LicianDragon wrote on 2018-07-05 13:08:11:
Does anyone have a strategy for how to beat it when you get have that one middle tile left? I get this outcome a lot.
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TOPIC | [G&G] Strategy Hub
[quote name="LicianDragon" date="2018-07-05 13:08:11" ]
[img]https://orig00.deviantart.net/711f/f/2018/186/6/5/capture_by_liciandragon-dcgdajz.png[/img]
Does anyone have a strategy for how to beat it when you get have that one middle tile left? I get this outcome a lot.
[/quote]
Do the rightmost middle one, then just the middle on the left. Or the reverse order, either works <:
[img]https://i.imgur.com/wkHQ5Gi.png[/img]
Does anyone have a strategy for how to beat it when you get have that one middle tile left? I get this outcome a lot.
@InkWings
I'm still struggling bit with the Special layouts, but I have been wondering, is the payout higher on the Special levels?
I'm still struggling bit with the Special layouts, but I have been wondering, is the payout higher on the Special levels?
@InkWings
I'm still struggling bit with the Special layouts, but I have been wondering, is the payout higher on the Special levels?
I'm still struggling bit with the Special layouts, but I have been wondering, is the payout higher on the Special levels?
@Evarinya
[url=http://www1.flightrising.com/forums/frd/2476523/2#post_34079997]Someone upthread[/url] said 2kt.
Piggybacking on a post above, this:
[img]https://i.imgur.com/aPjpeYS.png[/img]
turns to this:
[img]https://i.imgur.com/JuDaVTl.png[/img]
then this:
[img]https://i.imgur.com/xDJtNq3.png[/img]
Which you can complete with an inverse of @BezimiennyRaptor 's [url=http://www1.flightrising.com/forums/frd/2476523/2#post_34079752]solve here[/url]:
[img]https://i.imgur.com/D6qdxEh.png[/img]
@Evarinya
Someone upthread said 2kt.
Piggybacking on a post above, this:
turns to this:
then this:
Which you can complete with an inverse of @BezimiennyRaptor 's solve here:
Someone upthread said 2kt.
Piggybacking on a post above, this:
turns to this:
then this:
Which you can complete with an inverse of @BezimiennyRaptor 's solve here:
@/grimhound Working on it.
In the meantime, here's a solvable configuration:
[img]https://i.imgur.com/8M4BkIN.png[/img]
and several ways to get to it:
[img]https://i.imgur.com/IOIfwYX.png[/img]
[img]https://i.imgur.com/fNlWfd5.png[/img]
[img]https://i.imgur.com/2blFiuI.png[/img]
[img]https://i.imgur.com/rx5c5we.png[/img]
[img]https://i.imgur.com/EqYYTDI.png[/img]
[img]https://i.imgur.com/MJq4acD.png[/img]
[img]https://i.imgur.com/qYjp2Sy.png[/img]
[img]https://i.imgur.com/knhSOQK.png[/img]
[img]https://i.imgur.com/syBu5RW.png[/img]
@/grimhound Working on it.
In the meantime, here's a solvable configuration:
and several ways to get to it:
In the meantime, here's a solvable configuration:
and several ways to get to it:
@grimhound
Okay, got it! First four steps:
[img]https://i.imgur.com/EBuzrwm.png[/img]
[img]https://i.imgur.com/ZNKAPay.png[/img]
[img]https://i.imgur.com/heTPE3M.png[/img]
[img]https://i.imgur.com/n0YgCTQ.png[/img]
That gets you to this:
[img]https://i.imgur.com/wMPiRM4.png[/img]
Which is rotationally identical to one of my solves in the post above, so if you apply the same steps:
[img]https://i.imgur.com/w371HGY.png[/img]
(There's probably a faster way than 8 steps, but...)
@grimhound
Okay, got it! First four steps:
That gets you to this:
Which is rotationally identical to one of my solves in the post above, so if you apply the same steps:
(There's probably a faster way than 8 steps, but...)
Okay, got it! First four steps:
That gets you to this:
Which is rotationally identical to one of my solves in the post above, so if you apply the same steps:
(There's probably a faster way than 8 steps, but...)
For this configuration of special mode:
[img]https://i.imgur.com/tjIc3re.png[/img]
the middle hex should only take a few clicks at most, then since the outside half-hexes are all rotiationally identical, mess around with them until you get something symmetrical and then solve them with something like this:
[img]https://i.imgur.com/GbemMq5.png[/img]
or this:
[img]https://i.imgur.com/D4Df8Hz.png[/img]
or this:
[img]https://i.imgur.com/4sa7qvC.png[/img]
and so on:
[img]https://i.imgur.com/XQlW95m.png[/img]
[img]https://i.imgur.com/wKa94tc.png[/img]
[img]https://i.imgur.com/11GvX6d.png[/img]
For this configuration of special mode:
the middle hex should only take a few clicks at most, then since the outside half-hexes are all rotiationally identical, mess around with them until you get something symmetrical and then solve them with something like this:
or this:
or this:
and so on:
the middle hex should only take a few clicks at most, then since the outside half-hexes are all rotiationally identical, mess around with them until you get something symmetrical and then solve them with something like this:
or this:
or this:
and so on:
i'm not experienced in game theory or solving puzzles like these mathematically but i've got some ideas.... i've noticed on the hexagonal boards, you can change 7 spaces in the center, 5 spaces on an edge, or 4 spaces at a time. the amount of tiles on each board is as follows:
easy: 10 (3-4-3)
medium: 15 (3-4-5-4-3)
hard: 37 (4-5-6-7-6-5-4)
very hard: 61 (5-6-7-8-9-8-7-6-5)
the maths to solving, i would guess, would have something to do with these sums? something to do with flipping an even number of times = same tile, flipping an odd number of times = opposite tile. that might help someone to think about What Flipping A Tile Means... idk.
easy is unique in that it's an even amount of squares in the center row. each other difficulty has a single center tile. on medium difficulty, there is a win state where if you have the single center tile opposite the other tiles, you can click on two opposite edges and the other two opposite corners. i haven't found a similar win state for the other larger puzzles but they might exist. to fit like that, the tiles you want to change would have to be a sum using 4, 5 and 7 without any overlap.
i haven't found any particular win-states in the higher difficulties but there is this elimination pattern, which can happen on any of the corners:
[img]https://i.imgur.com/IqHKNqg.png[/img]
click B to turn the seven tiles light, click A-B or B-A to turn them shadow. below in the blue box is this pattern an extra step away from the edge. click C then A-B.
i generally try to chase tiles together two form a hexagon (7 tiles) or to the edges to be the partial-hexagons (4 or 5) but i don't have any reliable method :m i can't wrap my head around how to chase tiles to these sorts of positions tho without making everything else a mess, haha
EDIT: thought some more about the way shapes overlap across two clicks!
these are all possible overlap combinations of two clicks (rotations aren't unique). sorry it's crude as hell lol
[img]https://i.imgur.com/n0O0eMS.png[/img]
black dots/lines indicate edges, grey = non-overlap tiles i.e. tiles that are changed (flipped once)
the only generalized idea i can make from this is that, when using only two clicks, edges and corners are essential for eliminating an odd number of tiles. since all the three higher difficulty boards have an odd number of total tiles... hm.
i'm not experienced in game theory or solving puzzles like these mathematically but i've got some ideas.... i've noticed on the hexagonal boards, you can change 7 spaces in the center, 5 spaces on an edge, or 4 spaces at a time. the amount of tiles on each board is as follows:
easy: 10 (3-4-3)
medium: 15 (3-4-5-4-3)
hard: 37 (4-5-6-7-6-5-4)
very hard: 61 (5-6-7-8-9-8-7-6-5)
the maths to solving, i would guess, would have something to do with these sums? something to do with flipping an even number of times = same tile, flipping an odd number of times = opposite tile. that might help someone to think about What Flipping A Tile Means... idk.
easy is unique in that it's an even amount of squares in the center row. each other difficulty has a single center tile. on medium difficulty, there is a win state where if you have the single center tile opposite the other tiles, you can click on two opposite edges and the other two opposite corners. i haven't found a similar win state for the other larger puzzles but they might exist. to fit like that, the tiles you want to change would have to be a sum using 4, 5 and 7 without any overlap.
i haven't found any particular win-states in the higher difficulties but there is this elimination pattern, which can happen on any of the corners:
click B to turn the seven tiles light, click A-B or B-A to turn them shadow. below in the blue box is this pattern an extra step away from the edge. click C then A-B.
i generally try to chase tiles together two form a hexagon (7 tiles) or to the edges to be the partial-hexagons (4 or 5) but i don't have any reliable method :m i can't wrap my head around how to chase tiles to these sorts of positions tho without making everything else a mess, haha
EDIT: thought some more about the way shapes overlap across two clicks!
these are all possible overlap combinations of two clicks (rotations aren't unique). sorry it's crude as hell lol
black dots/lines indicate edges, grey = non-overlap tiles i.e. tiles that are changed (flipped once)
the only generalized idea i can make from this is that, when using only two clicks, edges and corners are essential for eliminating an odd number of tiles. since all the three higher difficulty boards have an odd number of total tiles... hm.
easy: 10 (3-4-3)
medium: 15 (3-4-5-4-3)
hard: 37 (4-5-6-7-6-5-4)
very hard: 61 (5-6-7-8-9-8-7-6-5)
the maths to solving, i would guess, would have something to do with these sums? something to do with flipping an even number of times = same tile, flipping an odd number of times = opposite tile. that might help someone to think about What Flipping A Tile Means... idk.
easy is unique in that it's an even amount of squares in the center row. each other difficulty has a single center tile. on medium difficulty, there is a win state where if you have the single center tile opposite the other tiles, you can click on two opposite edges and the other two opposite corners. i haven't found a similar win state for the other larger puzzles but they might exist. to fit like that, the tiles you want to change would have to be a sum using 4, 5 and 7 without any overlap.
i haven't found any particular win-states in the higher difficulties but there is this elimination pattern, which can happen on any of the corners:
click B to turn the seven tiles light, click A-B or B-A to turn them shadow. below in the blue box is this pattern an extra step away from the edge. click C then A-B.
i generally try to chase tiles together two form a hexagon (7 tiles) or to the edges to be the partial-hexagons (4 or 5) but i don't have any reliable method :m i can't wrap my head around how to chase tiles to these sorts of positions tho without making everything else a mess, haha
EDIT: thought some more about the way shapes overlap across two clicks!
these are all possible overlap combinations of two clicks (rotations aren't unique). sorry it's crude as hell lol
black dots/lines indicate edges, grey = non-overlap tiles i.e. tiles that are changed (flipped once)
the only generalized idea i can make from this is that, when using only two clicks, edges and corners are essential for eliminating an odd number of tiles. since all the three higher difficulty boards have an odd number of total tiles... hm.
••••••••
@snugglebug
#165006
he
@snugglebug
#165006
he
I'm working on a quick guide being built here. So far easy is solved for every permutation, and I'm working on getting images up.
Medium will follow, not sure about hard or special.
Medium will follow, not sure about hard or special.
I'm working on a quick guide being built here. So far easy is solved for every permutation, and I'm working on getting images up.
Medium will follow, not sure about hard or special.
Medium will follow, not sure about hard or special.
Wishlist collecting All the things!. Oh dear... Total quest on track!
Always looking for holiday toned UMAs.
Always looking for holiday toned UMAs.
