- Code: Select all
*-----------*
|8..|4.3|...|
|...|.89|..4|
|.64|57.|...|
|---+---+---|
|.4.|9..|5..|
|32.|...|.46|
|..5|..4|.7.|
|---+---+---|
|...|.41|89.|
|5..|39.|...|
|...|7.5|..1|
*-----------*
Play/Print this puzzle online
*-----------*
|8..|4.3|...|
|...|.89|..4|
|.64|57.|...|
|---+---+---|
|.4.|9..|5..|
|32.|...|.46|
|..5|..4|.7.|
|---+---+---|
|...|.41|89.|
|5..|39.|...|
|...|7.5|..1|
*-----------*
8 57 e127 | 4 f16 3 | 2679 256 29
b127 357 1237 | 16 8 9 | 267 256 4
9 6 4 | 5 7 2 | 13 18 38
------------------------+----------------------+---------------------
c17 4 d178 | 9 23 6 | 5 18 23
3 2 189 | 18 5 7 | 19 4 6
6 89 5 | 128 123 4 | 1239 7 2389
------------------------+----------------------+---------------------
a27 37 2367 | 6-2 4 1 | 8 9 5
5 1 26 | 3 9 8 | 4 26 7
4 89 89 | 7 g26 5 | 26 3 1
+-----------------------+--------------------+-----------------------+
| 8 h57* i127 | 4 6-1 3 | 2679 256 29 |
| 127 35 1237 | 16 8 9 | 267 256 4 |
| 9 6 4 | 5 7 2 | 13 18 38 |
+-----------------------+--------------------+-----------------------+
| 17 4 178 | 9 a123 6 | 5 18 238 |
| 3 2 189 | 18 5 7 | 19 4 6 |
| 6 89 5 | 128 a123 4 | 1239 7 2389 |
+-----------------------+--------------------+-----------------------+
|f27 g37 2367 | 26 4 1 | 8 9 5 |
| 5 1 e2*6 | 3 9 8 | 4 d26 7 |
| 4 89 89 | 7 b26 5 |c26 3 1 |
+-----------------------+--------------------+-----------------------+
*------------------------------------------------------------*
| 8 e57 127 | 4 d16 3 | 2679 d256 29 |
|f127 357 1237 |ce16 8 9 | 267 256 4 |
| 9 6 4 | 5 7 2 | 13 18 38 |
*-------------------+--------------------+-------------------|
| 17 4 178 | 9 23 6 | 5 18 23 |
| 3 2 189 | 18 5 7 | 19 4 6 |
| 6 89 5 | 128 123 4 | 1239 7 2389 |
*-------------------+--------------------+-------------------|
| 7-2 37 2367 |ab26 4 1 | 8 9 5 |
| 5 1 ab26 | 3 9 8 | 4 c26 7 |
| 4 89 89 | 7 26 5 | 26 3 1 |
*------------------------------------------------------------*
SteveG48 wrote: 2r7c4,r8c3 = (6)r7c4&r8c3 - (6=12)r2c4,r7c8 - (1|2=56)r1c58 - (5|6=71)r1c2,r2c4 - (1|7=2)r2c1 => -2 r7c1 ; stte
+---------------------+-------------------+----------------------+
| 8 57 127 | 4 16 3 | 2679 256 29 |
| 127 357 1237 | 16 8 9 | 267 256 4 |
| 9 6 4 | 5 7 2 | b13 c18 38 |
+---------------------+-------------------+----------------------+
| e17 4 178 | 9 23* 6 | 5 d18 23* |
| 3 2 189 | 18 5 7 | 19 4 6 |
| 6 89 5 | 128 123* 4 | a1239## 7 2389* |
+---------------------+-------------------+----------------------+
| f27 37 2367 | 26 4 1 | 8 9 5 |
| 5 1 g26 | 3 9 8 | 4 h26 7 |
| 4 89 89 | 7 26# 5 | 6-2 3 1 |
+---------------------+-------------------+----------------------+
(2)r7c1 = (2-1)r2c1 - (27=1)r1c3 - (1=6)r1c5 - (6=2)r9c5 => -2 r7c4; stte
\\ /
(1-7)r4c1 = r4c3
pjb wrote:
- Code: Select all
(2)r7c1 = (2-1)r2c1 - (27=1)r1c3 - (1=6)r1c5 - (6=2)r9c5 => -2 r7c4; stte
\\ /
(1-7)r4c1 = r4c3
If I wrote my solution like this one could call it "branching " rather than "memory". Is there any real difference?
(2)r7c1 = (2)r2c1 ------------ (27=1)r1c3 - (1=6)r1c5 - (6=2)r9c5 => -2 r7c4; stte
| |
(1)r2c1 = (1-7)r4c1 = (7)r4c3
Correct me please if I'm wrong, but doesn't Steve's solution have "branching"?
pjb wrote:(2)r7c1 = (2-1)r2c1* = (1-7)r4c1 = r4c3 - (27=1)r1c3* - (1=6)r1c5 - (6=2)r9c5 => -2 r7c4; stte
Ngisa wrote:(1=2)r46c5 - r9c5 = r9c6 - r8c8 = r8c3* - (2=7)r7c1 - r7c2 = r1c2* - (2*7*=1)r1c3 => - 1r1c5; stte
SpAce wrote:Yes, it does. However, the branching is internal to the nodes, which makes it a split-node AIC. The chain itself doesn't branch, and all the nodes are boolean variables, which makes it a valid AIC as per David's definitions. That doesn't change the fact that it branches.
SpAce wrote:I think my alternatives would make it more obvious where a memory is stored and where it's recalled. I've said it before but never received any comments. Is there a good argument for the prevailing, imho less readable, style? Anyway, nice solutions, everyone
SteveG48 wrote:I simply prefer to write bi-directional AICs per David's definitions whenever I can, even if that sometimes makes the chain more complex than it might be if other techniques were used. ... It's just a matter of taste, but I thank SteveC for noticing.
I agree. I prefer "markers next to the relevant candidates, on opposite sides (pointing to each other)". That's the way Danny Jones liked to do it.