Group minterms by number of 1s → Compare adjacent groups → Combine each Hamming-1 pair → Chart prime implicants against minterms → Choose essential PIs first, then a minimal cover for the rest. Anything never ticked is a prime implicant.

Dash vs tick — the two different marks (why each exists):

• You put a ==dash -== in the new, combined term at the one bit-position where the two parents disagreed. That dash records "this variable just cancelled via $X+\bar{X}=1$ and no longer appears." The dash lives on the child.
• You put a tick ✓ on each of the two parent terms you just used. The tick records "I was absorbed into a bigger group, so I am not prime by myself." The tick lives on the parents.

So every successful combine leaves three marks: one new dashed child + two ticked parents. Un-ticked survivors at the end are the prime implicants.

Count differing bits (Hamming distance). Combine only if distance $=1$; the differing position becomes -.

• (a) $0011$ vs $0111$ differ only in bit $B$ (position 2). Distance $1$ → YES → 0-11.
• (b) $0000$ vs $0011$ differ in bits $C$ and $D$. Distance $2$ → NO.
• (c) $1010$ vs $1011$ differ only in bit $D$. Distance $1$ → YES → 101-.
• (d) $0110$ vs $1001$ differ in all four bits. Distance $4$ → NO.

Read each column: $1\to$ plain letter, $0\to$ barred letter, -$\to$ drop it.

• (a) 1-0-: $A=1,C=0$ → $A\bar{C}$.
• (b) -011: $B=0,C=1,D=1$ → $\bar{B}CD$.
• (c) --00: $C=0,D=0$ → $\bar{C}\bar{D}$.

Each missing variable (a dropped one) doubles the coverage, because it can be $0$ or $1$ freely. With $n$ variables and $k$ literals, the number of dropped variables is $n-k$, so coverage $=2^{n-k}$. For $\bar{B}\bar{C}$: $n=4$, literals $k=2$, dropped $=2$ → covers $2^2=4$ minterms. (Indeed it covered $\{0,1,8,9\}$ in the parent note.)

Group by 1s: {0=000} | {1=001, 4=100} | {5=101}. Combine (size-2):

• $0,1$: $000$ vs $001$ → 00- ✓
• $0,4$: $000$ vs $100$ → -00 ✓
• $1,5$: $001$ vs $101$ → -01 ✓
• $4,5$: $100$ vs $101$ → 10- ✓

(Every one of the four size-2 terms got a ✓ — meaning each will be absorbed in the next pass, so none is prime yet.) Combine (size-4): 00- & 10- share dash in position $C$ and differ only in bit $A$ → -0- (covers $0,1,4,5$). Similarly -00 & -01 share dash in $A$, differ only in $C$ → -0- (same term). So the whole function collapses to one prime implicant. Decode -0- (order $ABC$): only $B=0$ survives → $\bar{B}$. $\boxed{f=\bar{B}}$ Sanity check: $0,1,4,5$ in binary are exactly the rows where $B=0$. ✔

Group by 1s: {0=0000} | {2=0010, 8=1000} | {10=1010}. Combine (size-2):

• $0,2$→00-0 ; $0,8$→-000 ; $2,10$→-010 ; $8,10$→10-0. Combine (size-4): 00-0 & 10-0 share dash in $C$, differ only in $A$ → -0-0. Also -000 & -010 share dash in $A$, differ only in $C$ → -0-0 (same). One PI. Decode -0-0 (order $ABCD$): $B=0,D=0$ → $\bar{B}\bar{D}$. $\boxed{f=\bar{B}\bar{D}}$ Check: $\{0,2,8,10\}$ are exactly the rows with $B=0$ and $D=0$. ✔

Group by 1s: {0=0000} | {1=0001, 2=0010, 4=0100} | {3=0011, 5=0101}. Combine (size-2):

• $0,1$→000- ; $0,2$→00-0 ; $0,4$→0-00
• $1,3$→00-1 ; $1,5$→0-01 ; $2,3$→001- ; $4,5$→010-

Combine (size-4):

• 000- & 001- share dash in $D$, differ only in $C$ → 00-- (covers $0,1,2,3$).
• 000- & 010- share dash in $D$, differ only in $B$ → 0-0- (covers $0,1,4,5$).
• 00-0 & 00-1? 00-0 covers $\{0,2\}$, 00-1 covers $\{1,3\}$; dash positions differ ($D$ vs $C$) — cannot merge.
• 0-01 & 00-1? dash positions differ — cannot.

Un-ticked survivors → prime implicants:

• 00-- $=\bar{A}\bar{B}$ covers $\{0,1,2,3\}$
• 0-0- $=\bar{A}\bar{C}$ covers $\{0,1,4,5\}$

PI chart (✗ = covered):

EPI hunt (scan columns): $m2,m3$ are covered only by $\bar{A}\bar{B}$ → essential. $m4,m5$ covered only by $\bar{A}\bar{C}$ → essential. Both PIs are essential. $\boxed{f=\bar{A}\bar{B}+\bar{A}\bar{C}}$ (You could also factor to $\bar{A}(\bar{B}+\bar{C})$ — same logic, see Boolean Algebra Laws.)

Group by 1s: {0=000} | {1=001, 2=010} | {5=101, 6=110} | {7=111}. Combine (size-2): compare adjacent groups only.

• $0,1$→00- ✓ ; $0,2$→0-0 ✓
• $1,5$→-01 ✓ ; $2,6$→-10 ✓
• $5,7$→1-1 ✓ ; $6,7$→11- ✓

Combine (size-4): for two size-2 terms to merge they need dashes in the same column and differ in one bit. Check pairs:

• 00- & 0-0? dash positions differ ($C$ vs $B$) → no.
• -01 & -10? both dash in $A$, but they differ in two remaining bits ($B$ and $C$) → no.
• 1-1 & 11-? dash positions differ ($B$ vs $C$) → no.
• every other pair fails one test similarly. No size-4 term exists.

So all six size-2 terms are prime implicants: 00-$=\bar{A}\bar{B}$, 0-0$=\bar{A}\bar{C}$, -01$=\bar{B}C$, -10$=B\bar{C}$, 1-1$=AC$, 11-$=AB$. PI chart:

EPI hunt: every column has $\ge 2$ ✗'s → no essential PI! This is a "cyclic" chart, solved with Prime Implicants and Petrick's Method or by inspection. Pick a minimal cover: choose $\bar{A}\bar{B}(0,1)$, $AC(5,7)$, $B\bar{C}(2,6)$ — union $=\{0,1,2,5,6,7\}$, all six. Three terms. $\boxed{f=\bar{A}\bar{B}+AC+B\bar{C}}$ Why a PI is dropped: $\bar{A}\bar{C}(0,2)$, $\bar{B}C(1,5)$, $AB(6,7)$ are all valid prime implicants, but the three we picked already cover everything, so the other three are redundant.

Include don't-cares while combining (they can enlarge groups) but do not give them chart columns. Minterms+dc in binary: $0=0000,1=0001,2=0010,3=0011,5=0101,7=0111,11=1011,15=1111$. Group by 1s: {0} | {1,2} | {3,5} | {7,11} | {15}. Combine (size-2):

• $0,1$→000- ; $0,2$→00-0 ; $1,3$→00-1 ; $1,5$→0-01 ; $2,3$→001-
• $3,7$→0-11 ; $3,11$→-011 ; $5,7$→01-1 ; $7,15$→-111 ; $11,15$→1-11

Combine (size-4):

• 000- & 001- → 00-- $(0,1,2,3)$
• 00-1 & 01-1? dash in $C$ both, differ only in $B$ → 0--1 $(1,3,5,7)$
• 0-11 & 1-11 → --11 $(3,7,11,15)$
• 0-11 & -011? dash positions differ — no.

Combine (size-8): none share a matching double-dash and differ by one bit. Prime implicants (un-ticked survivors): 00--$=\bar{A}\bar{B}$, 0--1$=\bar{A}D$, --11$=CD$. PI chart — real minterms only $\{1,3,7,11,15\}$:

EPI hunt: $m11,m15$ are covered only by $CD$ → $CD$ essential. After picking $CD$, it already covers $\{3,7,11,15\}$; the only remaining required minterm is $m1$. Now $m1$ is covered by two PIs, $\bar{A}\bar{B}$ and $\bar{A}D$ — each has exactly 2 literals, so neither is "cheaper" on literal count. Both give an equally minimal 2-term, 4-literal answer. We simply pick one; say $\bar{A}D$. $\boxed{f=CD+\bar{A}D}$ (The alternative $f=CD+\bar{A}\bar{B}$ is equally valid and equally minimal — this is a genuine tie, and that's a legitimate outcome of QM, not an error.) Check real minterms: $CD\to \{3,7,11,15\}$, $\bar{A}D\to \{1,3,5,7\}$; union over the required set $\{1,3,7,11,15\}$ ✔. (Don't-cares $0,2,5$ may or may not be $1$ — that's allowed.)

$ABC$ as a number ($A$ is MSB). Odd numbers: $1,3,5,7$. Numbers $\ge 6$: $6,7$. Union of required $=\{1,3,5,6,7\}$. Group by 1s: {1=001} | {3=011, 5=101, 6=110} | {7=111}. Combine (size-2):

• $1,3$→0-1 ; $1,5$→-01 ; $3,7$→-11 ; $5,7$→1-1 ; $6,7$→11-

Combine (size-4):

• 0-1 & 1-1 share dash in $B$, differ in $A$ → --1 $(1,3,5,7)$
• -01 & -11 share dash in $A$, differ in $B$ → --1 (same term)

Un-ticked survivors: --1$=C$ and 11-$=AB$. PI chart (columns = required minterms $\{1,3,5,6,7\}$):

EPI hunt (scan columns):

• $m1,m3,m5$ each have exactly one ✗ (only under $C$) → $C$ is essential.
• $m6$ has exactly one ✗ (only under $AB$) → $AB$ is essential.
• $m7$ is covered by both, but both are already forced in, so it is automatically covered.

Selection: both PIs are essential, so both go in; nothing is left to choose and no PI is redundant. Coverage check: $C\to \{1,3,5,7\}$, $AB\to \{6,7\}$; union $=\{1,3,5,6,7\}$ = all required. ✔ $\boxed{f=C+AB}$ Read plainly: "$1$ if the number is odd ($C=1$) or its top two bits are both $1$ (i.e. $6$ or $7$)" — exactly the spec. See Logic Gate Minimization for the 2-gate circuit.

Group by 1s: {1=001,2=010,4=100} | {3=011,5=101,6=110}. Combine (size-2):

• $1,3$→0-1 ; $1,5$→-01 ; $2,3$→01- ; $2,6$→-10 ; $4,5$→10- ; $4,6$→1-0

Combine (size-4): no two share a dash position and differ in one bit (e.g. 0-1 & 10- have dashes in different columns). All six size-2 terms are prime. Label the PIs (order $ABC$):

• $P_1$=0-1$=\bar{A}C(1,3)$
• $P_2$=-01$=\bar{B}C(1,5)$
• $P_3$=01-$=\bar{A}B(2,3)$
• $P_4$=-10$=B\bar{C}(2,6)$
• $P_5$=10-$=A\bar{B}(4,5)$
• $P_6$=1-0$=A\bar{C}(4,6)$

PI chart (columns $1,2,3,4,5,6$):

Every column has exactly two ✗'s → cyclic, no EPI. Petrick's method — the WHY. Each minterm must be covered, so for each column we write the sum (OR) of the PIs that cover it; the whole cover must satisfy all columns, so we take the product (AND) of those sums: $P=\underset{m1}{\underbrace{(P_1+P_2)}}\underset{m2}{\underbrace{(P_3+P_4)}}\underset{m3}{\underbrace{(P_1+P_3)}}\underset{m4}{\underbrace{(P_5+P_6)}}\underset{m5}{\underbrace{(P_2+P_5)}}\underset{m6}{\underbrace{(P_4+P_6)}}$ Multiply out, using $X\cdot X=X$ and absorption $X+XY=X$. The smallest surviving products each have 3 PIs; one is $P_1{P}_4{P}_5$. Check $P_1,P_4,P_5$ against the chart: $1\to P_1$✓, $3\to P_1$✓, $2\to P_4$✓, $6\to P_4$✓, $4\to P_5$✓, $5\to P_5$✓. All six columns covered with three PIs — and no 2-PI product survives (each PI covers only 2 columns, so 2 PIs reach at most 4 of the 6 minterms). Hence 3 is minimal. $\boxed{f=\bar{A}C+B\bar{C}+A\bar{B}}$ (By the cube's symmetry $P_2,P_3,P_6$ is an equally minimal alternative: $\bar{B}C+\bar{A}B+A\bar{C}$ — another genuine tie.)

Fill a 4-var K-map (rows $AB$, cols $CD$ in Gray order $00,01,11,10$):

Visual groups:

• Corners + edges $B=0,C=0$: cells $0,1,8,9$ → the four corners of the $B=0,C=0$ region → $\bar{B}\bar{C}$.
• $C=1,D=0$ column pattern: cells $2,6,10,14$ → $C\bar{D}$.
• $A=0,B=1,D=1$: cells $5,7$ → $\bar{A}BD$. Union $=\{0,1,8,9\}\cup \{2,6,10,14\}\cup \{5,7\}$ = all ten minterms. $\boxed{f=\bar{B}\bar{C}+C\bar{D}+\bar{A}BD}$ Identical to the QM result. ✔ The K-map "sees" the same groupings QM computes tabularly — two windows on the one combining law $X+\bar{X}=1$.

How many minterms does a 4-var term with 2 literals cover? ::: $2^{4-2}=4$. In L4.1, why does $CD$ turn out essential? ::: $m11$ and $m15$ are covered by no other PI. What makes a PI chart "cyclic"? ::: Every column is covered by $\ge 2$ PIs, so no essential PI exists. When are don't-cares dropped? ::: When building the PI chart — they get no columns; only required minterms do. Minimal SOP for $\sum m(0,1,4,5)$ (3 vars)? ::: $\bar{B}$. What does a tick ✓ next to a term mean? ::: It was absorbed into a bigger group, so it is not prime; only un-ticked terms are prime implicants. Difference between a dash and a tick? ::: A dash marks the cancelled bit on the NEW combined term; a tick marks each OLD parent term as "used, not prime".