Saare operation results ek saath exist karte hain; select lines ek MUX drive karti hain taaki sirf ekF par appear ho. Nayi operation add karna = uska circuit add karo + ek aur MUX input.
Recall Reveal se pehle predict karo (Forecast-then-Verify)
Ek control bit adder ko subtractor kyun bana sakta hai?
→ Kyunki A−B=A+B+1; B ko Ssub ke saath XOR karo aur Ssub ko c0 ke roop mein feed karo.
Overflow V=cn⊕cn−1 kyun hota hai?
→ Sign bit mein jaane wali aur bahar aane wali carry disagree karti hain ⇒ sign corrupt ho gayi.
Operation ka "selection" kahan hota hai?
→ Select lines se drive ek MUX ek parallel result ko pick karta hai.
Recall Feynman: 12-saal ke bachche ko explain karo
Imagine ek calculator jo already har jawab compute kar chuka hai — plus ka jawab, minus
ka jawab, "compare" ka jawab — sab ek saath, aur unhe lineup mein rakh diya. Phir ek chhota sa switch (select lines) simply us drawer ki taraf point karta hai jo tumne manga aur woh jawab deta hai. Yahi ALU hai: sab kuch karo, phir pick karo. Subtraction ek cheeky trick hai — B ke saare bits flip karo aur ek add karo, aur "add" ko "subtract" mein convert kar liya bina nayi machinery ke.
Ek combinational circuit jo do n-bit inputs par arithmetic + logic karta hai, control lines se select hota hai, aur result plus status flags (Z,C,N,V) produce karta hai.
Why is the ALU purely combinational?
Iska output sirf current inputs A, B, S par depend karta hai; iske andar koi internal memory ya clock nahi — state bahar ke registers mein hoti hai.
Sum bit formula of a full adder?
si=ai⊕bi⊕ci (parity: 1 jab odd number of inputs 1 hon).
Carry-out formula of a full adder?
ci+1=aibi+ci(ai⊕bi) (generate OR propagate).
How does one control bit make an adder subtract?
B ko S_sub ke saath XOR karo aur carry-in c0 = S_sub set karo; phir S_sub=1 deta hai A + ¬B + 1 = A − B.