1b) qntx159bdfh5=qntx159e2468 loop => both qnts exist =>
(qnt loop) => ai5<>1,ch5<>5,ci5<>9,e19<>1,e37<>5,e17<>9
alt: (p59=p1x)bh5-(1)df5=(1)e46-(p1x=p59)e28-(5)e46=(5)df5
1c) rsf or csf 9 => d2,h6<>9
1d)All puzzle conatining hp loops, (formally generalized as a simple interplay of 8 box sis), such that the four focal cells, in this case r5c28,r28c5, have these cells limited to the same 3 or fewer candidates, in this case 159, will have the EMZ property such that for example (8=9),(8-9)r56c2. (A true quantum cell) This puzzle is bit simpler than many hpl puzzles, as only depth 7 is required max to achieve the 16 relationships. Once (1) is determined, the other 15 follow the hp loop path around the grid.
One does not have to go as deep as the grid determination detail below, but given the symmetry, it takes almost no additional human effort:
[(9)'s r5c2&r8c2]-(9)r8c5&r5c8=loop15Box9-(15=8)r6c8&r2c4&(8)grid determined]=
[(9)'s r5c8&r3c4-(9)r2c5&r5c2=loop15Box1-(15=8)r2c6&r6c2&(8)grid] loop
=> r4c1,r1c4,r7c6,r6c7<>8 note we have also proven Z equiv EM.
Here is a sample chain: (915)r5c2,(815)r5c2, (1)Box 1, (5)Box1,
(915)r2c5,(9)r8c53,(9)r94c2 => (8=9)r56c2. By symmetry, one knows:
(8=9)r45c8. Then, (8=9)r56c2-(9=8)r45c8 =< r4c1,r6c7<>8. One can break this down into small chunks, or do it all at once. There are several ways to reach the same conclusions. There need be no T&E.
Probably from this point, if Denis could program in the 16 weak and strong links proven in 1d, the puzzle should solve with *t chains. These are new cells, and no inferences are required.
- Code: Select all
*-----------------------------------------------------------*
| 9 126 237 | 124 2467 78 | 3467 378 5 |
| 16 4 567 | 3 19 158 | 78 2 679 |
| 237 256 8 | 2459 2467 2457 | 1 39 3467 |
|-------------------+-------------------+-------------------|
| 124 7 2459 | 6 1245 3 | 2458 18 249 |
| 2346 59 2346 | 1245 8 124 | 2347 159 2347 |
| 18 158 2345 | 7 245 9 | 2345 6 1234 |
|-------------------+-------------------+-------------------|
| 246 68 1 | 2458 2347 2457 | 9 57 367 |
| 78 3 79 | 58 159 6 | 25 4 12 |
| 5 269 24 | 149 2347 1247 | 367 137 8 |
*-----------------------------------------------------------*
Above is the puzzle at the point my mind got tired. There should still be many steps, clearly not T&E, available. I have one more letter left in the alphabet.
Denis, it appears that your T&E is very, very deep. Piecewise shallow, but since many determinations build on each other, deep in total.
Edit: depth 7 required required for step 1d series. By symmetry, similar other depth 7 or less steps exist to prove all 16 relationships. All of those steps are purely AIC. None are T&E. If one chooses not to remember the newly proven pseudo cells, one can reprove whichever relationship one requires each time. Added one example of the series.
).