We've got a pair of 4's in hand
HTML-код
- Опубликовано: 29 июн 2024
- Logic Lemur solves 'Twin 4's' by Sir Moose
Two 4x4 sudokus with Arrow sums and Thermometers.
Try it out using Sven's SudokuPad sudokupad.app/njia60zkyb
Rules:
Place 8 of the digits 1-9 into the sudoku grids. Digits cannot repeat in any column, row or 2*2 box, and the left puzzle and right puzzle each have 4 unique digits.
Digits on an arrow sum to the digit in that arrows circle.
Digits increase from the bulb of the thermometer.
Video Contents:
00:00 Intro and Rules
01:28 Sudoku Solve
19:32 Outro
_________________________________
Twitter: / thelogiclemur
Contact me:
TheLogicLemur@gmail.com 
Игры
04:25 and I am very proud of myself
Very nice puzzle
Finished in 7:36. This was a really fun one to solve!
Thanks for the feature!
I like your selection of puzzles, good job!
Thanks, I try to keep things mixed up a bit....between rulesets and difficulty. Some days the brain doesn't function at high levels...ha
R2c1 must be even as yellow + blue = purple, so green = 2x(purple) which must therefore be even
And if LL pulled out the 2n value of green, he could have placed the 1 sooner.
Good logic..i didnt look at it in that way.
@@mikes_.5_cent Right you are, it didnt occur to me during the solve.
SPOILERS AHEAD!
The way I looked at it from the start is that the three two cell arrows on each side kind of see each other. ie the three cells on the left can't be the three on the right. Likewise the sums of the arrows must be different so must be a max of 17. So 6 cells must have a duplicate in and it must be the two cells you ended up identifying. The five unique digits you have on the line must add to at least 15 and so the duplicate must be at most 2. Then you can ask what would happen on the left if a 1 was on the right. That would put 2349 as the numbers on the left to satisfy the three cell arrow but then the two cell arrow couldn't be satisfied. So the right could be 224 = 8on the line but that would then leave 135 =9 on the left arrow which would again break the other left arrow. So now you know the left is 1348 and the right is 259 and something else which can quickly be concluded to be 7. It did take me a fair bit of thinking to get that all straight in my head though!
I started out by recognizing that the upper left arrow must be all different digits, i.e. at least 1+2+3 = 6. In general: A+B+C = D. Thereby that arrow defines the set of digits for the left grid. The 6 repeating in the top right. (I was coloring them in instead and goodliffed only the 6789 at the arrow bulb.) Also D at the end of row 1.
Then the other arrows bulb cannot be D, because it only sees two of the other digits. Also D is the largest digit in the grid, so it cannot be on the arrow part, either. That only leaves r4 on c2. Remaining D is trivial by sudoku. The bulb of the other arrow needs to be the next biggest digit (let's assume C) and the thermometer gives order to A and B, so A < B < C < D.
Colored the remaining grid and turned to the other.
After a bit, I realized that if C = A+B and D = A+B+C, that also means that D = 2×C, thus needs to be even, leaving only 6 and 8. 6 is simple to rule out:
Assuming E < F < G < H on the right grid, the lower arrow needs to be at best H = E + F + E. Choosing 6 will force A, B, C, D = 1, 2, 3, 6, leaving a minimum of 4, 5 for E and F, summing to 13, thus ruling out 1, 2, 3, 6 and instead fixing D = 8, directly followed by C = 4, quickly followed by the only combination of two DIFFERENT digits to sum to it being A, B = 1, 3.
Now the minimum on the right arrow would be E, F = 2, 5 resulting in the arrow being at least ? = 2 + 5 + 2 = 9. Being the highest possible digit, this forces the arrow to be an E+F+E arrow (instead of F+E+F or E+F+G in any order under the assumption that E < F < G < H). It also forces ? to be H, not G. With G being F+E on the other arrow. Choosing G is basic arithmetic now.
Despite the unnecessary early detour to the right grid, this was a nice little 5:20 solve.
Yes that is quite the dive into mind gymnastics. Thanks for the description.
10:10 2+3+5=10 which busts r2c1
right...wasnt thinking on the circle at that point and didnt see that.