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# Extended Thinking # Chain-of-Thought
This project demonstrates an extended thinking mode implementation that enables LLMs to solve complex reasoning problems by thinking step-by-step. It's designed to improve problem-solving accuracy through deliberate reasoning. This project demonstrates a Chain-of-Thought mode implementation that enables LLMs to solve complex reasoning problems by thinking step-by-step. It's designed to improve problem-solving accuracy through deliberate reasoning.
## Features ## Features
- Improves model reasoning on complex problems - Improves model reasoning on complex problems
- Works with models like Claude 3.7 Sonnet that support extended thinking - Works with models like Claude 3.7 Sonnet that support Chain-of-Thought
- Solves problems that direct prompting often fails on - Solves problems that direct prompting often fails on
- Provides detailed reasoning traces for verification - Provides detailed reasoning traces for verification
@ -33,7 +33,7 @@ This project demonstrates an extended thinking mode implementation that enables
By default, we will ask the example question: By default, we will ask the example question:
> Break a stick, then break the longer piece again. What's the probability of forming a triangle? > You keep rolling a fair die until you roll three, four, five in that order consecutively on three rolls. What is the probability that you roll the die an odd number of times?
4. Try your own reasoning problem: 4. Try your own reasoning problem:
```bash ```bash
@ -60,299 +60,589 @@ This approach helps LLMs solve problems that would be difficult with a single-pa
- **Standard prompting**: Telling the AI to "think step by step" or providing examples helps, but the thinking is usually not significant enough - **Standard prompting**: Telling the AI to "think step by step" or providing examples helps, but the thinking is usually not significant enough
- **Extended thinking models**: Models like Claude 3.7 Sonnet, GPT-4o, and Deepseek R1 natively support extended thinking with much better results - **Extended thinking models**: Models like Claude 3.7 Sonnet, GPT-4o, and Deepseek R1 natively support extended thinking with much better results
- **This implementation**: Explores how to achieve extended thinking with non-thinking models - **This implementation**: Explores how to achieve Chain-of-Thought with non-thinking models
## Example Thinking Process ## Example Thinking Process
> **Problem**: Break a stick, then break the longer piece again. What's the probability of forming a triangle? Let's try out this [Jane Street Quant Trading Interview Question](https://www.youtube.com/watch?v=gQJTkuEVPrU)
This problem demonstrates why extended thinking is valuable: > **Problem**: You keep rolling a fair die until you roll three, four, five in that order consecutively on three rolls. What is the probability that you roll the die an odd number of times?
This problem demonstrates why Chain-of-Thought is valuable:
- **Standard models without thinking**: Get the wrong answer - **Standard models without thinking**: Get the wrong answer
- **Models with extended thinking**: Find the correct answer (0.386) - **Models with extended thinking**: Find the correct answer (216/431 ≈ 0.5012.)
For comparison: For comparison:
- [Claude 3.7 Sonnet (without thinking)](https://claude.ai/share/31bf938c-94dd-42f6-bfac-e82ba3616dbc): Wrong answer - [Claude 3.7 Sonnet (without thinking)](https://claude.ai/share/da139326-42fe-42d9-9d7b-35870daa5c1b): Wrong answer
- [Claude 3.7 Sonnet with thinking](https://claude.ai/share/0863f9fd-ae75-4a0c-84ee-f7443d2fcf4a): Correct answer after 4.5 min - [Claude 3.7 Sonnet with thinking](https://claude.ai/share/6f4140ed-f33c-4949-8778-a57719498e40): Correct answer after 3m, 45s
- [GPT-o1 with thinking](https://chatgpt.com/c/67dcbad0-75c8-8000-a538-ee6df8083832): Correct answer after 0.5 min - [GPT-o1 with thinking](https://chatgpt.com/share/67fee0fd-2600-8000-bcdf-76e40a986ee4): Correct answer after 2m, 0s
- [GPT-o1 pro with thinking](https://chatgpt.com/share/67dcb1bf-ceb0-8000-823a-8ce894032e37): Correct answer after 1.5 min - [GPT-o1 pro with thinking](https://chatgpt.com/share/67fee11b-530c-8000-92d1-609b6ca49c9c): Correct answer after 4m, 24s
Below is an example of how Claude 3.7 Sonnet (without native thinking) to solve this complex problem, and get the correct result: Below is an example of how Claude 3.7 Sonnet (without native thinking) to solve this complex problem, and get the correct result:
> **Note:** Even with thinking mode, models don't always get the right answer, but their accuracy significantly improves on complex reasoning tasks.
``` ```
🤔 Processing question: Break a stick, then break the longer piece again. What's the probability of forming a triangle? 🤔 Processing question: You keep rolling a fair die until you roll three, four, five in that order consecutively on three rolls. What is the probability that you roll the die an odd number of times?
Let me understand the problem. We have a stick of some length, which we can call 1 for simplicity. We break it at a random point, creating two pieces. Then we break the longer piece again at a random point. This gives us three pieces. We want to find the probability that these three pieces can form a triangle. Thought 1:
Let me think through this problem by setting up a clear approach.
For three segments to form a triangle, the sum of the lengths of any two sides must be greater than the length of the third side. So we need to check: We need to find the probability that the total number of die rolls will be odd, given we stop when we roll "3,4,5" consecutively in that order.
a + b > c
a + c > b
b + c > a
Where a, b, and c are the lengths of the three pieces. This is a problem about a stochastic process that ends when a specific pattern occurs. To solve it, I'll:
1. Set up a Markov chain model to track progress toward the target sequence
2. Calculate expected lengths and probabilities for different outcomes
3. Determine if the total number of rolls has equal probability of being odd vs. even
Next thought needed: True First, I need to understand what states we can be in while rolling the die:
Total thoughts estimate: 8 - State 0: Haven't rolled any of the sequence yet
- State 1: Just rolled a 3, waiting for 4
- State 2: Rolled 3 followed by 4, waiting for 5
- State 3: Success! Rolled the full "3,4,5" sequence
When the die is rolled, we transition between these states with certain probabilities. Each state depends only on the previous 0-2 rolls, making this a Markov process.
Current Plan Status:
- [Done] Understand the problem structure: Identified that we need the probability of an odd number of total rolls until getting '3,4,5' consecutively. Set up a Markov chain with states representing progress toward target sequence.
- [Pending] Set up the Markov model with transition probabilities
- [Pending] Define relevant states in the Markov chain
- [Pending] Calculate transition probabilities between states
- [Pending] Calculate generating functions for the process
- [Pending] Determine probability of odd number of rolls
- [Pending] Extract coefficient information from generating function
- [Pending] Compute final probability value
- [Pending] Verify solution with alternative approach
- [Pending] Conclusion
-------------------------------------------------- --------------------------------------------------
Thought 2: Let me set up the problem mathematically. Let's say the original stick has length 1. Thought 2:
Evaluation of Thought 1: Correct. The approach of using a Markov chain to model this problem is appropriate. Identifying the states (0: no part of sequence, 1: just rolled 3, 2: rolled 3,4, 3: success with 3,4,5) correctly captures the progression toward our target sequence.
First break: We break the stick at a point X, where X is chosen uniformly from [0,1]. This gives us two pieces with lengths X and 1-X. I'll now set up the Markov model with transition probabilities. Each state represents our progress toward the target sequence "3,4,5".
Second break: We break the longer piece at a random point. If X ≤ 0.5, then 1-X is the longer piece. If X > 0.5, then X is the longer piece. State 0: Haven't rolled any relevant part of the sequence yet
- If we roll a 3: Move to State 1 (probability 1/6)
- If we roll anything else: Stay in State 0 (probability 5/6)
Case 1: If X ≤ 0.5, the longer piece is 1-X. We break it at point Y(1-X), where Y is chosen uniformly from [0,1]. This gives pieces of lengths: State 1: Just rolled a 3, waiting for a 4
- X (from first break) - If we roll a 4: Move to State 2 (probability 1/6)
- Y(1-X) (from second break) - If we roll a 3: Stay in State 1 (probability 1/6)
- (1-Y)(1-X) (remaining from second break) - If we roll anything else: Return to State 0 (probability 4/6)
Case 2: If X > 0.5, the longer piece is X. We break it at point YX, where Y is chosen uniformly from [0,1]. This gives pieces of lengths: State 2: Rolled 3 followed by 4, waiting for a 5
- 1-X (from first break) - If we roll a 5: Move to State 3 (success - probability 1/6)
- YX (from second break) - If we roll a 3: Move to State 1 (probability 1/6)
- (1-Y)X (remaining from second break) - If we roll anything else: Return to State 0 (probability 4/6)
For each case, I need to determine when these three pieces can form a triangle, then calculate the probability. State 3: Success! Rolled the sequence "3,4,5"
- This is an absorbing state; once we reach it, we stop rolling.
Next thought needed: True These transition probabilities can be represented as a matrix:
Total thoughts estimate: 8 P = [
[5/6, 1/6, 0, 0 ],
[4/6, 1/6, 1/6, 0 ],
[4/6, 1/6, 0, 1/6],
[0, 0, 0, 1 ]
]
This completes the setup of the Markov model with transition probabilities.
Current Plan Status:
- [Done] Understand the problem structure: Identified that we need the probability of an odd number of total rolls until getting '3,4,5' consecutively. Set up a Markov chain with states representing progress toward target sequence.
- [Done] Set up the Markov model with transition probabilities: Defined 4 states (0: no sequence, 1: just rolled 3, 2: rolled 3,4, 3: success with 3,4,5) and calculated transition probabilities between all states in matrix form.
- [Pending] Calculate generating functions for the process
- [Pending] Set up generating function equations for each state
- [Pending] Solve the system of equations to find generating functions
- [Pending] Determine probability of odd number of rolls
- [Pending] Extract coefficient information from generating function
- [Pending] Compute final probability value
- [Pending] Verify solution with alternative approach
- [Pending] Conclusion
-------------------------------------------------- --------------------------------------------------
Thought 3: Let's analyze when the three pieces can form a triangle for both cases. Thought 3:
Evaluation of Thought 2: Correct. The Markov chain setup with the four states and the transition probability matrix accurately represents our problem. Now I'll proceed to calculating the generating functions for this process.
For a triangle to form, each side must be less than the sum of the other two sides. So we need: To calculate the generating functions, I'll use the variable z to represent a single die roll. For each state, the generating function gives us information about the number of rolls until we reach our success state.
Case 1 (X ≤ 0.5): We have pieces of lengths X, Y(1-X), and (1-Y)(1-X) Let G₀(z), G₁(z), G₂(z), and G₃(z) be the generating functions for reaching the success state (State 3) when starting from State 0, 1, 2, and 3 respectively.
The triangle inequality conditions are: G₃(z) = 1 (since we're already in the success state)
- X + Y(1-X) > (1-Y)(1-X)
- X + (1-Y)(1-X) > Y(1-X)
- Y(1-X) + (1-Y)(1-X) > X
Simplifying: For the other states, we can write equations based on the first roll and subsequent state:
- X + Y(1-X) > (1-Y)(1-X)
- X + (1-Y)(1-X) > Y(1-X)
- (1-X) > X (since Y(1-X) + (1-Y)(1-X) = 1-X)
The third condition is always true when X ≤ 0.5 (which is our assumption for Case 1). G₀(z) = z·(5/6·G₀(z) + 1/6·G₁(z))
This means: we roll once (factor z), then with probability 5/6 we stay in State 0, or with probability 1/6 we move to State 1.
Let's work on the first condition: G₁(z) = z·(4/6·G₀(z) + 1/6·G₁(z) + 1/6·G₂(z))
X + Y(1-X) > (1-Y)(1-X) When in State 1, we roll once (factor z), then either move to State 0, stay in State 1, or advance to State 2.
X + Y(1-X) > (1-X) - Y(1-X)
X + 2Y(1-X) > (1-X)
X + 2Y(1-X) > 1-X
2X + 2Y(1-X) > 1
2X + 2Y - 2YX > 1
Similarly for the second condition: G₂(z) = z·(4/6·G₀(z) + 1/6·G₁(z) + 1/6·G₃(z))
X + (1-Y)(1-X) > Y(1-X) = z·(4/6·G₀(z) + 1/6·G₁(z) + 1/6·1)
X + (1-X) - Y(1-X) > Y(1-X) = z·(4/6·G₀(z) + 1/6·G₁(z) + 1/6)
X + (1-X) > 2Y(1-X)
1 > 2Y(1-X)
1/(2(1-X)) > Y
So for Case 1, we need: Now I can solve this system of equations for G₀(z):
- 2X + 2Y - 2YX > 1
- Y < 1/(2(1-X))
Next thought needed: True From G₀(z) = z·(5/6·G₀(z) + 1/6·G₁(z)):
Total thoughts estimate: 8 G₀(z) - 5z/6·G₀(z) = z/6·G₁(z)
G₀(z)(1 - 5z/6) = z/6·G₁(z)
G₀(z) = (z/6·G₁(z))/(1 - 5z/6) ... (1)
From G₁(z) = z·(4/6·G₀(z) + 1/6·G₁(z) + 1/6·G₂(z)):
G₁(z) - z/6·G₁(z) = 4z/6·G₀(z) + z/6·G₂(z)
G₁(z)(1 - z/6) = 4z/6·G₀(z) + z/6·G₂(z) ... (2)
From G₂(z) = z·(4/6·G₀(z) + 1/6·G₁(z) + 1/6):
G₂(z) = 4z/6·G₀(z) + z/6·G₁(z) + z/6 ... (3)
Substituting (3) into (2):
G₁(z)(1 - z/6) = 4z/6·G₀(z) + z/6·(4z/6·G₀(z) + z/6·G₁(z) + z/6)
G₁(z)(1 - z/6) = 4z/6·G₀(z) + 4z²/36·G₀(z) + z²/36·G₁(z) + z²/36
G₁(z)(1 - z/6 - z²/36) = 4z/6·G₀(z)(1 + z/6) + z²/36
G₁(z) = (4z/6·G₀(z)(1 + z/6) + z²/36)/(1 - z/6 - z²/36) ... (4)
Substituting (4) into (1) gives us G₀(z) in terms of z only, which I can then simplify to find the closed form.
This is getting algebraically intricate. Let me take a step back and consider a different approach using first-step analysis with parity.
Current Plan Status:
- [Done] Understand the problem structure: Identified that we need probability of an odd number of total rolls until getting '3,4,5' consecutively. Set up a Markov chain with states representing progress toward target sequence.
- [Done] Set up the Markov model with transition probabilities: Defined 4 states (0: no sequence, 1: just rolled 3, 2: rolled 3,4, 3: success with 3,4,5) and calculated transition probabilities between all states in matrix form.
- [Verification Needed] Calculate generating functions for the process (The generating function approach became algebraically complicated. Consider a more direct approach using first-step analysis with parity.)
- [Done] Set up generating function equations for each state: Established equations G₀(z), G₁(z), G₂(z), and G₃(z) to track rolls until success.
- [Verification Needed] Solve the system of equations to find generating functions (Equations become unwieldy; may not be most efficient approach.)
- [Pending] Use first-step analysis with parity
- [Pending] Define probability variables with parity information
- [Pending] Set up equations relating odd/even outcome probabilities
- [Pending] Solve the system of equations
- [Pending] Determine probability of odd number of rolls
- [Pending] Extract final probability from analysis
- [Pending] Verify the mathematical result
- [Pending] Verify solution with alternative approach
- [Pending] Conclusion
-------------------------------------------------- --------------------------------------------------
Thought 4: I need to continue analyzing Case 1 and then move on to Case 2. Thought 4:
Evaluation of Thought 3: Correct, but pivoting is wise. The generating function approach, while theoretically sound, is becoming algebraically complex. The decision to switch to a first-step analysis with parity is appropriate for this problem.
For Case 1 (X ≤ 0.5), we have the constraints: I'll now use first-step analysis focusing specifically on the parity of the number of rolls.
- 2X + 2Y - 2YX > 1
- Y < 1/(2(1-X))
Let's rewrite the first constraint: Let's define:
2X + 2Y - 2YX > 1 - P₀: Probability of rolling an odd number of times when starting in State 0
2Y(1-X) > 1 - 2X - P₁: Probability of rolling an odd number of times when starting in State 1
Y > (1 - 2X)/(2(1-X)) - P₂: Probability of rolling an odd number of times when starting in State 2
So for Case 1, we need: Using first-step analysis, we can write equations for each state:
- Y > (1 - 2X)/(2(1-X))
- Y < 1/(2(1-X))
This means Y must be in the range ((1 - 2X)/(2(1-X)), 1/(2(1-X))). For this range to be valid, we need: For State 0:
(1 - 2X)/(2(1-X)) < 1/(2(1-X)) - If we roll a 3 (prob 1/6), we go to State 1 with 1 roll used (odd). The total number of rolls will be odd if the remaining rolls from State 1 are even.
1 - 2X < 1 - If we roll anything else (prob 5/6), we stay in State 0 with 1 roll used (odd). The total will be odd if the remaining rolls from State 0 are even.
-2X < 0
X > 0
Which is true for our domain. So we have a valid range for Y when X ≤ 0.5. Therefore: P₀ = (1/6)*(1-P₁) + (5/6)*(1-P₀) = 5/6 - 5P₀/6 + 1/6 - P₁/6
Now let's analyze Case 2 (X > 0.5): We have pieces of lengths (1-X), YX, and (1-Y)X For State 1:
- If we roll a 4 (prob 1/6), we go to State 2 with 1 roll used (odd). The total will be odd if remaining rolls from State 2 are even.
- If we roll a 3 (prob 1/6), we stay in State 1 with 1 roll used (odd). The total will be odd if remaining rolls from State 1 are even.
- If we roll anything else (prob 4/6), we go to State 0 with 1 roll used (odd). The total will be odd if remaining rolls from State 0 are even.
The triangle inequality conditions are: Therefore: P₁ = (1/6)*(1-P₂) + (1/6)*(1-P₁) + (4/6)*(1-P₀) = 1 - P₂/6 - P₁/6 - 4P₀/6
- (1-X) + YX > (1-Y)X
- (1-X) + (1-Y)X > YX
- YX + (1-Y)X > (1-X)
Simplifying: For State 2:
- (1-X) + YX > (1-Y)X - If we roll a 5 (prob 1/6), we're done with 1 more roll (odd). Total is definitely odd.
- (1-X) + (1-Y)X > YX - If we roll a 3 (prob 1/6), we go to State 1 with 1 roll used (odd). The total will be odd if remaining rolls from State 1 are even.
- X > (1-X) (since YX + (1-Y)X = X) - If we roll anything else (prob 4/6), we go to State 0 with 1 roll used (odd). The total will be odd if remaining rolls from State 0 are even.
The third condition is always true when X > 0.5 (which is our assumption for Case 2). Therefore: P₂ = (1/6)*1 + (1/6)*(1-P₁) + (4/6)*(1-P₀) = 1/6 + 1/6 - P₁/6 + 4/6 - 4P₀/6 = 6/6 - P₁/6 - 4P₀/6
Next thought needed: True Now I have a system of 3 equations with 3 unknowns:
Total thoughts estimate: 8 P₀ = 1 - 5P₀/6 - P₁/6
P₁ = 1 - P₁/6 - P₂/6 - 4P₀/6
P₂ = 1 - P₁/6 - 4P₀/6
Simplifying:
(6/6)P₀ = 6/6 - 5P₀/6 - P₁/6
(6/6)P₁ = 6/6 - P₁/6 - P₂/6 - 4P₀/6
(6/6)P₂ = 6/6 - P₁/6 - 4P₀/6
Further simplifying:
11P₀/6 + P₁/6 = 1
4P₀/6 + 7P₁/6 + P₂/6 = 1
4P₀/6 + P₁/6 + 6P₂/6 = 1
This is a linear system of equations that I can solve.
Current Plan Status:
- [Done] Understand the problem structure: Identified that we need probability of an odd number of total rolls until getting '3,4,5' consecutively. Set up a Markov chain with states representing progress toward target sequence.
- [Done] Set up the Markov model with transition probabilities: Defined 4 states (0: no sequence, 1: just rolled 3, 2: rolled 3,4, 3: success with 3,4,5) and calculated transition probabilities between all states in matrix form.
- [Verification Needed] Calculate generating functions for the process (Approach became algebraically complicated; shifted to more direct first-step analysis with parity.)
- [Done] Set up generating function equations for each state: Established equations G₀(z), G₁(z), G₂(z), and G₃(z) to track rolls until success.
- [Verification Needed] Solve the system of equations to find generating functions (Equations became unwieldy; more efficient to use first-step analysis with parity instead.)
- [Done] Use first-step analysis with parity: Defined variables P₀, P₁, P₂ for probability of odd rolls from each state. Set up system of equations tracking odd/even outcomes.
- [Done] Define probability variables with parity information: Defined P₀, P₁, P₂ as probabilities of odd number of rolls starting from respective states.
- [Done] Set up equations relating odd/even outcome probabilities: Derived equations: 11P₀/6 + P₁/6 = 1, 4P₀/6 + 7P₁/6 + P₂/6 = 1, 4P₀/6 + P₁/6 + 6P₂/6 = 1.
- [Pending] Solve the system of equations
- [Pending] Determine probability of odd number of rolls
- [Pending] Extract final probability from analysis
- [Pending] Verify the mathematical result
- [Pending] Verify solution with alternative approach
- [Pending] Conclusion
-------------------------------------------------- --------------------------------------------------
Thought 5: Continuing with Case 2 (X > 0.5), let's simplify the remaining triangle inequality conditions: Thought 5:
Evaluation of Thought 4: Correct. The first-step analysis with parity is well set up, with appropriate variables P₀, P₁, and P₂ representing the probability of rolling an odd number of times from each state. The system of equations is correctly derived, though I notice it's written in slightly different forms across the thought. Let me solve this system now.
First condition: Taking the system of equations from the end of Thought 4:
(1-X) + YX > (1-Y)X 11P₀/6 + P₁/6 = 1
(1-X) + YX > X - YX 4P₀/6 + 7P₁/6 + P₂/6 = 1
(1-X) + 2YX > X 4P₀/6 + P₁/6 + 6P₂/6 = 1
1 - X + 2YX > X
1 + 2YX > 2X
2YX > 2X - 1
Y > (2X - 1)/(2X)
Second condition: Let's multiply all equations by 6 to eliminate fractions:
(1-X) + (1-Y)X > YX 11P₀ + P₁ = 6
(1-X) + X - YX > YX 4P₀ + 7P₁ + P₂ = 6
(1-X) + X > 2YX 4P₀ + P₁ + 6P₂ = 6
1 > 2YX
1/(2X) > Y
So for Case 2, we need: Subtracting the third equation from the second:
- Y > (2X - 1)/(2X) (4P₀ + 7P₁ + P₂) - (4P₀ + P₁ + 6P₂) = 6 - 6
- Y < 1/(2X) 6P₁ - 5P₂ = 0
6P₁ = 5P₂
P₁ = 5P₂/6
This means Y must be in the range ((2X - 1)/(2X), 1/(2X)). For this range to be valid, we need: From the first equation:
(2X - 1)/(2X) < 1/(2X) 11P₀ + P₁ = 6
2X - 1 < 1 11P₀ + 5P₂/6 = 6
2X < 2 66P₀ + 5P₂ = 36
X < 1
Which is true for our domain. So we have a valid range for Y when X > 0.5. From the third equation:
4P₀ + P₁ + 6P₂ = 6
4P₀ + 5P₂/6 + 6P₂ = 6
24P₀ + 5P₂ + 36P₂ = 36
24P₀ + 41P₂ = 36
Now I need to calculate the probability. For both cases, I need to find the area of the region in the X-Y plane where the triangle can be formed, divided by the total area. Now solving for P₀ in terms of P₂:
24P₀ + 41P₂ = 36
24P₀ = 36 - 41P₂
P₀ = (36 - 41P₂)/24 = 3/2 - 41P₂/24
Next thought needed: True Substituting back into 66P₀ + 5P₂ = 36:
Total thoughts estimate: 8 66(3/2 - 41P₂/24) + 5P₂ = 36
99/2 - 66*41P₂/24 + 5P₂ = 36
99/2 - 2706P₂/24 + 5P₂ = 36
99/2 - 2706P₂/24 + 120P₂/24 = 36
99/2 - (2706-120)P₂/24 = 36
99/2 - 2586P₂/24 = 36
99 - 2586P₂/12 = 72
99 - 72 = 2586P₂/12
27 = 2586P₂/12
27*12 = 2586P₂
324 = 2586P₂
P₂ = 324/2586 = 108/862 = 54/431
Now we can find P₁:
P₁ = 5P₂/6 = 5(54/431)/6 = 270/2586 = 45/431
And P₀:
P₀ = 3/2 - 41P₂/24 = 3/2 - 41(54/431)/24 = 3/2 - 2214/10344 = 3/2 - 1107/5172
Let's calculate:
P₀ = 3/2 - 1107/5172 = 7758/5172 - 1107/5172 = 6651/5172 = 2217/1724
Now we have:
P₀ = 2217/1724
P₁ = 45/431
P₂ = 54/431
Wait, I see an error in my calculation. Let me restart with the system of equations.
From the equations:
11P₀/6 + P₁/6 = 1
4P₀/6 + 7P₁/6 + P₂/6 = 1
4P₀/6 + P₁/6 + 6P₂/6 = 1
Multiplying by 6:
11P₀ + P₁ = 6
4P₀ + 7P₁ + P₂ = 6
4P₀ + P₁ + 6P₂ = 6
From the first: P₁ = 6 - 11P₀
Substituting into the second:
4P₀ + 7(6 - 11P₀) + P₂ = 6
4P₀ + 42 - 77P₀ + P₂ = 6
4P₀ - 77P₀ + P₂ = 6 - 42
-73P₀ + P₂ = -36
P₂ = -36 + 73P₀
Substituting P₁ and P₂ into the third equation:
4P₀ + (6 - 11P₀) + 6(-36 + 73P₀) = 6
4P₀ + 6 - 11P₀ - 216 + 438P₀ = 6
4P₀ - 11P₀ + 438P₀ = 6 - 6 + 216
431P₀ = 216
P₀ = 216/431 = 8/16 = 1/2
Now we can find:
P₁ = 6 - 11P₀ = 6 - 11(216/431) = 6 - 2376/431 = 2586/431 - 2376/431 = 210/431
P₂ = -36 + 73P₀ = -36 + 73(216/431) = -36 + 15768/431 = -15552/431 + 15768/431 = 216/431
The answer to our original question is P₀, which is 216/431.
Actually, I notice the fraction simplified to 1/2 above is incorrect (216/431 ≠ 8/16). Let's double-check with the equations:
From 11P₀ + P₁ = 6 and P₁ = 210/431, we get:
11(216/431) + 210/431 = 6
2376/431 + 210/431 = 6
2586/431 = 6
2586 = 2586
This confirms our answer: P₀ = 216/431, which is the probability of rolling the die an odd number of times.
Current Plan Status:
- [Done] Understand the problem structure: Identified that we need probability of an odd number of total rolls until getting '3,4,5' consecutively. Set up a Markov chain with states representing progress toward target sequence.
- [Done] Set up the Markov model with transition probabilities: Defined 4 states (0: no sequence, 1: just rolled 3, 2: rolled 3,4, 3: success with 3,4,5) and calculated transition probabilities between all states in matrix form.
- [Verification Needed] Calculate generating functions for the process (The generating function approach was algebraically complex; pivoted to first-step analysis with parity instead.)
- [Done] Set up generating function equations for each state: Established equations G₀(z), G₁(z), G₂(z), and G₃(z) to track rolls until success.
- [Verification Needed] Solve the system of equations to find generating functions (Equations became unwieldy; more efficient to use first-step analysis with parity instead.)
- [Done] Use first-step analysis with parity: Successfully applied first-step analysis with parity to solve the problem directly.
- [Done] Define probability variables with parity information: Defined P₀, P₁, P₂ as probabilities of odd number of rolls starting from respective states.
- [Done] Set up equations relating odd/even outcome probabilities: Derived equations: 11P₀/6 + P₁/6 = 1, 4P₀/6 + 7P₁/6 + P₂/6 = 1, 4P₀/6 + P₁/6 + 6P₂/6 = 1.
- [Done] Solve the system of equations: Found P₀ = 216/431, P₁ = 210/431, P₂ = 216/431.
- [Done] Determine probability of odd number of rolls: The probability of rolling the die an odd number of times is 216/431 ≈ 0.501
- [Done] Extract final probability from analysis: Identified P₀ = 216/431 as the answer to our original question.
- [Done] Verify the mathematical result: Verified solution by substituting back into original equations.
- [Pending] Verify solution with alternative approach
- [Pending] Conclusion
-------------------------------------------------- --------------------------------------------------
Thought 6: Now I need to calculate the probability by finding the area in the X-Y plane where the triangle can be formed. Thought 6:
Evaluation of Thought 5: Correct. The calculation was methodical and accurate, resulting in the solution P₀ = 216/431. The self-correction when an arithmetic error was spotted was appropriate. The substitution back into the original equation verified that the answer is correct.
For Case 1 (X ≤ 0.5): Let me now verify our solution using an alternative approach. The problem asks for the probability of an odd number of rolls when stopping at the sequence "3,4,5". An alternative way to approach this is to consider the complementary probability - the probability of an even number of rolls.
- Y must be between (1 - 2X)/(2(1-X)) and 1/(2(1-X))
The probability for this case is the area of the region divided by the total area of the X-Y plane for X ≤ 0.5, which is 0.5 × 1 = 0.5. Since P₀ = 216/431 is the probability of an odd number of rolls, the probability of an even number of rolls must be 1 - 216/431 = 215/431.
Area = ∫₀^(0.5) [1/(2(1-X)) - (1 - 2X)/(2(1-X))] dX Let me verify this with the equations we derived:
= ∫₀^(0.5) [(1 - (1 - 2X))/(2(1-X))] dX - Let Q₀, Q₁, Q₂ be the probabilities of rolling an even number of times from states 0, 1, and 2 respectively.
= ∫₀^(0.5) [2X/(2(1-X))] dX - Since the events "odd number of rolls" and "even number of rolls" are complementary: Q₀ = 1-P₀, Q₁ = 1-P₁, Q₂ = 1-P₂
= ∫₀^(0.5) [X/(1-X)] dX - Therefore: Q₀ = 1 - 216/431 = 215/431, Q₁ = 1 - 210/431 = 221/431, Q₂ = 1 - 216/431 = 215/431
This integral equals -X - ln(1-X) evaluated from 0 to 0.5, which is: Now, I can derive the equations for Q₀, Q₁, Q₂ directly:
[-0.5 - ln(0.5)] - [0 - ln(1)] = -0.5 - (-0.693) = -0.5 + 0.693 = 0.193
So the probability for Case 1 is 0.193/0.5 = 0.386. For State 0:
- If we roll a 3 (prob 1/6), we transition to State 1 after 1 roll (odd). The total will be even if the remaining rolls from State 1 are odd.
- If we roll anything else (prob 5/6), we stay in State 0 after 1 roll (odd). The total will be even if the remaining rolls from State 0 are odd.
For Case 2 (X > 0.5): Therefore: Q₀ = (1/6)(P₁) + (5/6)(P₀) = P₁/6 + 5P₀/6 = 210/431/6 + 5(216/431)/6 = 210/2586 + 1080/2586 = 1290/2586 = 215/431
- Y must be between (2X - 1)/(2X) and 1/(2X)
The probability for this case is the area of the region divided by the total area of the X-Y plane for X > 0.5, which is 0.5 × 1 = 0.5. For State 1:
- If we roll a 4 (prob 1/6), we go to State 2 after 1 roll (odd). The total will be even if the remaining rolls from State 2 are odd.
- If we roll a 3 (prob 1/6), we stay in State 1 after 1 roll (odd). The total will be even if the remaining rolls from State 1 are odd.
- If we roll anything else (prob 4/6), we go to State 0 after 1 roll (odd). The total will be even if the remaining rolls from State 0 are odd.
Area = ∫₀.₅^1 [1/(2X) - (2X - 1)/(2X)] dX Therefore: Q₁ = (1/6)(P₂) + (1/6)(P₁) + (4/6)(P₀) = P₂/6 + P₁/6 + 4P₀/6 = 216/431/6 + 210/431/6 + 4(216/431)/6 = 216/2586 + 210/2586 + 864/2586 = 1290/2586 = 215/431
= ∫₀.₅^1 [(1 - (2X - 1))/(2X)] dX
= ∫₀.₅^1 [(2 - 2X)/(2X)] dX
= ∫₀.₅^1 [(1 - X)/X] dX
= ∫₀.₅^1 [1/X - 1] dX
= [ln|X| - X]₀.₅^1
= [ln(1) - 1] - [ln(0.5) - 0.5] = -1 - (-0.693 - 0.5) = -1 + 0.693 + 0.5 = 0.193
So the probability for Case 2 is also 0.193/0.5 = 0.386. Wait, that doesn't match our expected Q₁ = 221/431. Let me double-check...
The total probability is the sum of the probabilities for both cases: 0.386 + 0.386 = 0.772. The direct calculation of Q₁ gives a different result than 1-P₁. This suggests an error somewhere. Let me reconsider our system:
Wait, I should double-check my calculations. Let me reconsider the integrals. If P₀ + Q₀ = 1, P₁ + Q₁ = 1, P₂ + Q₂ = 1, then our system should be consistent. Let me verify our values from Thought 5:
Next thought needed: True From 11P₀ + P₁ = 6 and P₀ = 216/431:
Total thoughts estimate: 8 11(216/431) + P₁ = 6
2376/431 + P₁ = 6
P₁ = 6 - 2376/431 = 2586/431 - 2376/431 = 210/431
From 4P₀ + P₁ + 6P₂ = 6:
4(216/431) + 210/431 + 6P₂ = 6
864/431 + 210/431 + 6P₂ = 6
1074/431 + 6P₂ = 6
6P₂ = 6 - 1074/431 = 2586/431 - 1074/431 = 1512/431
P₂ = 1512/431/6 = 1512/2586 = 252/431
This doesn't match our calculated P₂ = 216/431 from Thought 5. Let me solve the system again, more carefully.
From the system:
11P₀ + P₁ = 6
4P₀ + 7P₁ + P₂ = 6
4P₀ + P₁ + 6P₂ = 6
Subtracting the first from the third:
(4P₀ + P₁ + 6P₂) - (11P₀ + P₁) = 6 - 6
4P₀ - 11P₀ + 6P₂ = 0
-7P₀ + 6P₂ = 0
6P₂ = 7P₀
P₂ = 7P₀/6
Subtracting the first from the second:
(4P₀ + 7P₁ + P₂) - (11P₀ + P₁) = 6 - 6
4P₀ - 11P₀ + 6P₁ + P₂ = 0
-7P₀ + 6P₁ + P₂ = 0
Substituting P₂ = 7P₀/6:
-7P₀ + 6P₁ + 7P₀/6 = 0
-42P₀ + 36P₁ + 7P₀ = 0
-35P₀ + 36P₁ = 0
36P₁ = 35P₀
P₁ = 35P₀/36
Now, from the first equation:
11P₀ + P₁ = 6
11P₀ + 35P₀/36 = 6
396P₀ + 35P₀ = 216
431P₀ = 216
P₀ = 216/431
Therefore:
P₁ = 35P₀/36 = 35(216/431)/36 = 35(6)/36(431/6) = 35/6 ÷ 431/6 = 35/431
P₂ = 7P₀/6 = 7(216/431)/6 = 7(6)/6(431/6) = 7/431/6 × 6 = 7/431 × 6 = 42/431
Hmm, these values are different from what we calculated in Thought 5. Let me check these values in the original equations:
11(216/431) + 35/431 = 6?
2376/431 + 35/431 = 6?
2411/431 = 6?
2411 = 2586?
This doesn't check out. Let me solve the system one more time using a different approach.
From the original system:
11P₀ + P₁ = 6 (1)
4P₀ + 7P₁ + P₂ = 6 (2)
4P₀ + P₁ + 6P₂ = 6 (3)
From (3) - (1):
4P₀ + P₁ + 6P₂ - 11P₀ - P₁ = 6 - 6
-7P₀ + 6P₂ = 0
6P₂ = 7P₀
P₂ = 7P₀/6 (4)
From (2) - (1):
4P₀ + 7P₁ + P₂ - 11P₀ - P₁ = 6 - 6
-7P₀ + 6P₁ + P₂ = 0
Substituting (4):
-7P₀ + 6P₁ + 7P₀/6 = 0
-42P₀ + 36P₁ + 7P₀ = 0
-35P₀ + 36P₁ = 0
36P₁ = 35P₀
P₁ = 35P₀/36 (5)
Substituting (5) into (1):
11P₀ + 35P₀/36 = 6
(11×36)P₀ + 35P₀ = 6×36
396P₀ + 35P₀ = 216
431P₀ = 216
P₀ = 216/431
Now calculate:
P₁ = 35P₀/36 = 35(216/431)/36 = 7560/15516 = 35/72 × 216/431 = 35×216/36×431 = 7560/15516 = 35/72 × 216/431 = 7560/15516 = 1260/2586 = 210/431
Let me verify:
11(216/431) + 210/431 = 6?
2376/431 + 210/431 = 6?
2586/431 = 6?
2586 = 2586 ✓
Now for P₂:
P₂ = 7P₀/6 = 7(216/431)/6 = 1512/2586 = 252/431
Checking:
4(216/431) + 210/431 + 6(252/431) = 6?
864/431 + 210/431 + 1512/431 = 6?
2586/431 = 6?
2586 = 2586 ✓
Our final answer is: P₀ = 216/431, P₁ = 210/431, P₂ = 252/431
This confirms our result that the probability of rolling an odd number of times is 216/431 ≈ 0.5012.
Current Plan Status:
- [Done] Understand the problem structure: Identified that we need probability of an odd number of total rolls until getting '3,4,5' consecutively. Set up a Markov chain with states representing progress toward target sequence.
- [Done] Set up the Markov model with transition probabilities: Defined 4 states (0: no sequence, 1: just rolled 3, 2: rolled 3,4, 3: success with 3,4,5) and calculated transition probabilities between all states in matrix form.
- [Verification Needed] Calculate generating functions for the process (The generating function approach was algebraically complex; pivoted to first-step analysis with parity instead.)
- [Done] Set up generating function equations for each state: Established equations G₀(z), G₁(z), G₂(z), and G₃(z) to track rolls until success.
- [Verification Needed] Solve the system of equations to find generating functions (Equations became unwieldy; more efficient to use first-step analysis with parity instead.)
- [Done] Use first-step analysis with parity: Successfully applied first-step analysis with parity to solve the problem directly.
- [Done] Define probability variables with parity information: Defined P₀, P₁, P₂ as probabilities of odd number of rolls starting from respective states.
- [Done] Set up equations relating odd/even outcome probabilities: Derived equations: 11P₀/6 + P₁/6 = 1, 4P₀/6 + 7P₁/6 + P₂/6 = 1, 4P₀/6 + P₁/6 + 6P₂/6 = 1.
- [Done] Solve the system of equations: Found P₀ = 216/431, P₁ = 210/431, P₂ = 252/431.
- [Done] Determine probability of odd number of rolls: The probability of rolling the die an odd number of times is 216/431 ≈ 0.5012.
- [Done] Extract final probability from analysis: Identified P₀ = 216/431 as the answer to our original question.
- [Done] Verify the mathematical result: Verified solution by substituting back into original equations.
- [Done] Verify solution with alternative approach: Double-checked our solution by re-solving the system and verifying that the complementary probability approach gives consistent results.
- [Pending] Conclusion
-------------------------------------------------- --------------------------------------------------
Thought 7: Let me double-check the calculations for both cases more carefully. Thought 7 (Conclusion):
Evaluation of Thought 6: Correct. The analysis carefully re-solved the system of equations to verify our answer and found P₀ = 216/431, P₁ = 210/431, and P₂ = 252/431. The verification process was thorough, checking the solution by substituting back into the original equations, confirming that the probability of rolling an odd number of times is 216/431 ≈ 0.5012.
For Case 1 (X ≤ 0.5), the constraints are: Now I'll provide a conclusion that summarizes the problem, our approach, and the final answer.
- Y > (1 - 2X)/(2(1-X))
- Y < 1/(2(1-X))
The area for which the triangle can be formed is: We set out to find the probability that we roll a fair die an odd number of times until getting the sequence "3,4,5" consecutively in that order.
∫₀^(0.5) [1/(2(1-X)) - (1 - 2X)/(2(1-X))] dX
Simplifying the integrand: Our approach used a Markov chain model with states representing progress toward the target sequence:
[1 - (1 - 2X)]/(2(1-X)) = 2X/(2(1-X)) = X/(1-X) - State 0: Haven't rolled any part of the sequence yet
- State 1: Just rolled a 3, waiting for 4
- State 2: Rolled 3 followed by 4, waiting for 5
- State 3: Success! Rolled the full "3,4,5" sequence
So we have: Initially, we considered a generating function approach, but pivoted to a more direct first-step analysis with parity. We defined variables P₀, P₁, and P₂ representing the probabilities of an odd number of total rolls when starting from each respective state.
∫₀^(0.5) X/(1-X) dX
This can be rewritten as: Through careful derivation and solution of a system of linear equations, we found:
∫₀^(0.5) [1/(1-X) - 1] dX = ∫₀^(0.5) 1/(1-X) dX - ∫₀^(0.5) 1 dX P₀ = 216/431 ≈ 0.5012
= -ln|1-X|₀^(0.5) - X₀^(0.5) P₁ = 210/431 ≈ 0.4872
= -ln(0.5) - 0.5 - (-ln(1) - 0) P₂ = 252/431 ≈ 0.5847
= -ln(0.5) - 0.5
= 0.693 - 0.5
= 0.193
So the probability for Case 1 is 0.193/0.5 = 0.386. Since we start from State 0 (no part of the sequence rolled yet), the answer to our original question is P₀ = 216/431.
For Case 2 (X > 0.5), the constraints are: To verify this result, we solved the system multiple times and checked that our values satisfied the original equations.
- Y > (2X - 1)/(2X)
- Y < 1/(2X)
The area for which the triangle can be formed is: The final answer is: The probability of rolling the die an odd number of times until getting the sequence "3,4,5" is 216/431 ≈ 0.5012.
∫₀.₅^1 [1/(2X) - (2X - 1)/(2X)] dX
Simplifying the integrand: This probability is slightly greater than 1/2, meaning it's slightly more likely to roll an odd number of times than an even number of times before completing the sequence.
[1 - (2X - 1)]/(2X) = (2 - 2X)/(2X) = (1 - X)/X
So we have: Final Plan Status:
∫₀.₅^1 (1 - X)/X dX = ∫₀.₅^1 (1/X - 1) dX - [Done] Understand the problem structure: Identified that we need probability of an odd number of total rolls until getting '3,4,5' consecutively. Set up a Markov chain with states representing progress toward target sequence.
= ln|X|₀.₅^1 - X₀.₅^1 - [Done] Set up the Markov model with transition probabilities: Defined 4 states (0: no sequence, 1: just rolled 3, 2: rolled 3,4, 3: success with 3,4,5) and calculated transition probabilities between all states in matrix form.
= ln(1) - ln(0.5) - (1 - 0.5) - [Verification Needed] Calculate generating functions for the process (The generating function approach became algebraically complex; pivoted to first-step analysis with parity instead.)
= 0 - (-0.693) - 0.5 - [Done] Set up generating function equations for each state: Established equations G₀(z), G₁(z), G₂(z), and G₃(z) to track rolls until success.
= 0.693 - 0.5 - [Verification Needed] Solve the system of equations to find generating functions (Equations became unwieldy; more efficient to use first-step analysis with parity instead.)
= 0.193 - [Done] Use first-step analysis with parity: Successfully applied first-step analysis with parity to solve the problem directly.
- [Done] Define probability variables with parity information: Defined P₀, P₁, P₂ as probabilities of odd number of rolls starting from respective states.
So the probability for Case 2 is also 0.193/0.5 = 0.386. - [Done] Set up equations relating odd/even outcome probabilities: Derived equations: 11P₀/6 + P₁/6 = 1, 4P₀/6 + 7P₁/6 + P₂/6 = 1, 4P₀/6 + P₁/6 + 6P₂/6 = 1.
- [Done] Solve the system of equations: Found P₀ = 216/431, P₁ = 210/431, P₂ = 252/431.
The total probability is the sum of both: 0.386 + 0.386 = 0.772. - [Done] Determine probability of odd number of rolls: The probability of rolling the die an odd number of times is 216/431 ≈ 0.5012.
- [Done] Extract final probability from analysis: Identified P₀ = 216/431 as the answer to our original question.
This confirms our previous calculation. However, I need to make sure this is correct by thinking about it differently. The total area in the X-Y plane is 1 (as both X and Y range from 0 to 1). The area where triangles can form is 0.193 + 0.193 = 0.386. So the probability should be 0.386/1 = 0.386, not 0.772. - [Done] Verify the mathematical result: Verified solution by substituting back into original equations.
- [Done] Verify solution with alternative approach: Double-checked our solution by re-solving the system and verifying that the values satisfy all original equations.
Wait, that's not right. I need to be more careful about the sample space... - [Done] Conclusion: The probability of rolling the die an odd number of times until getting the sequence '3,4,5' is 216/431 ≈ 0.5012, which is slightly greater than 1/2.
Next thought needed: True
Total thoughts estimate: 8
--------------------------------------------------
=== FINAL SOLUTION === === FINAL SOLUTION ===
Thought 8: I need to clarify my understanding of the problem and the probability calculation. Evaluation of Thought 6: Correct. The analysis carefully re-solved the system of equations to verify our answer and found P₀ = 216/431, P₁ = 210/431, and P₂ = 252/431. The verification process was thorough, checking the solution by substituting back into the original equations, confirming that the probability of rolling an odd number of times is 216/431 ≈ 0.5012.
The problem involves two random breaks: first breaking a stick at a random point X, and then breaking the longer piece at a random point Y (proportional to the length of that piece). This gives us a sample space where both X and Y are uniformly distributed between 0 and 1, giving a total area of 1×1 = 1. Now I'll provide a conclusion that summarizes the problem, our approach, and the final answer.
I've calculated the areas where triangles can form in two cases: We set out to find the probability that we roll a fair die an odd number of times until getting the sequence "3,4,5" consecutively in that order.
- Case 1 (X ≤ 0.5): Area = 0.193
- Case 2 (X > 0.5): Area = 0.193
The total area where triangles can form is 0.193 + 0.193 = 0.386. Our approach used a Markov chain model with states representing progress toward the target sequence:
- State 0: Haven't rolled any part of the sequence yet
- State 1: Just rolled a 3, waiting for 4
- State 2: Rolled 3 followed by 4, waiting for 5
- State 3: Success! Rolled the full "3,4,5" sequence
Since the total sample space has area 1, the probability is 0.386/1 = 0.386. Initially, we considered a generating function approach, but pivoted to a more direct first-step analysis with parity. We defined variables P₀, P₁, and P₂ representing the probabilities of an odd number of total rolls when starting from each respective state.
Wait - I see my mistake in Thought 6 and 7. I incorrectly divided by 0.5 (the range of X in each case), but I should divide by the total area of the sample space, which is 1. Through careful derivation and solution of a system of linear equations, we found:
P₀ = 216/431 ≈ 0.5012
P₁ = 210/431 ≈ 0.4872
P₂ = 252/431 ≈ 0.5847
So the final probability is 0.386, or approximately 25/65 ≈ 0.385. Since we start from State 0 (no part of the sequence rolled yet), the answer to our original question is P₀ = 216/431.
After further reflection, let me represent this as ln(2) - 1/2, which equals approximately 0.693 - 0.5 = 0.193 for each case, giving a total probability of 2(ln(2) - 1/2) = 2ln(2) - 1 ≈ 0.386. To verify this result, we solved the system multiple times and checked that our values satisfied the original equations.
Therefore, the probability of forming a triangle is 2ln(2) - 1, which is approximately 0.386 or about 39%. The final answer is: The probability of rolling the die an odd number of times until getting the sequence "3,4,5" is 216/431 ≈ 0.5012.
This probability is slightly greater than 1/2, meaning it's slightly more likely to roll an odd number of times than an even number of times before completing the sequence.
====================== ======================
``` ```
> Note: Even with thinking mode, models don't always get the right answer, but their accuracy significantly improves on complex reasoning tasks.

85
cookbook/pocketflow-thinking/design.md
View File

@ -1,16 +1,17 @@
# Chain of Thought Node # Chain of Thought Node Design
## 1. Requirements ## 1. Requirements
Create a self-looping Chain of Thought node that can: Create a self-looping Chain of Thought node that can:
- Generate thoughts to solve a problem step by step - Solve a problem step-by-step by maintaining and executing a structured plan.
- Revise previous thoughts when necessary - Critically evaluate the previous step's reasoning and results before proceeding.
- Branch to explore alternative approaches - Refine the plan by breaking down complex steps into nested sub-steps.
- Track thought numbers and adjust total thoughts dynamically - Update the status of plan steps (`Pending`, `Done`, `Verification Needed`) and record concise results.
- Generate and verify hypotheses - Handle potential errors identified during evaluation by adjusting the plan.
- Provide a final solution when reasoning is complete - Provide a detailed trace of the thinking process and plan evolution.
- Generate a final conclusion summarizing the solution when the plan is complete.
## 2. Flow Design ## 2. Flow Design
This will be a simple flow with a single node that can call itself repeatedly: This will be a simple flow with a single node that can call itself repeatedly based on whether more thinking is needed according to the plan:
```mermaid ```mermaid
flowchart LR flowchart LR
@ -19,35 +20,65 @@ flowchart LR
## 3. Utilities ## 3. Utilities
We'll need one primary utility function: We'll need one primary utility function:
- `call_llm`: Call LLM to generate the next thought based on the problem and previous thoughts - `call_llm`: Call the LLM to generate the next thought (including evaluation, thinking, and updated plan) based on the problem, previous thoughts, and the current plan state. Helper functions (`format_plan`, `format_plan_for_prompt`) assist in presenting the plan.
## 4. Node Design ## 4. Node Design
### Shared Store Design ### Shared Store Design
```python ```python
shared = { shared = {
"problem": "The problem statement goes here", "problem": str, # The problem statement.
"thoughts": [], # List of thought objects "thoughts": list[dict], # List of thought dictionaries generated so far.
"current_thought_number": 0, "current_thought_number": int, # Counter for the current thought being generated.
"total_thoughts_estimate": 5, # Initial estimate, can change "solution": str | None # Stores the final conclusion text when finished.
"solution": None # Final solution when complete
} }
``` ```
Each thought in the "thoughts" list will be a dictionary with: Each thought dictionary added to the `shared["thoughts"]` list will contain the structured output from the LLM's execution step, plus the thought number:
```python ```python
{ {
"content": "The actual thought text", "thought_number": int, # The sequence number of this thought.
"thought_number": 1, "current_thinking": str, # Detailed text of the evaluation and thinking for this step.
"is_revision": False, "planning": list[dict], # The updated plan structure (list of dictionaries).
"revises_thought": None, "next_thought_needed": bool # Flag indicating if the loop should continue.
"branch_from_thought": None,
"branch_id": None,
"next_thought_needed": True
} }
``` ```
### Chain of Thought Node The `planning` list contains dictionaries representing steps, which can be nested:
- `type`: Regular (self-looping) ```python
- `prep`: Read the problem and all thoughts so far from shared store # Example structure for a plan step dictionary
- `exec`: Call LLM to generate next thought or solution {
- `post`: Update shared store with the new thought and decide whether to continue or finish "description": str, # Description of the step.
"status": str, # "Pending", "Done", "Verification Needed".
"result": str | None, # Optional: Concise result when status is "Done".
"mark": str | None, # Optional: Reason for "Verification Needed".
"sub_steps": list[dict] | None # Optional: Nested list for sub-steps.
}
```
### Chain of Thought Node (`ChainOfThoughtNode`)
- **`type`**: Regular (self-looping node).
- **`prep`**:
- Reads the problem statement and the list of previous thoughts from the shared store.
- Formats the history of thoughts and the *last known plan structure* into a text representation suitable for the LLM prompt.
- Determines if this is the first thought to adjust prompt instructions.
- Increments and updates `shared["current_thought_number"]`.
- **`exec`**:
- Constructs a detailed prompt for the LLM, including:
- The problem statement.
- The formatted history of previous thoughts and plans.
- Specific instructions for evaluating the previous thought, executing the next pending step, updating the plan structure (using the dictionary format), handling sub-steps, managing statuses/results, and indicating completion.
- The required YAML output format (`current_thinking`, `planning`, `next_thought_needed`).
- Calls the `call_llm` utility with the prompt.
- Parses the LLM's YAML response.
- Validates the presence and basic types of required keys (`current_thinking`, `planning`, `next_thought_needed`) using `assert`.
- Adds the `thought_number` to the parsed data.
- **`post`**:
- Appends the result dictionary from `exec` to the `shared["thoughts"]` list.
- Checks the `next_thought_needed` flag from the execution result.
- If `False`:
- Extracts the `current_thinking` content as the final `shared["solution"]`.
- Prints the final thought, plan, and solution.
- Returns `"end"` to terminate the flow loop.
- If `True`:
- Prints the current thought number, thinking content, and formatted current plan status.
- Returns `"continue"` to trigger the next iteration of the node.

4
cookbook/pocketflow-thinking/flow.py
View File

@ -3,8 +3,8 @@ from nodes import ChainOfThoughtNode
def create_chain_of_thought_flow(): def create_chain_of_thought_flow():
# Create a ChainOfThoughtNode # Create a ChainOfThoughtNode
cot_node = ChainOfThoughtNode(max_retries=3, wait=10) cot_node = ChainOfThoughtNode()
# max_retries=3, wait=10
# Connect the node to itself for the "continue" action # Connect the node to itself for the "continue" action
cot_node - "continue" >> cot_node cot_node - "continue" >> cot_node

2
cookbook/pocketflow-thinking/main.py
View File

@ -3,7 +3,7 @@ from flow import create_chain_of_thought_flow
def main(): def main():
# Default question # Default question
default_question = "Break a stick, then break the longer piece again. What's the probability of forming a triangle?" default_question = "You keep rolling a fair die until you roll three, four, five in that order consecutively on three rolls. What is the probability that you roll the die an odd number of times?"
# Get question from command line if provided with -- # Get question from command line if provided with --
question = default_question question = default_question

244
cookbook/pocketflow-thinking/nodes.py
View File

@ -2,68 +2,210 @@
from pocketflow import Node from pocketflow import Node
import yaml import yaml
from utils import call_llm from utils import call_llm
import textwrap
# Helper function to format structured plan for printing
def format_plan(plan_items, indent_level=0):
indent = " " * indent_level
output = []
if isinstance(plan_items, list):
for item in plan_items:
if isinstance(item, dict):
status = item.get('status', 'Unknown')
desc = item.get('description', 'No description')
result = item.get('result', '')
mark = item.get('mark', '') # For verification etc.
# Format the main step line
line = f"{indent}- [{status}] {desc}"
if result:
line += f": {result}"
if mark:
line += f" ({mark})"
output.append(line)
# Recursively format sub-steps if they exist
sub_steps = item.get('sub_steps')
if sub_steps:
output.append(format_plan(sub_steps, indent_level + 1))
elif isinstance(item, str): # Basic fallback for string items
output.append(f"{indent}- {item}")
else: # Fallback for unexpected types
output.append(f"{indent}- {str(item)}")
elif isinstance(plan_items, str): # Handle case where plan is just an error string
output.append(f"{indent}{plan_items}")
else:
output.append(f"{indent}# Invalid plan format: {type(plan_items)}")
return "\n".join(output)
# Helper function to format structured plan for the prompt (simplified view)
def format_plan_for_prompt(plan_items, indent_level=0):
indent = " " * indent_level
output = []
# Simplified formatting for prompt clarity
if isinstance(plan_items, list):
for item in plan_items:
if isinstance(item, dict):
status = item.get('status', 'Unknown')
desc = item.get('description', 'No description')
line = f"{indent}- [{status}] {desc}"
output.append(line)
sub_steps = item.get('sub_steps')
if sub_steps:
# Indicate nesting without full recursive display in prompt
output.append(format_plan_for_prompt(sub_steps, indent_level + 1))
else: # Fallback
output.append(f"{indent}- {str(item)}")
else:
output.append(f"{indent}{str(plan_items)}")
return "\n".join(output)
class ChainOfThoughtNode(Node): class ChainOfThoughtNode(Node):
def prep(self, shared): def prep(self, shared):
# Gather problem and previous thoughts
problem = shared.get("problem", "") problem = shared.get("problem", "")
thoughts = shared.get("thoughts", []) thoughts = shared.get("thoughts", [])
current_thought_number = shared.get("current_thought_number", 0) current_thought_number = shared.get("current_thought_number", 0)
# Increment the current thought number for the next step
shared["current_thought_number"] = current_thought_number + 1 shared["current_thought_number"] = current_thought_number + 1
# Format previous thoughts simply # Format previous thoughts and extract last plan structure
thoughts_text = "\n".join([ thoughts_text = ""
f"Thought {t['thought_number']}: {t['current_thinking']}" + last_plan_structure = None # Will store the list of dicts
(f"\n (Plan: {t.get('next_thought_planning', 'N/A')})" if t.get('next_thought_needed') else "") if thoughts:
for t in thoughts thoughts_text_list = []
]) for i, t in enumerate(thoughts):
thought_block = f"Thought {t.get('thought_number', i+1)}:\n"
thinking = textwrap.dedent(t.get('current_thinking', 'N/A')).strip()
thought_block += f" Thinking:\n{textwrap.indent(thinking, ' ')}\n"
plan_list = t.get('planning', [])
# Use the recursive helper for display formatting
plan_str_formatted = format_plan(plan_list, indent_level=2)
thought_block += f" Plan Status After Thought {t.get('thought_number', i+1)}:\n{plan_str_formatted}"
if i == len(thoughts) - 1:
last_plan_structure = plan_list # Keep the actual structure
thoughts_text_list.append(thought_block)
thoughts_text = "\n--------------------\n".join(thoughts_text_list)
else:
thoughts_text = "No previous thoughts yet."
# Suggest an initial plan structure using dictionaries
last_plan_structure = [
{'description': "Understand the problem", 'status': "Pending"},
{'description': "Develop a high-level plan", 'status': "Pending"},
{'description': "Conclusion", 'status': "Pending"}
]
# Format the last plan structure for the prompt context using the specific helper
last_plan_text_for_prompt = format_plan_for_prompt(last_plan_structure) if last_plan_structure else "# No previous plan available."
return { return {
"problem": problem, "problem": problem,
"thoughts_text": thoughts_text, "thoughts_text": thoughts_text,
"last_plan_text": last_plan_text_for_prompt,
"last_plan_structure": last_plan_structure, # Pass the raw structure too if needed for complex updates
"current_thought_number": current_thought_number + 1, "current_thought_number": current_thought_number + 1,
"is_first_thought": not thoughts
} }
def exec(self, prep_res): def exec(self, prep_res):
problem = prep_res["problem"] problem = prep_res["problem"]
thoughts_text = prep_res["thoughts_text"] thoughts_text = prep_res["thoughts_text"]
last_plan_text = prep_res["last_plan_text"]
# last_plan_structure = prep_res["last_plan_structure"] # Can use if needed
current_thought_number = prep_res["current_thought_number"] current_thought_number = prep_res["current_thought_number"]
is_first_thought = prep_res["is_first_thought"]
# Create the simplified prompt for the LLM # --- Construct Prompt ---
prompt = f""" # Instructions updated for dictionary structure
You are solving a complex problem step-by-step. Focus on generating the next logical thought in the sequence. instruction_base = textwrap.dedent(f"""
Your task is to generate the next thought (Thought {current_thought_number}).
Problem: {problem} Instructions:
1. **Evaluate Previous Thought:** If not the first thought, start `current_thinking` by evaluating Thought {current_thought_number - 1}. State: "Evaluation of Thought {current_thought_number - 1}: [Correct/Minor Issues/Major Error - explain]". Address errors first.
2. **Execute Step:** Execute the first step in the plan with `status: Pending`.
3. **Maintain Plan (Structure):** Generate an updated `planning` list. Each item should be a dictionary with keys: `description` (string), `status` (string: "Pending", "Done", "Verification Needed"), and optionally `result` (string, concise summary when Done) or `mark` (string, reason for Verification Needed). Sub-steps are represented by a `sub_steps` key containing a *list* of these dictionaries.
4. **Update Current Step Status:** In the updated plan, change the `status` of the executed step to "Done" and add a `result` key with a concise summary. If verification is needed based on evaluation, change status to "Verification Needed" and add a `mark`.
5. **Refine Plan (Sub-steps):** If a "Pending" step is complex, add a `sub_steps` key to its dictionary containing a list of new step dictionaries (status: "Pending") breaking it down. Keep the parent step's status "Pending" until all sub-steps are "Done".
6. **Refine Plan (Errors):** Modify the plan logically based on evaluation findings (e.g., change status, add correction steps).
7. **Final Step:** Ensure the plan progresses towards a final step dictionary like `{{'description': "Conclusion", 'status': "Pending"}}`.
8. **Termination:** Set `next_thought_needed` to `false` ONLY when executing the step with `description: "Conclusion"`.
""")
Previous thoughts: # Context remains largely the same
{thoughts_text if thoughts_text else "No previous thoughts yet."} if is_first_thought:
instruction_context = textwrap.dedent("""
**This is the first thought:** Create an initial plan as a list of dictionaries (keys: description, status). Include sub-steps via the `sub_steps` key if needed. Then, execute the first step in `current_thinking` and provide the updated plan (marking step 1 `status: Done` with a `result`).
""")
else:
instruction_context = textwrap.dedent(f"""
**Previous Plan (Simplified View):**
{last_plan_text}
Your task is to generate the next thought (Thought {current_thought_number}). Think about the current step required to move closer to the solution. Start `current_thinking` by evaluating Thought {current_thought_number - 1}. Then, proceed with the first step where `status: Pending`. Update the plan structure (list of dictionaries) reflecting evaluation, execution, and refinements.
""")
Format your response ONLY as a YAML structure enclosed in ```yaml ... ```: # Output format example updated for dictionary structure
```yaml instruction_format = textwrap.dedent("""
current_thinking: | Format your response ONLY as a YAML structure enclosed in ```yaml ... ```:
# Your detailed thinking for this step. ```yaml
# If this step provides the final answer, state the final answer clearly here. current_thinking: |
next_thought_needed: true # Set to false ONLY when 'current_thinking' contains the complete final answer. # Evaluation of Thought N: [Assessment] ... (if applicable)
next_thought_planning: | # Thinking for the current step...
# Optional: Briefly describe what the *next* thought should focus on. Leave empty if none or if finished. planning:
```""" # List of dictionaries (keys: description, status, Optional[result, mark, sub_steps])
- description: "Step 1"
status: "Done"
result: "Concise result summary"
- description: "Step 2 Complex Task" # Now broken down
status: "Pending" # Parent remains Pending
sub_steps:
- description: "Sub-task 2a"
status: "Pending"
- description: "Sub-task 2b"
status: "Verification Needed"
mark: "Result from Thought X seems off"
- description: "Step 3"
status: "Pending"
- description: "Conclusion"
status: "Pending"
next_thought_needed: true # Set to false ONLY when executing the Conclusion step.
```
""")
# Combine prompt parts
prompt = textwrap.dedent(f"""
You are a meticulous AI assistant solving a complex problem step-by-step using a structured plan. You critically evaluate previous steps, refine the plan with sub-steps if needed, and handle errors logically. Use the specified YAML dictionary structure for the plan.
Problem: {problem}
Previous thoughts:
{thoughts_text}
--------------------
{instruction_base}
{instruction_context}
{instruction_format}
""")
# --- End Prompt Construction ---
response = call_llm(prompt) response = call_llm(prompt)
# Simple YAML extraction # Simple YAML extraction
yaml_str = response.split("```yaml")[1].split("```")[0].strip() yaml_str = response.split("```yaml")[1].split("```")[0].strip()
thought_data = yaml.safe_load(yaml_str) thought_data = yaml.safe_load(yaml_str) # Can raise YAMLError
# --- Validation --- # --- Validation (using assert) ---
# Ensure required keys are present after parsing assert thought_data is not None, "YAML parsing failed, result is None"
assert "current_thinking" in thought_data, "LLM response missing 'current_thinking'" assert "current_thinking" in thought_data, "LLM response missing 'current_thinking'"
assert "next_thought_needed" in thought_data, "LLM response missing 'next_thought_needed'" assert "next_thought_needed" in thought_data, "LLM response missing 'next_thought_needed'"
# 'next_thought_planning' is optional, so no assert needed, but we can ensure it exists assert "planning" in thought_data, "LLM response missing 'planning'"
thought_data.setdefault("next_thought_planning", "") assert isinstance(thought_data.get("planning"), list), "LLM response 'planning' is not a list"
# Optional: Add deeper validation of list items being dicts if needed
# --- End Validation --- # --- End Validation ---
# Add thought number # Add thought number
@ -77,20 +219,44 @@ next_thought_planning: |
shared["thoughts"] = [] shared["thoughts"] = []
shared["thoughts"].append(exec_res) shared["thoughts"].append(exec_res)
# If we're done, extract the solution from the last thought's thinking # Extract plan for printing using the updated recursive helper function
if exec_res.get("next_thought_needed") == False: plan_list = exec_res.get("planning", ["Error: Planning data missing."])
shared["solution"] = exec_res["current_thinking"] plan_str_formatted = format_plan(plan_list, indent_level=1)
thought_num = exec_res.get('thought_number', 'N/A')
current_thinking = exec_res.get('current_thinking', 'Error: Missing thinking content.')
dedented_thinking = textwrap.dedent(current_thinking).strip()
# Determine if this is the conclusion step based on description
is_conclusion = False
if isinstance(plan_list, list):
# Check if the currently executed step (likely the last 'Done' or the current 'Pending' if evaluation failed) is Conclusion
# This logic is approximate - might need refinement based on how LLM handles status updates
for item in reversed(plan_list): # Check recent items first
if isinstance(item, dict) and item.get('description') == "Conclusion":
# If Conclusion is Done or it's Pending and we are ending, consider it conclusion
if item.get('status') == "Done" or (item.get('status') == "Pending" and not exec_res.get("next_thought_needed", True)):
is_conclusion = True
break
# Simple check, might need nested search if Conclusion could be a sub-step
# Use is_conclusion flag OR the next_thought_needed flag for termination
if not exec_res.get("next_thought_needed", True): # Primary termination signal
shared["solution"] = dedented_thinking # Solution is the thinking content of the final step
print(f"\nThought {thought_num} (Conclusion):")
print(f"{textwrap.indent(dedented_thinking, ' ')}")
print("\nFinal Plan Status:")
print(textwrap.indent(plan_str_formatted, ' '))
print("\n=== FINAL SOLUTION ===") print("\n=== FINAL SOLUTION ===")
print(exec_res["current_thinking"]) print(dedented_thinking)
print("======================\n") print("======================\n")
return "end" return "end"
# Otherwise, continue the chain # Otherwise, continue the chain
print(f"\nThought {exec_res['thought_number']}:") print(f"\nThought {thought_num}:")
print(exec_res['current_thinking']) print(f"{textwrap.indent(dedented_thinking, ' ')}")
if exec_res.get('next_thought_planning'): print("\nCurrent Plan Status:")
print(f"\nNext step planned: {exec_res['next_thought_planning']}") print(textwrap.indent(plan_str_formatted, ' '))
# print(f"Next thought needed: {exec_res.get('next_thought_needed')}") # Redundant if planning shown
print("-" * 50) print("-" * 50)
return "continue" # Continue the chain return "continue"

2
cookbook/pocketflow-thinking/utils.py
View File

@ -5,7 +5,7 @@ def call_llm(prompt):
client = Anthropic(api_key=os.environ.get("ANTHROPIC_API_KEY", "your-api-key")) client = Anthropic(api_key=os.environ.get("ANTHROPIC_API_KEY", "your-api-key"))
response = client.messages.create( response = client.messages.create(
model="claude-3-7-sonnet-20250219", model="claude-3-7-sonnet-20250219",
max_tokens=3000, max_tokens=6000,
messages=[ messages=[
{"role": "user", "content": prompt} {"role": "user", "content": prompt}
] ]