Luminations
Rick Nungester
by Rick Nungester, April 28, 2002
Luminations is a 5-inch high battery operated pyramid with a red,
yellow, and green LED on each of its 4 points. It senses which point is
up. As a new point is raised, the LEDs on 1 to 4 points change state
among red, yellow, green, flashing red, flashing yellow, flashing green,
or all off, abbreviated rygRYGx respectively. The goal of each of the 5
difficulty levels is to make all points red. 5 characters give puzzle
state, for example 2xrgx means point 2 is up, and points 1,2,3,4 are
off, red, green, off.
To start Levels 1 through 4, turn on the power with point 1 through 4
up. Level 5 can only be started by first completing Level 4.
--------------------------------------------------------------------
Level 1 (1xxxx, poweron, tilt for random start state). Points cycle
rgxyG (a cycle of 5). A point only changes color when brought to the up
position.
Algorithm:
- Pick one of the 3 down points as an intermediate target
color. Keep it down and "rock" the puzzle until an up
point is the same color.
- Make the two same-color points both down. Keeping these two
down, rock more until an up point is the same color.
- Now 3 or 4 points are the same color. If 3, make all 3 the
same color as the 4th by keeping the 4th down while "rolling"
the puzzle so the other 3 each cycle through their states to
match the 4th.
- Once all 4 points are the same color, if red, you're done.
If all 4 are the same color but not red, raise each one in
sequence (for example 2431) how ever many times is necessary
to move all 4 to red.
--------------------------------------------------------------------
Level 2 (2xxxx, poweron, tilt for random start state). Points cycle
rYxgy (a cycle of 5). A point changes color when moving either up or
down.
Algorithm:
- Make any two down points the same color.
- Leave these two down and rock until the top point is also the
same color.
- If all 4 are the same color and red, you're done. If all 4
are the same color, not red, skip the next step.
- Make the 3 like-colored points (abc) match the 4th down-point
(d) by raising a,b,c, repeatedly. If all 4 are red, you're
done. If not, continue.
- All 4 points are the same color, not red. Say a is up, bcd
are down. Raise bcda. The color of all 4 points has advanced
by 2 colors. Repeat until all are red.
----------------------------------------------------------------------
Level 3 (3xxxx, poweron, tilt for random start state). Points cycle
rgRYGy (a cycle of 6). A point changes color only when it moves up, and
only if was not the previous 2 up points. Example sequences of up
points:
- 234 (2 or 3 next will not change color. 1 next will.)
- 121 (2 next will not change color. 3 or 4 will.)
Algorithm:
- Get a red point on top. Call this is point A.
- For all future moves, do BA, CA, or DA. This leaves A red.
Never repeat BA twice, CA twice, or DA twice. This means
as A, B, or C is raised it will always change color. This
allows cycling B, C, or D colors in a controlled manner.
- Make B and C the same color.
- Make D match B and C by the sequence BADACADA as many times as
needed. (This sequence changes D 2 colors for every 1-color
change in B and C.)
- Now B, C, D are the same color. Do the sequence BACADA as many
times as needed to make all points red.
----------------------------------------------------------------------
Level 4 (4xxxx, poweron, tilt for random start state). Points cycle
ryxYgRG (a cycle of 7). If point A is raised, A doesn't change color,
but BCD do.
Algorithm:
- If A is up, move BABABABA... until any two low points are the
same color.
- Keep the two some-color low points low, and alternate the two
other points in the up position until all 3 low points are
the same color. If all 4 points are red, you're done. If all
4 are the same color but not red, skip the next step.
- Say A is up. BCD are the same color, and different than
A. Moves BACADA change A forward 3 colors (= back 4) and
change BCD forward 5 colors (= back 2). Repeat this move
sequence until ABCD the same color. If all are red, you're
done, else continue.
- All 4 points are the same color, not red. Say A is up. Repeat
the moves BCDA until all points are red. Each sequence moves
all 4 points forward 3 colors (= back 4).
----------------------------------------------------------------------
Level 5 (solve Level 4, hear beeps, tilt for random start state).
Points cycle rGYgRxygY (a cycle of 9). Each move, each point steps its
color forward in the 9-cycle the number of times it was the up-point in
the current and previous 3 states. (This means each point changes by 0,
1, or 2 steps in the 9-cycle each move. Each move, total color steps on
all points is always 4.)
Notation: XYZ ABCD indicates a puzzle state. X, Y, and Z represent
up-points (value 1 through 4) two moves ago, one move ago, and now,
respectively. A, B, C, D represent the 9-cycle color position
(rGYgRxygY = 012345678) of point 1 through 4 respectively. Example: 124
0836 means point 1 was up, then point 2, then point 4, and the colors on
points 1 through 4 are red, flashing Yellow (following green), green
(following flashing Yellow), and yellow.
Algorithm:
This algorithm is easiest to implement using a spreadsheet, but can
also be done with pencil and paper. Level 5 may be solved in at most
23 moves.
Determine start state: Notice the up-point 1 step before the solution
to Level 4 (A). (Yes, this actually affects Level 5.) Solve Level 4
and notice the up-point (B, B <> A). Start Level 5 by raising point C
(C <> B and C <> A). Notice the point colors. If none are Y or g, you
have enough information to solve the puzzle (in 22 steps max) -- current
and 2 previous up-points, and position on the 9-cycle of colors for each
point. If one or more points are Y or g, raise point D (D <> C and D <>
B and D <> A). This causes all point colors to advance in the 9-cycle
by 1. Now any ambiguity due to reuse of Y and g is gone, and enough is
known to solve the puzzle. (This also uses 1 move to determine complete
starting state, resulting in 23 moves maximum.) Example:
- Point 3 is up 1 move before the end of Level 4.
- Point 1 is up at the solution to Level 4.
- Start Level 5 by raising point 2 or 4. Say we choose 2. Level 5
has started, and point colors are rxYg = 0 5 (2 or 8) (3 or 7).
- Ambiguity of Y and g must be removed by another move. Up-point
history so far is 312, so raise point 4, which will advance all
points by 1 color. Now colors are GygR = 1634, with no ambiguity.
- Start state is 124 1634, with one move being spent determining it.
See the addendum "1- and 2-move solutions". The rest of this solution
process ignores these few trivial cases, and will still solve them, just
with more solution steps than required.
Generate point colors EFGH, that account for existing previous
history, by incrementing (mod 9) point X, Y, Z by 1, 2, 3 respectively.
For example, start state 124 1634 would imply incrementing point 1 by 1,
point 2 by 2, and point 4 by 3, resulting in EFGH = 2837.
Determine the minimum number of color steps each of EFGH must advance
to be red. This is e=9-E, f=9-F, g=9-G, h=9-H, for points 1 through 4
respectively, with the "-" being done mod 9 (9 - 0 = 0, not 9).
A solution is achieved when points 1 through 4 advance in unison by
e+9*i, f+9*j, g+9*k, h+9*m, where i, j, k, m are whole numbers (0, 1, 2,
3, ...).
Example: EFGH = 2837 means efgh = 7162.
Use e, f, g, h, and the following table to pick solution moves:
Ways to step forward n (0-8) colors
((n+9*i)/4, quotient & remainder)
n= 0 1 2 3 4 5 6 7 8
i q r q r q r q r q r q r q r q r q r
= ===== ===== ===== ===== ===== ===== ===== ===== =====
0 0 0 0* 1* 0* 2* 0 3 1 0 1 1 1 2 1 3 2 0
1 2 1 2 2 2 3 3 0 3 1 3 2 3* 3* 4* 0* 4 1
2 4 2 4 3 5 0 5 1 5 2 5 3 6 0 6 1 6 2
3 6 3 7 0 7 1 7 2 7 3 8 0 8 1 8 2 8 3
4 9 0 9 1 9 2 9 3 10 0 10 1 10 2 10 3 11 0
5 11 1 11 2 11 3 12 0 12 1 12 2 12 3 13 0 13 1
6 13 2 13 3 14 0 14 1 14 2 14 3 15 0 15 1 15 2
^^^^^ ^^^^^ ^^^^^ ^^^^^
Point: 2 4 3 1
e, f, g, and h each choose a column, for the example being developed, 7,
1, 6, and 2. (The same column can be used twice if e, f, g, or h have
the same value.) Under each column, write the point corresponding to e,
f, g, h as shown. Pick from each column an r,q pair such that the 4
picks have remainder ("r") value 0, 1, 2, and 3. Choose toward the top
of the table as much as possible to keep total solution length low. "*"
indicates a valid choice for the example problem.
The last 3 moves of the solution are the points beneath the columns with
remainder 3, 2, 1 respectively. In the example, points 3, 4, 2. (This
is because the last, prev, prev-prev move has its point incremented only
1, 2, 3 times, while any other solution moves contribute 4 increments to
its point.) The "q" values indicate how many more times the point
beneath the column has to be included in the solution: 0 2s, 0 4s, 3 3s,
4 1s. Lastly, the solution can't have the same up-point appear twice in
a row. Remember we're solving 124 1634, so 4 can't be the first
solution move. A solution is 1313131342 (3 3s, 4 1s, then 342, with no
value repeating and the first move different than the current up-point).
Another more complicated example: Solve 343 3881.
point
1234
====
3881 starting colors
3833 colors after incrementing 3 by 1, 4 by 2, 3 by 3, mod 9
6166 color steps to solution = (9 - <color above>), 9-0=0 when
applicable
n= 0 1 2 3 4 5 6 7 8
i q r q r q r q r q r q r q r q r q r
= ===== ===== ===== ===== ===== ===== ===== ===== =====
0 0 0 0* 1* 0 2 0 3 1 0 1 1 1* 2* 1 3 2 0
1 2 1 2 2 2 3 3 0 3 1 3 2 3* 3* 4 0 4 1
2 4 2 4 3 5 0 5 1 5 2 5 3 6* 0* 6 1 6 2
3 6 3 7 0 7 1 7 2 7 3 8 0 8 1 8 2 8 3
4 9 0 9 1 9 2 9 3 10 0 10 1 10 2 10 3 11 0
5 11 1 11 2 11 3 12 0 12 1 12 2 12 3 13 0 13 1
6 13 2 13 3 14 0 14 1 14 2 14 3 15 0 15 1 15 2
^^^^^ ^^^^^
Points: 2 1,3,4
"*" = first solution attempt (remainders 0,1,2,3), but it won't work.
Notice the q=6 value, and other q values are only 0, 1, and 3. Whatever
point is associated with the q=6 value can't repeat, which requires 5
"separator" values. But there are only 0+1+3=4. So try another
solution:
n= 0 1 2 3 4 5 6 7 8
i q r q r q r q r q r q r q r q r q r
= ===== ===== ===== ===== ===== ===== ===== ===== =====
0 0 0 0 1 0 2 0 3 1 0 1 1 1* 2* 1 3 2 0
1 2 1 2 2 2 3 3 0 3 1 3 2 3 3 4 0 4 1
2 4 2 4* 3* 5 0 5 1 5 2 5 3 6* 0* 6 1 6 2
3 6 3 7 0 7 1 7 2 7 3 8 0 8* 1* 8 2 8 3
4 9 0 9 1 9 2 9 3 10 0 10 1 10 2 10 3 11 0
5 11 1 11 2 11 3 12 0 12 1 12 2 12 3 13 0 13 1
6 13 2 13 3 14 0 14 1 14 2 14 3 15 0 15 1 15 2
^^^^^ ^^^^^
Points: 2 1,3,4
Which of points 1,3,4 goes with remainder 2,0,1 is arbitrary. Say point
1,3,4 goes with remainder 2,0,1 respectively. The last 3 solution moves
are r=3, r=2, r=1, or points 2,1,4. Preceding these steps are 4 2s, 1
1, 6 3s, and 8 4s. Remember we're solving 343 3881, so the first
solution step can't be a 3. One solution:
2424243434343434321214
| | | | 4 2s
| 1 1
| | | | | | 6 3s
| | | | | | | | 8 4s
||| ending in 214
343 3881 was intentionally chosen, as one of the few starting states
requiring at least 22 moves.
It is important to note that for this technique to work, it must be
known that any start state is solvable, and in under 22 moves. The
table above need not be developed further since the most any single move
can appear in the solution is 11 times (every other position of a
22-move solution). This was proven by a program, with results given
below.
----------------------------------------------------------------------
Addendum: 1- and 2-move solutions to Level 5
The algorithm above assumes a solution of 3 or more steps. A method of
determining if the start state has a 1- or 2-move solution follows,
which is simply "try the 3 next moves and 9 next move pairs".
1-move solution: Given XYZ abcd (abcd = desired color moves of points
1234), try next-move T as the the 3 non-Z values. If incrementing
points XYZT by 1 totals abcd (a=9-A, b=9-B, c=9-C, d=9-D, mod 9, meaning
9-0=0), you have a solution. Example: Given XYZ ABCD = 343 0077, then
abcd = 0022. (Points 1 and 2 are already red. Points 3 and 4 are 2
steps back.) Try 3431, 3432, (3433 is invalid because 3 is repeated),
3434:
3431 increments points 1234 by 1021, not 0022. Not a solution.
3432 increments points 1234 by 0121, not 0022. Not a solution.
3434 increments points 1234 by 0022. A solution.
2-move solution: Given XYZ abcd, try all 9 possible next move pairs T
and U. If incrementing point X,Y,Z,T,U by 1,2,2,2,1 totals abcd, you
have a solution. Example: Given XYZ ABCD = 343 8065, then abcd = 1034.
34312 increments points 1234 by 2132, not 1034. Not a solution.
34313 increments points 1234 by 2042, not 1034. Not a solution.
34314 increments points 1234 by 2033, not 1034. Not a solution.
34321 increments points 1234 by 1232, not 1034. Not a solution.
34323 increments points 1234 by 0242, not 1034. Not a solution.
34324 increments points 1234 by 0233, not 1034. Not a solution.
34341 increments points 1234 by 1034. A solution.
34342 increments points 1234 by 0134, not 1034. Not a solution.
34343 increments points 1234 by 0044, not 1034. Not a solution.
----------------------------------------------------------------------
Addendum: How Many Level 5 Start States?
There are 4 possible up-points. Given current up-point, there are 3
possible previous up-points. Given previous up-point, there are 3
possible previous previous up-points. There are 9 colors (actually
"positions on the 9-cycle of colors") for each of the 4 points. This
makes 3*3*4*9*9*9*9 = 236,196 starting states total. If positions on
the 9-cycle of colors (rGYgRxygY) are assigned the digits 0 through 8, a
puzzle state can be represented as 7 digits: previous previous up-point,
previous up-point, current up-point, position in the 9-cycle of colors
of points 1, 2, 3, and 4.
A program was written to start at the solution and work backwards to
generate all states that can lead to it. Results:
36 games at level 0, example: 434 0000
108 games at level 1, example: 343 0077
324 games at level 2, example: 343 8065
972 games at level 3, example: 343 8754
2700 games at level 4, example: 343 8723
6084 games at level 5, example: 343 8710
10944 games at level 6, example: 343 8778
14832 games at level 7, example: 343 8855
19044 games at level 8, example: 343 8824
21384 games at level 9, example: 343 8811
23724 games at level 10, example: 343 8870
24552 games at level 11, example: 343 5446
24840 games at level 12, example: 343 8835
23364 games at level 13, example: 343 0002
20160 games at level 14, example: 343 8881
15300 games at level 15, example: 434 7878
11412 games at level 16, example: 434 7847
7200 games at level 17, example: 434 7852
4824 games at level 18, example: 434 7857
2412 games at level 19, example: 434 7853
1368 games at level 20, example: 434 7858
432 games at level 21, example: 434 6207
180 games at level 22, example: 434 3818
236196 games of 236196 total were found
All counts above are evenly divisible by 36, due to puzzle symmetry.
Given starting state XYZ ABCD, there are 36 values for XYZ: 4 for Z, 3
for Y (Y <> Z), 3 for X (X <> Y). 3*3*4 = 36. But since point number
assignments are arbitrary, these 36 possibilities can be reduced to 3 --
Take any starting state and number the up-point 1. Hold the previous
up-point toward yourself, and number it 2. Number the back-left 3 and
the back-right 4. Using this point numbering system, XYZ = 121, 321, or
421. But these 3 can be further reduced to 1, by noticing these 3 start
games have the same solution:
121 A B C D
321 A+1 B C-1 D
421 A+1 B C D-1
where the +/- operation is mod 9 (8+1=0, 0-1=8). If all 121 ABCD
solutions are known, all 321 ABCD and 421 ABCD solutions can be
inferred.
Example: Solve 213 7308. Point 3 is currently up. Point 1 was
previously up, so point it at yourself. Point 2 is back-left and point
4 is back-right. (These last 2 are due to how the puzzle is physically
labeled.) Renumber point 3 to 1 (up), 1 to 2 (front), 2 to 3
(back-left), and 4 to 4 (back-right). Now solve puzzle 321 0738. But
these have the same solution:
121 8748
321 0738 <<<
421 0747
By convention, I solve the 121 history set, and extrapolate to other
start histories, so would solve 121 8748. Minimum solution length for
this starting state is 18, and one solution is:
212121212121434243 (a solution to 121 8748 and 321 0738 and 421 0747)
131313131313424142 (the same solution mapped back to original point
numbers)
A detail: 121 0000 need not be considered solved, and can be moved
through to another solved 0000-color state, via solution 212121212.
This solution provides the basis for 35 other symmetrical start states.
So eliminating symmetries, there are only 236,196/36 = 9^4 = 6561
starting states: 121 0000, 121 0001, 121 0002, ..., 121 8888.
----------------------------------------------------------------------
Addendum: Understanding Progression Toward a Level 5 Solution
The following table shows what is happening during a solution, and helps
understand the solution algorithm. A solution to 121 0836 is 2123431:
121 0836 starting state
212 2136 2 up
121 4336 1 up
212 6536 2 up
123 7746 3 up
234 8857 4 up
343 8078 3 up
431 0000 1 up, solution, all points red
Notice the number of color increments due to each move. Moves -2, -1, 0
increment their up-points a total of 1, 2, 3 times respectively. Moves
last-2, last-1, last, increment their up-points a total of 3, 2, 1 times
respectively. All mid-solution up-points (between the first 3 and last
3) increment their up-points a total of 4 times:
0836 starting colors
1 1 prev prev up-point increments
2 2 prev up-point increments
1 3 starting up-point increments
2 4 first solution move increments
1 4 ...
2 4 ...
3 4 ...
4 3 (last - 2) up-point increments
3 2 (last - 1) up-point increments
1 1 last up-point increments
====
0000 sum of column, mod 9 (all red, a solution)
----------------------------------------------------------------------
Addendum: How many Level 5 paths to a solution at depth n?
The number of paths to a solution goes up as the minimum number of
solution steps goes up, because the number of ways of rearranging the
mid-solution steps goes up. Here are some test cases:
min
sol n-step
start steps solutions paths (3^n) ratio
======== ===== ========= =========== =====
121 0836 7 5 2187 0.229%
343 8881 14 9354 4782969 0.196%
343 7860 15 29646 14348907 0.207%
343 7820 16 83252 43046721 0.193%
It is interesting that the 0.2% ratio seems fairly constant. I didn't
investigate this further.
----------------------------------------------------------------------
Addendum: Original notes that led to understanding the Level 5 "rules"
move number
| puzzle state
| | color increments
| ===== ====
-1 4 \_ Both can be inferred after state 3 is known
0 2 / color increments
1 1gRxg ----
2 2RyxY 1201 <<< 2nd logic) state -1 had point 4 up
3 3xYyY 1210 <<< 1st logic) state 0 had point 2 up
4 4yrgr 1111 with 3 history, 12341234... changes each point by 1 on
9-cycle
5 1gGYG 1111
6 2YYrY 1111
7 3rgGg 1111
8 4GRYR 1111
9 1Yxgx 1111
10 2gyRy 1111
11 3Rgxg 1111
12 4xYyY 1111 = 3, 9 moves back
13 1yrgr 1111 = 4
14 2gGYG 1111 = 5... 12341234... changes each point by 1
15 4YYYg 1102 change pattern to 414141...
16 1GgYR 2101
17 4YRYy 1102
18 1RRYY 2002 4141... changes 1 and 4 by 2
19 4yRYG 2002 "
20 1YRYg 2002 "
21 2GxYR 2101 change pattern to 123123...
22 3Yyrx 1111
23 1RgGx 2110
24 2xrYx 1210
25 3yGRx 1120
26 1YYxx 2110 123123... results in 1-2-3 change by 2 each.
27 2rRyx 1210
28 3GxYx 1120
*** Each point is incremented the number of times it was up in the
current and previous 3 states. ***
----------------------------------------------------------------------
Addendum: Lotus 123 analysis of one 22-step solution to 343 3881
# u 1 2 3 4 point number of column
-2 3 1 past history (343) pending color increments
-1 4 2
0 3 3
-------------
3 8 8 1 start colors
3 8 3 3 start+pending colors (mod 9)
6 1 6 6 required color steps to solution
6 1 6 6 color steps below, mod 9 (SOLUTION CHECK)
-------------22 solution steps below
1 2 4 solution step color increments
2 4 4
3 2 4
4 4 4
5 2 4
6 4 4
7 2 4
8 4 4
9 2 4
10 4 4
11 2 4
12 4 4
13 2 4
14 3 4
15 2 4
16 3 4
17 2 4
18 3 4
19 1 4
20 3 3
21 1 2
22 2 1
Jaap
Rick and I shared our results last weekend, it turned out we had been
thinking along the same lines.
Although I had been planning to do a similar computer analysis as
Rick's, I don't think I will bother doing it now. Using these ideas I
did find an easy solution to the level 5 game. It is far from being
shortest of course.
Label the four corners A-D in any order. Any sequence of moves can be
simply denoted by listing which corners are brought to the top
position. The solution is as follows:
1. Do BCDA.
2. Repeat B CBDA BCDA until corners B and A are the same.
3. Repeat C BCDA until corner C is the same as A and B.
4. Repeat D BCDA until corner D is the same as A, B, and C.
5. Repeat BCDA until solved.
This uses the BCDA sequence repeatedly to make sure the history of the
up-corners is always the same, and does each corner one extra time
until it matches. You can do extra BCDA's in between if you have an
ambiguous colour (green or flashing yellow) to make sure the corners
really match. Also, you can actually combine many steps, e.g. do CD
BCDA if both C and D do not match A.
Jaap
Rick Nungester
I have the following feedback regarding the web page you sent me for
review.
First and foremost: Good job! As usual, you did a summary that will be
very useful in the future.
Regarding Level 5 starting positions:
Jaap: "Each of the four corners is in one of 9 states, so there are at
most 9^4=6561 visibly different positions."
Because of the repetition of green and flashing yellow, they aren't all
visibly different. "At most" can be omitted since the value is exact.
Consider replacing "visibly different positions" with "unique positions
in the 9-cycle of colors, on the 4 points" or similar.
Jaap: "It is more complex however due to the fact that it matters which
corners were up. We can assume that the puzzle is held so that the
previous up corner lies at the back. The up corner before that can be
one of three possibilities, so there are really 3ยท94=19683 positions. If
reflections are considered the same, then there are 10206."
"Position" isn't defined, so we have different numbers. Considering the
points as numbered on the puzzle, there are 236196 combinations of
previous-previous up-point, previous up-point, current up-point, and
"positions in the 9-cycle" (not colors) on the 4 points. Your 19683
number is already reduced some.
Jaap: "It is interesting that the level 5 column contains only multiples
of 3. This is because the positions come in sets of three which look
different, but if the effect of the previous up corners are taken into
account they are actually the same. If any particular move is performed
on such a triplet, the result will be the same on all three."
This is unclear unless more explanation is included, as is in my
previous post. In summary, your Level 5 "19683" number seems low, and
"10206" seems high. I think it is more correct to start with 236196 and
reduce it by a factor of 36, using the reasoning in my posting.
Defining
"position" might be adequate.
Jaap: "Solution to Level 5: Label the up corner A, and the other three
B, C, D."
This step isn't necessary, since the puzzle points are already labeled
1, 2, 3, 4. Since the number of steps in your solution doesn't matter,
it would be shorter and clearer to refer to the points by 1, 2, 3, 4
here, as they are labeled. Then map "BCDA" to "1234".
Jaap: "The solutions to levels 1 to 4 are by Rick Nungester."
That's all the credit you gave me, yet much more of the Level 5 content
on your page first appeared in my earlier posting. There is also no
mention of my entire Level 5 algorithm, which leads to much shorter
solutions than yours or Courtney McFarren's (which is mentioned). Was
this intentional?
Your website is of course yours to do with as you please, but I would
feel better about all the effort I put into this if the above things
were changed.
Thanks again for your website -- It is the best of its kind. I would
really appreciate having my Luminations work posted there.
Rick
Jaap
Rick,
Thanks for your comments. I agree with them, and will incorporate the
changes before putting the webpage up.
> That's all the credit you gave me, yet much more of the Level 5 content
> on your page first appeared in my earlier posting. There is also no
> mention of my entire Level 5 algorithm, which leads to much shorter
> solutions than yours or Courtney McFarren's (which is mentioned). Was
> this intentional?
Sorry, I forgot to include the link to the newsgroup post(s). I am
still debating whether or not to include your full solution as is. I
feel that if I do that, I should also include similar solutions for
the other levels. This would change the explanation a bit, seeing as
getting a solution from a set of numbers in a table is something that
occurs on earlier levels as well if you are trying to find a shortest
method. I'm still mulling things over.
Jaap
Rick Nungester
Jaap:
I've been busy with work/home things, but plan to clean up and finish
the Level 5 solution I posted, including a better explanation of WHY it
works and how it can be used to find a minimum-step solution.
I'm not a web designer -- How does linking to a newsgroup post work?
Don't newsgroup posts "go away" months after being posted? That's one
of the reasons I was hoping to get my work on a real, maintained website
like yours.
Rick
"Jaap" <jaa...@hotmail.com> wrote in message
news:30fbb89.02050...@posting.google.com...
Michael Mendelsohn
> I'm not a web designer -- How does linking to a newsgroup post work?
> Don't newsgroup posts "go away" months after being posted? That's one
> of the reasons I was hoping to get my work on a real, maintained website
> like yours.
Currently, news articles are archived on groups.google.com; you can link
to that archive. This will work only as long as google keeps the same
search query syntax, and maintains the archive (dejanews used to offer
this service, but does no more). So having them on a maintained website
(with a real domain, not some "free" space) is certainly preferable.
Have fun with new slinks
Michael
--
When the tongue or the pen is let loose in a phrenzy of passion,
it is the man, and not the subject, that becomes exhausted.
-- Thomas Paine, "On Usenet"
Rick Nungester
easy to remember, avoids all point color ambiguities due to the
repeated sequencing of green and flashing yellow, and requires only
checking the color of the current top point. It takes 112 moves
maximum, 70 moves average. Reaching a solution can happen anywhere
during the algorithm. The 4 points of the pyramid are numbered
1,2,3,4. A move consists of raising the named point. For example,
"21, then repeat 21 until..." means raise point 2, raise point 1,
then repeat raising point 2 and raising point 1, one or more times,
until... In other words, don't look for the "until" condition
until after the "repeat" action has been done at least once.
- If point 2 is up, then raise point 1.
- 21, then repeat 21 until point 1 is flashing red.
- 32, then repeat 32 until point 2 is all-off.
- 43, then repeat 43 until point 3 is yellow.
- Repeat 1234 until point 4 is yellow.
- Repeat 123 until solved.
Rick Nungester
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