 2, but the only further cell you could add is a 1 cell. 2 cells can't connect r2c6 and r4c9, but also the total would need to be in r2c6, or you'd need a cell in box 3, row 1, pushing the necessary number of cells even above 6, which is the maximum for 123.
It should be mentioned to complete the logic, but I think this can be deduced in several ways.
That's where I got stuck. You can certainly put a 3 into the available cells in boxes 2, 3, and 6. But to connect them, you need 4 cells, which requires a 4 in the cage. That would require using 4 boxes, but we don't have access to a 4th box. The conclusion was correct, but the reasoning behind it got cut a bit short.
@@oldguydoesstuff120 no, a 4 would not be required, if a cage has the digits 123 it has 6 cells, enough to connect directly.
One problem with that is the two 2s would both be in box 2.
The other problem is, the total would need to be seen in the already visible leftmost cell of the cage in box 2, but it isn't.
So there needs to be one cell in row 1 in box 3. And that in turn makes it impossible to reach box 6, even with a total of 6 cells each being 1,2, or 3.
@@OlafDoschke I think a 4 would be required. Connecting the known cage cell in box 2 to the potential cage cell in box 6 requires using 4 cells in box 3. The smallest those digits could be is 1234. Remember, cages are connected orthogonally - that is, edge to edge. Touching just by a corner isn't enough. But I now see your observation as well. Ignoring the connection problem, you also need two 2s, and there's no place for a second 2.
@@oldguydoesstuff120 Of course that thought is valid, too. When only concentrating on the cell count first, 6 is enough, but since 2s can only go in box 3, that's making 4 necessary, true. It depends on what you think of first, how you get to the conclusion, but you always come out at this not working.
Ahaha awesome thumbnail
Agreed, that made me laugh!