Self-consistency and tree-of-thought — Hinglish notes

4.4.8 · AI-ML › Alignment, Prompting & RAG

YEH EXIST KYU KARTA HAI?

Plain Chain-of-Thought (CoT) prompting model ko "step by step sochne" ke liye kehta hai aur ek hi reasoning ki line produce karta hai. Problem yeh hai: ek single chain possible reasonings ke ek bade space mein ek greedy walk hai. Ek early mistake poori answer ko barbaad kar deti hai, aur greedy decoding ke paas recover karne ka koi tarika nahi hai.

DO METHODS KAUN SE HAIN?

Axis	Chain-of-Thought	Self-Consistency	Tree-of-Thought
Structure	1 linear chain	$N$ linear chains	branching tree
Combine by	—	majority vote	search + evaluation
Backtracking?	No	No	Yes
Cost	1×	$N$ ×	can be $≫ N$ ×

Figure — Self-consistency and tree-of-thought

SELF-CONSISTENCY KAISE KAAM KARTA HAI — FIRST PRINCIPLES SE

Goal. Hum sabse zyada probable answer $a$ chahte hain, hidden reasoning path $r$ par marginalize karte hue:

p (a ∣ prompt) = r \sum p (a, r ∣ prompt) = r \sum p (a ∣ r) p (r ∣ prompt) .

Monte Carlo estimate. Hum sab $r$ enumerate nahi kar sakte, isliye hum $r^{(i)} \sim p (r ∣ prompt)$ ke liye $i = 1.. N$ tak sample karte hain (yahi exactly temperature sampling karta hai). Har chain deterministically ek answer $a^{(i)}$ deta hai. Tab:

\overset{p}{^} (a) = \frac{1}{N} i = 1 \sum N 1 [a^{(i)} = a] .

Mode ek sample se better kyun hai (math)

Maano har independent chain probability $p > \frac{1}{2}$ se correct hai aur har specific galat answer kam likely hai. Majority-vote (Condorcet) logic se, majority ke correct hone ki probability $N$ badhne par 1 ki taraf jaati hai:

P (majority correct) = k = ⌈(N + 1) /2 ⌉ \sum N (k N) p^{k} (1 - p)^{N - k} N \to \infty 1 (p > \frac{1}{2}) .

TREE-OF-THOUGHT KAISE KAAM KARTA HAI — FIRST PRINCIPLES SE

ToT, CoT ko state-space search ki tarah reframe karta hai. Define karo:

State $s$ = prompt + abhi tak ka partial reasoning.
Thought generator $G (s)$ = LLM $k$ candidate next thoughts propose karta hai.
State evaluator $V (s)$ = LLM (ya heuristic) score karta hai "yeh state kitna promising hai?" (jaise "sure / maybe / impossible", ya ek numeric value).
Search = BFS har depth par top- $b$ states rakhta hai (beam); DFS deep jaata hai aur backtrack karta hai jab $V$ kehta hai koi branch dead hai.

Cost. Depth $d$ , branching $k$ , width $b$ wale tree ka cost lagbhag hai

calls \approx d \cdot b \cdot k (generate) + d \cdot b \cdot k (evaluate),

jo SC ke $N$ calls se kaafi zyada hai. Trade-off: ToT harder-problem capability khareedta hai; SC saste accuracy gains khareedta hai.

Common mistakes (steel-manned)

The 80/20 (jo actually marks dilata hai)

SC = sample many CoT chains + majority vote, temperature $> 0$ chahiye.
Yeh answer marginal ka ek Monte-Carlo estimate hai $p (a ∣ prompt) = \sum_{r} p (a ∣ r) p (r ∣ prompt)$ .
ToT = reasoning as a searchable tree with generate + evaluate + prune + backtrack.
SC cost $N \times$ ; ToT cost $≫ N \times$ lekin harder search problems solve karta hai.

Flashcards

Chain-of-thought ke upar self-consistency kaunsa decoding trick add karta hai?

Temperature > 0 par N independent CoT chains sample karo, phir majority-vote final answer lo.

Self-consistency ke liye temperature > 0 kyun hona chahiye?

Temperature 0 par sabhi chains identical hoti hain, isliye vote mein koi diversity nahi aur koi benefit nahi milta.

Self-consistency kis quantity ka Monte-Carlo estimate hai?

Answer marginal

p (a ∣ prompt) = \sum_{r} p (a ∣ r) p (r ∣ prompt)

; samples ka mode uska argmax estimate karta hai.

Correct answers vote kyun jeette hain?

Correct reasoning paths ek hi answer par converge karte hain jabki galat paths idiosyncratic, scattered galtiyan karte hain.

Tree-of-Thought ke teen functional components kya hain?

Ek thought generator G(s), ek state evaluator V(s), aur ek search algorithm (BFS/DFS/beam) with pruning aur backtracking.

ToT ki woh key capability jo CoT aur SC mein nahi hai?

Backtracking — yeh dead-end partial reasoning chhodkar doosra branch try kar sakta hai.

CoT vs SC vs ToT ka cost comparison?

CoT 1×, SC N×, ToT roughly d·b·k generate + evaluate calls (≫ N).

Agar single-chain accuracy p>0.5 hai, toh N badhne par majority accuracy ka kya hota hai?

Yeh 1 ki taraf badhti hai (Condorcet/majority-vote theorem).

Recall Feynman: ek 12-saal ke bachche ko samjhao

Ek mushkil homework problem imagine karo. Agar aap ek dost se poochho, toh woh galti kar sakta hai. Isliye aap chaalees dosto se poochho jo har ek apne tarike se solve kare, phir woh answer lo jo unme se zyada ne diya — correct answer baar baar dikhta hai, jabki galat answers sab alag hote hain, isliye woh haarte hain. Yahi self-consistency hai. Tree-of-thought ek maze solve karne jaisa hai: har fork par aap kuch directions try karte ho, jaldi check karo "kya yeh kahin acchi jagah ja raha hai?", aur agar koi path dead end hai toh wapas aao aur doosra try karo. Computer branch karta hai, check karta hai, aur seedha blindly chalne ki jagah backtrack karta hai.

Connections

Chain-of-Thought Prompting — woh base technique jo dono methods extend karti hain.
Monte Carlo Estimation — SC, reasoning paths par MC marginalization hai.
Beam Search and BFS/DFS — ToT ka search backbone.
Ensemble Methods and Majority Voting — vote ka statistical justification.
Temperature and Sampling in LLMs — diversity ke liye temp > 0 kyun chahiye.
Retrieval-Augmented Generation (RAG) — har branch/chain ko verified facts supply kar sakta hai.