5.1 3D Block Stacks, Hidden Cube Counting & Painted Cube Formulas (Küp Sayma)

5.1 3D Block Stacks, Hidden Cube Counting & Painted Cube Formulas (Küp Sayma)

Cube-counting exercises can ask for occupied blocks, an enclosing prism or painted faces. Hidden space makes a supplied plan or explicit support rule valuable. The methods below count specified structures exactly; a view with insufficient information may require a minimum or maximum rather than a unique total.

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1. Method 1 — The Column-Height Plan Map (Sütun Yüksekliği / Tepe Haritası Yöntemi)

In a standard gravity-supported stack of identical unit cubes (özdeş birim küpler), every vertical $1 \times 1$ column of stacked blocks has exactly one exposed top roof tile (üst yatay yüzey). A rear column can be entirely concealed. Use a supplied height map, multiple views, or explicit constraints; a single perspective drawing need not determine every hidden column uniquely.

Step-by-Step Procedure

1. Establish the Ground Reference ($z = 1$): Locate the lowest frontmost or outermost cubes resting directly on the base plane. Their top roofs sit at height $h = 1$.
2. Label Every Exposed Roof Tile with Its Column Height $h$: Step upward from terrace to terrace. Each vertical step up adds $+1$ to the column height ($h = 1, 2, 3, 4, \dots$). Write the integer $h$ directly inside each visible top roof square on your exam booklet.
3. Tally by Height and Multiply: If $n_k$ denotes the number of roof tiles labeled with height $k$, the total number of unit cubes (visible plus hidden) is:

$$N_{\text{total}} = \sum_{k=1}^{h_{\max}} k \cdot n_k = 1 \cdot n_1 + 2 \cdot n_2 + 3 \cdot n_3 + 4 \cdot n_4 + \dots$$

Counting from a Fully Specified Column Map

Suppose a $3 \times 3$ ground footprint has column heights arranged from back-left to front-right as shown in the top-down plan map (kuşbakışı sütun haritası) below:

$$\begin{bmatrix} 4 & 3 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 1 \end{bmatrix}$$

• A column labeled $4$ in the back-left corner automatically accounts for 1 visible top cube and 3 hidden supporting cubes beneath it ($z = 1, 2, 3$).
• Grouping equal numbers gives: one $4$, two $3$s, three $2$s, and three $1$s:

$$N_{\text{total}} = (4 \times 1) + (3 \times 2) + (2 \times 3) + (1 \times 3) = 4 + 6 + 6 + 3 = 19\text{ unit cubes}$$

Overhang Exception (Askıda / Çıkıntılı Küpler): If a figure explicitly depicts a glued cantilever or bridge where a cube projects outward with empty space underneath it, label that column with the number of actual occupied blocks in that vertical line rather than its top elevation.

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2. Method 2 — Horizontal Floor Slice Counting (Katman / Kat Yöntemi)

When a 3D figure is terraced like a wedding cake or staircase, slicing the solid horizontally into floors (katlar) from bottom to top provides an instant cross-check.

1. Floor 1 (1. Kat — Base Footprint): Count the total number of $(x, y)$ grid squares covered by the base of the structure. Using the $3 \times 3$ matrix above, every cell has $h \ge 1$, so Floor 1 contains $F_1 = 9$ cubes.
2. Floor 2 (2. Kat): Count how many columns reach height $h \ge 2$ (or simply subtract the $h = 1$ columns from Floor 1: $F_2 = F_1 - n_1$). Above, three cells have $h = 1$, so $F_2 = 9 - 3 = 6$ cubes.
3. Floor 3 (3. Kat): Subtract the $h = 2$ columns from Floor 2 ($F_3 = F_2 - n_2$). Since three cells have $h = 2$, $F_3 = 6 - 3 = 3$ cubes.
4. Floor 4 (4. Kat): Subtract the $h = 3$ columns from Floor 3 ($F_4 = F_3 - n_3$). Since two cells have $h = 3$, $F_4 = 3 - 2 = 1$ cube.
5. Sum the Floor Slices:

$$N_{\text{total}} = F_1 + F_2 + F_3 + F_4 = 9 + 6 + 3 + 1 = 19\text{ unit cubes}$$

Under standard gravity support, floor counts must decrease monotonically ($F_1 \ge F_2 \ge F_3 \ge \dots \ge F_{h_{\max}}$), and the difference between two consecutive floors $F_k - F_{k+1}$ equals $n_k$, the exact count of exposed roof tiles at height $k$.

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3. Completing a Block Stack to a Prism or Cube (Prizmaya / Küpe Tamamlama)

1. Count Existing Cubes ($N_{\text{existing}}$): Use the Column-Height Plan Map to find the exact number of cubes already present.
2. Measure the Three Maximum Spans ($A, B, C$):
   • $A = x_{\max}$ (maximum width in unit cubes from left to right)
   • $B = y_{\max}$ (maximum depth in unit cubes from front to back)
   • $C = z_{\max}$ (maximum height in unit cubes from bottom to top)
3. Apply the Target Volume Formula:
   • To complete to the minimum enclosing Rectangular Prism (Dikdörtgenler Prizması):

$$\text{Missing Cubes} = (A \times B \times C) - N_{\text{existing}}$$

• To complete to the minimum enclosing Cube (En Küçük Küp): A cube must have equal edge lengths along all three axes, so its side length must equal the largest span, $L = \max(A, B, C)$:

$$\text{Missing Cubes for a Cube} = L^3 - N_{\text{existing}}, \quad \text{where } L = \max(A, B, C)$$

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4. Painted Cube & Cuboid Dissection Formulas (Boyalı Küpler ve Prizmalar)

Suppose a solid $n \times n \times n$ cube ($n \ge 2$) is dipped into paint so that all 6 exterior faces are painted, and is then sliced by $3(n - 1)$ planes parallel to its faces into $n^3$ unit cubes. Each unit cube's painted-face count depends strictly on its geometric location in the original block:

• 3 Painted Faces (3 Yüzü Boyalı — Corner Cubes / Köşe Küpleri): Only the cubes sitting at the 8 vertices of the large block touch three exterior faces. Regardless of how large $n \ge 2$ is, there are always 8 such cubes.
• 2 Painted Faces (2 Yüzü Boyalı — Edge Cubes / Kenar / Ayrıt Küpleri): Located along the 12 edges of the cube, excluding the 2 corner cubes at the ends of each edge. Each edge of length $n$ contains $(n - 2)$ interior edge cubes, yielding $12(n - 2)$ cubes.
• 1 Painted Face (1 Yüzü Boyalı — Face-Center Cubes / Yüzey Merkez Küpleri): Located in the interior of the 6 square faces after peeling off the 1-unit outer border along all four sides of each face. Each face leaves an $(n - 2) \times (n - 2)$ inner square, yielding $6(n - 2)^2$ cubes.
• 0 Painted Faces (Boyasız — Interior Core Cubes / İç Küpler): Peeling off the 1-unit painted skin from left/right, front/back, and top/bottom reduces each dimension by $2$, leaving an unpainted inner cube of dimensions $(n - 2) \times (n - 2) \times (n - 2)$, or $(n - 2)^3$ cubes.

Partial Painting Variant: Cube Resting on an Unpainted Floor (5 Faces Painted)

A classic TR-YÖS twist states: "An $n \times n \times n$ wooden cube resting on a flat table is spray-painted on all exposed surfaces (5 faces painted, bottom face unpainted) and then cut into $n^3$ unit cubes." Because the bottom face receives no paint, every cube in the bottom layer ($z = 1$) loses one painted face compared to the 6-face formula:

• 3 Painted Faces: Only the 4 top corners meet three painted faces $\implies 4$.
• 2 Painted Faces: The 4 bottom corners (which touch two painted vertical walls and the unpainted floor) plus the 4 top horizontal edges ($4(n - 2)$) plus the 4 vertical edges ($4(n - 2)$) $\implies 8(n - 2) + 4$.
• 1 Painted Face: The 1 top face interior ($(n - 2)^2$) plus the 4 vertical side face interiors ($4(n - 2)^2$) plus the 4 bottom horizontal edge interiors ($4(n - 2)$, which touch only one vertical painted wall) $\implies 5(n - 2)^2 + 4(n - 2)$.
• 0 Painted Faces: The inner core $(n - 2)^3$ plus the unpainted bottom face interior $(n - 2)^2$ $\implies (n - 2)^2(n - 1)$.

Summary Reference Table: Painted Cube & Cuboid Formulas ($n, a, b, c \ge 2$)

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Worked Examples

Worked Example 5.1.1 — Column-Height Map & Completing to a Minimum Cube

Problem: A corner block structure rests on a $4 \times 3$ base (width $x_{\max} = 4$, depth $y_{\max} = 3$). Its top-down column-height plan map is given by:

$$\begin{bmatrix} 5 & 4 & 2 & 1 \\ 3 & 2 & 1 & 1 \\ 2 & 1 & 1 & 0 \end{bmatrix}$$

(where $0$ indicates an empty ground square). Find: (a) the total number of unit cubes in the structure, (b) the minimum number of additional unit cubes needed to complete it into a $4 \times 3 \times 5$ rectangular prism, and (c) the minimum number of additional unit cubes needed to complete it into a solid cube.

Solution:

1. Count existing unit cubes ($N_{\text{existing}}$) via height tally:
   • Height $5$: $n_5 = 1$ column $\implies 5 \times 1 = 5$
   • Height $4$: $n_4 = 1$ column $\implies 4 \times 1 = 4$
   • Height $3$: $n_3 = 1$ column $\implies 3 \times 1 = 3$
   • Height $2$: $n_2 = 3$ columns $\implies 2 \times 3 = 6$
   • Height $1$: $n_1 = 5$ columns $\implies 1 \times 5 = 5$
   • Total existing cubes: $N_{\text{existing}} = 5 + 4 + 3 + 6 + 5 = 23$.
   • (Floor-slice verification: $F_1 = 11,\; F_2 = 6,\; F_3 = 3,\; F_4 = 2,\; F_5 = 1 \implies 11 + 6 + 3 + 2 + 1 = 23$.)
2. Complete to a $4 \times 3 \times 5$ rectangular prism:
   • Target prism volume $= 4 \times 3 \times 5 = 60$.
   • Missing cubes $= 60 - 23 = \mathbf{37}$.
3. Complete to the smallest enclosing cube:
   • Since the maximum coordinate spans are $A = 4,\; B = 3,\; C = 5$, the smallest cube that can enclose the figure without removing any existing blocks has edge length $L = \max(4, 3, 5) = 5$.
   • Target cube volume $= 5^3 = 125$.
   • Missing cubes $= 125 - 23 = \mathbf{102}$.

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Worked Example 5.1.2 — Painted Cuboid Dissection ($4 \times 6 \times 7$)

Problem: A $4 \times 6 \times 7$ rectangular wooden prism is painted red on all 6 exterior faces and then sawed into $1 \times 1 \times 1$ unit cubes. How many unit cubes have (a) exactly 2 red faces, (b) exactly 1 red face, and (c) no red faces?

Solution:

1. Identify reduced inner dimensions: With $a = 4,\; b = 6,\; c = 7$, subtract $2$ from each edge:

$$a - 2 = 2, \qquad b - 2 = 4, \qquad c - 2 = 5$$

1. (a) Exactly 2 red faces (Edge cubes):

$$N_2 = 4\bigl[(a - 2) + (b - 2) + (c - 2)\bigr] = 4(2 + 4 + 5) = 4 \times 11 = \mathbf{44}$$

1. (b) Exactly 1 red face (Face-center cubes):

$$N_1 = 2\bigl[(2)(4) + (2)(5) + (4)(5)\bigr] = 2(8 + 10 + 20) = 2 \times 38 = \mathbf{76}$$

1. (c) Exactly 0 red faces (Interior core cubes):

$$N_0 = (a - 2)(b - 2)(c - 2) = 2 \times 4 \times 5 = \mathbf{40}$$

1. Sanity Check: Adding the $N_3 = 8$ corner cubes gives $8 + 44 + 76 + 40 = 168 = 4 \times 6 \times 7$.

Review the Method

When one exterior face is left unpainted (such as a cube resting on a table), boundary cubes touching that unpainted face lose one painted surface, shifting 4 corners to 2-face, 4 edges to 1-face, and 1 face interior to 0-face.