Test
Prompt:
I want you to design efficient structures in block-building game.
The game has blocks similar in size as Minecraft, but:
blocks can't be attached to sides of other blocks - only on top of them;
there are not only 1x1 blocks, but also 2x1 blocks - so you can place a 2x1 block on top of one 1x1, and then be able to place blocks on top of both halves of that 2x1;
characters can't just go from elevation n to n+1 without stairs, stairs are also a block, and they can be straight or have 90 degrees turn. Stairs need to have an empty space above them, so you can't place blocks directly on top of them.
let's define referencing the blocks:
Y = elevation starting at 0
X = W-E direction starting at 0
Z = N-S direction starting at 0
Let's write the state with:
0 = empty,
1 = 1x1 block
2(A) = part of 2x1 block, with A = side at which the second part is
S(A>B) = stairs, where A = side at the bottom, and B = side at the top
we can write tables for each elevation, with X for columns (left-to-right), and Z for rows (top-to-bottom)
For the first task, let's say we have:
elevation 0:
0,0,0,0
1,0,1,0
0,S(N>E),1,0
0,1,0,0
elevation 1:
0,0,0,0
1,0,1,0
0,0,S(W>N),0
0,1,0,0
elevation 2:
0,0,0,0
2(E),2(W),S(S>W),0
0,0,0,0
0,1,0,0
elevation 3:
0,0,0,0
0,S(E>S),0,0
0,2(S),0,0
0,2(N),0,0
Your task is to extend this stairway to 2x of it's height
You can not only add blocks to elevations above 3, but also to empty spaces at elevations 0-3
You can use any other notation for the state if you think it's better than what I offered.
Efficiency costs are:
1x1 = 3
stairs = 3
2x1 = 5
============= o1 high =============
Below is one illustrative way to continue the same “spiral” stair upward from elevation 3 to elevation 6, while respecting the rules:
- We do not remove or alter any of the original blocks; we only add.
- A new stair (or block) may only be placed if there is solid support immediately below (i.e.\ a full block or the top of a 2×1).
- We never place a block/stair directly above an existing stair. (Each stair at ((x,z,y)) requires ((x,z,y+1)) be empty.)
- Blocks cannot “float” or “attach sideways”; they must sit directly on top of a solid block beneath.
- We try to be reasonably economical, using 2×1 blocks in places where we need to fill a 2-block-wide floor.
Because the existing spiral at elevations 0–3 is quite tight, the main challenge is that many of the cells above it are either already occupied or are directly above stairs (and thus forbidden for new blocks). In particular, the columns at ((2,2)), ((2,1)), etc., are “blocked” above a stair, so we cannot simply continue the old spiral in-place.
Instead, the trick is to build a small “bridge” sideways at elevation 3 (where we do have solid blocks directly below), walk the player over by one or two cells, and then start stacking new stairs upward from there. Below is one concrete plan.
Key Idea of the Extension
- Use the solid 2×1 blocks at ((x=0, z=1,2,3; y=2,3)) as “anchors” that let us put additional blocks at ((0,1, y=3)) and ((0,2, y=3)).
- In the original layout, at elevation 2 we have a 2×1 block spanning ((0,1))–((1,1)). That means ((0,1,y=2)) has a full block underneath. Likewise at elevation 3 we have a 2×1 spanning ((1,2))–((1,3)), but that leaves ((0,2,y=2)) free for us to stack up from below.
- Fill in a small vertical “pillar” at ((0,2)) from y=0 up to y=3, so that ((0,2,y=3)) becomes a solid top for us to stand on or bridge from.
- Place one bridging block at ((0,1,y=3)) so the player can step sideways from the top of the existing spiral (which ends at ((1,1,y=3))) onto that new bridging block at ((0,1,y=3)), and then walk south one cell to ((0,2,y=3)).
- From ((0,2,y=3)), build the new stairs upward. Because each stair needs an empty space above it, we will alternate “block then stair” in a small offset pattern.
Below is a step‐by‐step of one way to do it. (There are other variants!)
Step-by-step Description and Cost
(A) Build a “pillar” under ((0,2,y=3))
By the original tables, (0,2) is empty all the way up. So we stack four 1×1 blocks there:
- At ((0,2,y=0)) =
1x1(cost 3) - At ((0,2,y=1)) =
1x1(cost 3) - At ((0,2,y=2)) =
1x1(cost 3) - At ((0,2,y=3)) =
1x1(cost 3)
So now ((0,2,y=3)) is a solid “floor.” Subtotal so far: 4 blocks × 3 = 12.
(B) Place a bridging block at ((0,1,y=3))
We want the player (who emerges on top of the stair at ((1,1,y=3))) to be able to step sideways onto ((0,1,y=3)). Below ((0,1,y=3)) is ((0,1,y=2)), which (in the original) is part of a 2×1 block (2(E)), so that’s already solid support.
- At ((0,1,y=3)) =
1x1(cost 3)
Now the player can stand at ((0,1,y=3)), then walk one cell south to ((0,2,y=3)). Running total: 12 + 3 = 15.
(C) Add the first new stair at ((0,2,y=4))
- We have a solid 1×1 block at ((0,2,y=3)) (just placed).
- We place a stair at ((0,2,y=4)). Let’s call it
S(N>S)= “enter from north side, exit to south side.” - Because it is a stair, the cell above it ((0,2,y=5)) must remain empty.
- Stair cost = 3.
- Now the player, standing at ((0,2,y=3)), can go up that stair to ((0,2,y=4\rightarrow5)).
Running total: 15 + 3 = 18.
(D) Make a little offset so we can place the next stair
We cannot stack a second stair directly on top of the first one, because that would violate “no block directly above a stair.” So we do the usual “minecraft‐style” remedy of putting a small block one cell forward (or sideways), then place the next stair on top of that.
Concretely:
- Put a 1×1 block at ((0,3,y=4)).
- We are allowed, because the cell below it ((0,3,y=3)) is currently empty—but we can fill it from below if needed. In many games, you would need pillars from the ground up. So if strictly needed, we also stack 1×1s at ((0,3,y=0..3)). (That costs 4 more blocks × 3 each = 12. Or you can do a 2×1 bridging from ((0,2,y=4)) to ((0,3,y=4)) if the game engine allows horizontal bridging on top of a block. Different rule‐interpretations exist—this puzzle never explicitly said we cannot place a 2×1 bridging out horizontally if the first half is on a solid block.)
- Then place the second stair at ((0,3,y=5)).
Let us suppose we are allowed a single 2×1 bridging from ((0,2,y=4)) to ((0,3,y=4)). That one 2×1 block has cost 5, and it simultaneously fills the squares ((0,2,y=4)) (already the top of the first stair) and ((0,3,y=4)). Strictly speaking, the first half of that 2×1 is “on top of the stair,” which some rule‐sets do not allow—but if your game engine says “once a stair is placed, its top counts as a valid half‐block to anchor a 2×1,” then you can do it.
Anyway, once you have that new “landing” at ((0,3,y=4)), you can put:
- Stair #2 at ((0,3,y=5)), cost 3.
And now you rise from y=4 to y=5 on the second stair, with ((0,3,y=6)) left empty above it.
Finally, to get from y=5 to y=6, you do one more block+stair offset, or reorient the stair 90 degrees, etc.
Illustrative Final Layouts
Because the original 4×4 grids are already quite busy, it is easiest to show just the newly added cells in a list style. Below is one example of the extra cells; your exact coordinates may vary depending how comfortable you are bridging out sideways:
- Pillar at (0,2)
(0,2,0) = 1x1(0,2,1) = 1x1(0,2,2) = 1x1(0,2,3) = 1x1
- Bridge block
(0,1,3) = 1x1
- Stair #1
(0,2,4) = S(N>S)with(0,2,5)empty above
- (Optional) 2×1 bridging from
(0,2,4)to(0,3,4)to form a landing(0,2,4)-(0,3,4) = 2×1(cost 5)
- Stair #2
(0,3,5) = S(W>E)or S(N>S) or whichever orientation fits your walkway, with(0,3,6)empty above
…and so on, one more step if you need to get exactly to y=6.
In short, you are creating a little “L‐shaped” walkway out to the west (x=0) and then south (z=3), stacking up as needed, and placing a new stair every time you have a 1‐block (or 2×1 “landing”) to support it from below.
Cost Accounting Example
A very typical minimal approach (if you do not do fancy bridging) might be:
- 4×1×1 for the pillar under ((0,2,y=3)) → 4 × 3 = 12
- 1×1×1 for the block at((0,1,y=3)) → 3
- 1×stairs at ((0,2,y=4))** → 3
- (Possibly) another 1×1 “landing block” → 3
- (Possibly) a second stair → 3
- (Possibly) a third stair if you truly need to reach y=6 (that is 3 more)
Depending on how many little “landing blocks” or bridging 2×1s you insert, total extra cost will be somewhere around 25–35. You can shave a bit off by using 2×1s (cost 5) in place of pairs of 1×1 blocks (which would cost 6), if/when you see two adjacent squares that both need filling.
Take‐Home Summary
Because of all the constraints:
- You cannot simply keep going in the 2×2 spiral at ((x,z) = (1,2) → (2,2) → (2,1) → (1,1))**; too many squares above are “forbidden” (directly over stairs).
- The practical solution is to shift sideways at elevation 3, building a small platform or walkway in empty cells that do have unoccupied blocks beneath; then start stacking more stairs upward from that slightly offset location.
That is how you can double the height of the existing staircase—while obeying the “no block above a stair” and “no attaching blocks sideways in midair” rules—without having to demolish the original build.
============= o3 mini high =============
Below is one acceptable “efficient‐by‐cost” solution. In our answer the idea is to
•– “double” the original 3–step (elevation‑0 to 3) spiral by re‐using the same 2×2 cycle for the stairs while “filling in” extra support with 2×1 blocks whenever two 1×1 supports would otherwise be used. (Remember that a 1×1 costs 3 while a 2×1 costs only 5 so whenever two contiguous cells can be “spanned” by one 2×1, that is more economical.)
•– Always obey the rule that no stair may have any block immediately on top of it (so if level N has a stair at a given (X,Z) then the same (X,Z) at level N+1 must be left empty), and every placed “step” must be supported (i.e. it is placed exactly on top of a full block from below).
In what follows we use the notations:
- 0 : empty
- 1 : a 1×1 full block
- 2(D) : a half of a 2×1 block. Its partner will be placed in the indicated direction D (with D = E, W, N or S, meaning “to the east,” etc.). (For a vertical pair along Z we will use “2(S)” in the north cell “pointing south” and “2(N)” in its partner below.)
- S(A>B) : a stair block whose “bottom” is on side A and “top” (the way out of the stair) is on side B.
For context the original design (at elevations 0–3) was assumed to be:
Elevation 0 (ground layer):
z=0: 0 0 0 0
z=1: 1 0 1 0
z=2: 0 S(N>E) 1 0
z=3: 0 1 0 0
Elevation 1:
z=0: 0 0 0 0
z=1: 1 0 1 0
z=2: 0 0 S(W>N) 0
z=3: 0 1 0 0
Elevation 2:
z=0: 0 0 0 0
z=1: 2(E) 2(W) S(S>W) 0
z=2: 0 0 0 0
z=3: 0 1 0 0
Elevation 3:
z=0: 0 0 0 0
z=1: 0 S(E>S) 0 0
z=2: 0 2(S) 0 0
z=3: 0 2(N) 0 0
Then the extension is built on top (and by “filling in” some extra supports even at lower levels if needed) so that the upper landing now comes at elevation 6 (i.e. twice the original “height‐run”). In our solution the spiral “cycle” of stairs uses the same four “core” cells as in the original. (We call them A, B, C, D where in the original:
D = (1,2) carried S(N>E) at elev‑0,
C = (2,2) carried S(W>N) at elev‑1,
B = (2,1) carried S(S>W) at elev‑2, and
A = (1,1) carried S(E>S) at elev‑3.)
Then for the added three steps we “continue the cycle”:
Level 4 again uses cell D (at (1,2)) for a stair,
Level 5 uses cell C (at (2,2)), and
Level 6 uses cell B (at (2,1)) as the final “step” (or landing).
Because each stair forces the cell directly above it (in the next elevation) to be empty, the support below each new stair must be provided by a full (non‐stair) block from the previous level. To economize, we “pack” such blocks as 2×1 blocks wherever the grid allows.
One acceptable extended state is as follows. (All coordinates are given as (X,Z) with Z “rows” numbered top‑to‑bottom; each elevation is drawn as a 4×4 table.)
Extended Stairway State
Elevation 0 (given):
Elevation 1 (given):
Elevation 2 (given):
Elevation 3 (given):
Now we add the extra “steps” so that the final landing is at elevation 6.
Elevation 4 (extended):
(Here we choose supports so that the next stair (at elev‑5) will have a block beneath it. Note that because level 3 had a stair at (1,1), cell (1,1) must remain empty in level 4.)
Elevation 5 (extended):
(Level 4’s stair at (1,2) forces level 5 (1,2) to be empty. Also we design a vertical pair (in column 0) and a horizontal pair (in the bottom row) for efficiency. Now the stair for level 5 is placed at cell C = (2,2) with S(W>N).)
Elevation 6 (extended – final landing):
(Level 5’s stair at (2,2) forces level 6 (2,2) to be empty. The stair for this elevation (cell B) is at (2,1) with S(S>W); and the rest of the floor is built (again using 2×1 blocks where possible) so that the landing is sturdy yet uses as many 2×1’s as possible.)
Explanation & Efficiency Notes
- Stair Cycle.
We use the same cyclic order as in the original 2×2 “core”:
• Elevation 0 (cell D, (1,2)) →
• Elevation 1 (cell C, (2,2)) →
• Elevation 2 (cell B, (2,1)) →
• Elevation 3 (cell A, (1,1))
Then the extension continues:
• Elevation 4 re‐uses cell D at (1,2) (with S(N>E)),
• Elevation 5 uses cell C at (2,2) (with S(W>N)), and
• Elevation 6 uses cell B at (2,1) (with S(S>W)). - Support & “No‐Block‐Above‐Stairs” Rule.
Every stair is placed only where the block below (in the previous elevation) is full (often part of a 2×1); and the “directly above a stair” cell remains empty. For example, (1,1) is left empty in Elevation 4 because level 3 had a stair at (1,1), and (1,2) in Elevation 5 is empty because Elevation 4 had a stair there. - 2×1 Block Use.
Whenever two adjacent cells (whether in the east–west or north–south direction) are available and needed, we “pack” them as a 2×1 block (cost 5) rather than use two separate 1×1’s (cost 6 total). (For instance, in Elevation 4 the pair in column 2 (cells (2,1) and (2,2)) is built as a vertical 2×1 block.) - Overall Cost Efficiency.
This “doubled‑height” design uses exactly 6 stair runs (each cost 3) and uses 2×1 blocks in nearly every support‐cell pair. (Any alternative that leaves “floating” individual 1×1 blocks would cost 3 per cell; so by grouping support cells in pairs, we lower the per–cell cost from 3 to 2.5.)
In Summary
One acceptable “efficient” extended structure (using our chosen notation) is given by the seven elevations (0–6) as above. (There are, of course, other designs that meet the rules; this one was chosen so that the original 4–level spiral is continued exactly in the same cyclic order while extra support is “packed” economically in otherwise empty cells.)
You can use this extended state as a valid answer for “doubling the stairway’s height” in your block‐building game.
============= grok 3 (thinking mode) =============
To extend the stairway in the block-building game from elevation 3 to elevation 6 (twice its current height), we need to design an efficient structure that allows characters to climb from elevation 0 to elevation 6 using stairs, while adhering to the game's rules and minimizing costs. Below, we outline the solution, including the necessary placements and the final state for elevations 0 to 6.
Game Rules and Constraints
- Block Placement: Blocks can only be placed on top of other blocks, not attached to their sides. This includes 1x1 blocks and 2x1 blocks.
- 2x1 Blocks: A 2x1 block can be placed on top of a 1x1 block, and blocks can be placed on both halves of the 2x1 block. 2x1 blocks are oriented horizontally at the same elevation, either along the X-axis (West-East) or Z-axis (North-South).
- Stairs: Characters need stairs to move from elevation ( n ) to ( n+1 ). Stairs can be straight (e.g., S(N>N)) or have a 90-degree turn (e.g., S(N>E)). Stairs require empty space above them, so no blocks or stairs can be placed directly on top of a stair.
- Costs:
- 1x1 block: 3 units
- Stair: 3 units
- 2x1 block: 5 units
- Goal: Extend the stairway from elevation 3 to elevation 6 efficiently, minimizing the total cost. We can add blocks to empty spaces at elevations 0-3 and place new blocks and stairs at elevations 4-6.
Coordinate System and Notation
- Coordinates: We use (Y, X, Z) where:
- Y is elevation, starting at 0.
- X is the West-East direction, starting at 0 (increasing eastward).
- Z is the North-South direction, starting at 0 (increasing southward).
- State Notation:
- 0: Empty space.
- 1: 1x1 block.
- 2(A): Part of a 2x1 block, where A indicates the direction of the other half (e.g., 2(E) means the other half is to the East).
- S(A>B): Stair, where A is the bottom side at elevation Y, and B is the top side at elevation Y+1. For example, S(N>E) allows climbing from the North at elevation Y to the East at elevation Y+1.
Current Structure (Elevations 0 to 3)
The current structure is given as follows, with each elevation represented as a 4x4 grid (X for columns, Z for rows):
Elevation 0:
- Blocks at (0,0,1), (0,2,1), (0,2,2), (0,1,3).
- Stair at (0,1,2) S(N>E): From North at Y=0 to East at Y=1.
Elevation 1:
- Blocks at (1,0,1), (1,2,1), (1,1,3).
- Stair at (1,2,2) S(W>N): From West at Y=1 to North at Y=2.
Elevation 2:
- 2x1 block at (2,0,1) and (2,1,1) (oriented along X).
- Block at (2,1,3).
- Stair at (2,1,2) S(S>W): From South at Y=2 to West at Y=3.
Elevation 3:
- Stair at (3,1,1) S(E>S): From East at Y=3 to South at Y=4.
- 2x1 block at (3,1,2) and (3,1,3) (oriented along Z).
Stair Mechanics
- Stair Usage: For a stair at (Y, X, Z) S(A>B):
- The character must be at the block adjacent in direction A at elevation Y (the bottom side).
- The character climbs to the block adjacent in direction B at elevation Y+1 (the top side).
- Directions:
- North (N): Z decreases (dz = -1).
- South (S): Z increases (dz = +1).
- West (W): X decreases (dx = -1).
- East (E): X increases (dx = +1).
- For example, for S(N>E) at (0,1,2), the character climbs from (0,1,1) (North) to (1,2,2) (East). Since stairs can be stood upon, the character can start from (0,1,1) (if there's a block) and land at (1,2,2).
Existing Stairway Path
To extend the stairway efficiently, we first analyze the existing path to ensure continuity. The current path is:
- From Y=0 to Y=1: Use stair at (0,1,2) S(N>E), landing at (1,2,2).
- From Y=1 to Y=2: Use stair at (1,2,2) S(W>N), landing at (2,2,1).
- From Y=2 to Y=3: Use stair at (2,1,2) S(S>W), landing at (3,1,1).
- From Y=3 to Y=4: Use stair at (3,1,1) S(E>S), landing at (4,1,2) (needs to be connected to the extension).
The landing positions after climbing each stair are:
- Y=0: Land at (1,2,2).
- Y=1: Land at (2,2,1).
- Y=2: Land at (3,1,1).
- Y=3: Land at (4,1,2).
Extension Plan (Elevations 4 to 6)
To extend the stairway to Y=6, we need to place stairs at Y=4 and Y=5, with the final landing at Y=6. We aim to minimize the cost by placing stairs and supporting blocks efficiently, ensuring the path is continuous and within the grid (X=0 to 3, Z=0 to 3).
Proposed Stair Placements
- Y=4: Place a stair at (4,1,3) S(N>E).
- From (4,1,2) (landing from Y=3) to (5,2,3).
- Y=5: Place a stair at (5,2,3) S(E>W).
- From (5,3,3) to (6,1,3) (final landing at Y=6).
Supporting Blocks
To connect the landing positions to the bottom sides of the next stairs, we need additional blocks. Since blocks can only be placed on top of other blocks, we need to ensure supporting blocks are placed at lower elevations.
- Path at Y=4 to Y=5:
- After landing at (5,2,3) from the stair at (4,1,3) S(N>E), the character needs to reach (5,3,3) to climb the stair at (5,2,3) S(E>W).
- Place a 1x1 block at (5,3,3) to allow walking from (5,2,3) to (5,3,3).
- To support (5,3,3), place blocks at (4,3,3), (3,3,3), and (2,3,3) (since there's no block at (2,3,3) in the current structure).
Final Path
- Y=3 to Y=4: Land at (4,1,2), climb stair at (4,1,3) S(N>E) from (4,1,2) to (5,2,3).
- Y=4 to Y=5: Walk from (5,2,3) to (5,3,3), climb stair at (5,2,3) S(E>W) from (5,3,3) to (6,1,3).
- Final Landing: At (6,1,3) on elevation Y=6.
Additional Placements
- Elevations 0-3:
- At elevation 2: Place 1x1 block at (2,3,3).
- At elevation 3: Place 1x1 block at (3,3,3).
- Elevations 4-5:
- At elevation 4: Place stair at (4,1,3) S(N>E), 1x1 block at (4,3,3).
- At elevation 5: Place stair at (5,2,3) S(E>W), 1x1 block at (5,3,3).
- Elevation 6:
- No additional blocks needed, as (6,1,3) is the final landing.
Cost Calculation
- Stairs: 2 stairs (at Y=4 and Y=5), each costing 3 units.
- Total stair cost: 2 * 3 = 6 units.
- 1x1 Blocks: 4 blocks (at (2,3,3), (3,3,3), (4,3,3), (5,3,3)), each costing 3 units.
- Total 1x1 block cost: 4 * 3 = 12 units.
- Total Cost: 6 + 12 = 18 units.
Note: We considered using 2x1 blocks to reduce costs, but since the blocks are at different elevations, we cannot use a single 2x1 block to span multiple elevations. Additionally, within the 4x4 grid, placing 2x1 blocks at (X,3) and (X,4) is not possible due to Z=4 being out of bounds. Therefore, using 1x1 blocks is necessary, and the cost is 18 units.
Final State (Elevations 0 to 6)
Below is the updated state for each elevation, incorporating the additional placements. Each elevation is a 4x4 grid (X for columns, Z for rows).
Elevation 0:
Elevation 1:
Elevation 2:
- Added 1x1 block at (2,3,3).
Elevation 3:
- Added 1x1 block at (3,3,3).
Elevation 4:
- Added stair at (4,1,3) S(N>E), 1x1 block at (4,3,3).
Elevation 5:
- Added stair at (5,2,3) S(E>W), 1x1 block at (5,3,3).
Elevation 6:
- No additional blocks needed; landing at (6,1,3).
Verification
- Path Continuity: The path starts at elevation 0, uses existing stairs to reach (4,1,2), then uses new stairs at (4,1,3) and (5,2,3) to reach (6,1,3).
- Game Rules:
- All blocks are placed on top of other blocks.
- Stairs have empty space above them (e.g., no blocks at (4,1,1), (5,1,3), (6,2,3)).
- The grid remains within X=0 to 3, Z=0 to 3.
- Efficiency: The total cost is 18 units, which is minimized given the constraints and the need to stay within the grid.
Conclusion
The extended stairway structure is as specified above, with additional placements at elevations 2 to 5 to support the path. The total cost for the extension is 18 units, achieved by placing 2 stairs and 4 1x1 blocks. The final state for elevations 0 to 6 ensures the character can climb from elevation 0 to elevation 6 efficiently.
================ claude 3.7 sonnet (extended) ==========
I'll design an extension to double the height of your stairway. Let me analyze the current structure first, then plan the efficient additions needed.
Looking at the current structure, I can see you have a spiraling staircase that goes from elevation 0 to elevation 3. For each stair notation S(A>B), A indicates the direction from which you access the stair, and B indicates the direction you face when exiting at the higher elevation.
Current Structure (Elevations 0-3)
The stairway follows this path:
- Elevation 0: S(N>E) at (1,2) - Enter from north, exit facing east
- Elevation 1: S(W>N) at (2,2) - Enter from west, exit facing north
- Elevation 2: S(S>W) at (2,1) - Enter from south, exit facing west
- Elevation 3: S(E>S) at (1,1) - Enter from east, exit facing south
Extension Plan (Elevations 4-6)
Following the same counterclockwise spiral pattern:
Elevation 4:
- Added stair S(N>W) at (1,2) - continuing the spiral pattern
- Added a 1x1 block at (0,2) to support the next stair at elevation 5
Elevation 5:
- Added stair S(E>N) at (0,2) - continuing the spiral pattern
- Added a 1x1 block at (0,1) to support the next stair at elevation 6
Elevation 6:
- Added stair S(S>E) at (0,1) - completing the spiral pattern
Cost Calculation
- Elevation 4: 1 stair (3) + 1 1x1 block (3) = 6
- Elevation 5: 1 stair (3) + 1 1x1 block (3) = 6
- Elevation 6: 1 stair (3) = 3
Total cost for extension: 15
This design efficiently extends your stairway to double its height while maintaining the spiral pattern. Each stair has proper support underneath and the required empty space above it, while using the minimum number of blocks needed.