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
- 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
- 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:
> 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:
```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
- **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
> **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
- **Models with extended thinking**: Find the correct answer (0.386)
- **Models with extended thinking**: Find the correct answer (216/431 ≈ 0.5012.)
- [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/share/67fee0fd-2600-8000-bcdf-76e40a986ee4): Correct answer after 2m, 0s
- [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:
> **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:
a + b > c
a + c > b
b + c > a
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.
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
Total thoughts estimate: 8
First, I need to understand what states we can be in while rolling the die:
- 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
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:
- X (from first break)
- Y(1-X) (from second break)
- (1-Y)(1-X) (remaining from second break)
State 1: Just rolled a 3, waiting for a 4
- If we roll a 4: Move to State 2 (probability 1/6)
- If we roll a 3: Stay in State 1 (probability 1/6)
- 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:
- 1-X (from first break)
- YX (from second break)
- (1-Y)X (remaining from second break)
State 2: Rolled 3 followed by 4, waiting for a 5
- If we roll a 5: Move to State 3 (success - probability 1/6)
- If we roll a 3: Move to State 1 (probability 1/6)
- 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
Total thoughts estimate: 8
These transition probabilities can be represented as a matrix:
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
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:
- X + Y(1-X) > (1-Y)(1-X)
- X + (1-Y)(1-X) > Y(1-X)
- Y(1-X) + (1-Y)(1-X) > X
G₃(z) = 1 (since we're already in the success state)
For the other states, we can write equations based on the first roll and subsequent state:
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.
G₁(z) = z·(4/6·G₀(z) + 1/6·G₁(z) + 1/6·G₂(z))
When in State 1, we roll once (factor z), then either move to State 0, stay in State 1, or advance to State 2.
G₂(z) = z·(4/6·G₀(z) + 1/6·G₁(z) + 1/6·G₃(z))
= z·(4/6·G₀(z) + 1/6·G₁(z) + 1/6·1)
= z·(4/6·G₀(z) + 1/6·G₁(z) + 1/6)
Now I can solve this system of equations for G₀(z):
From G₀(z) = z·(5/6·G₀(z) + 1/6·G₁(z)):
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)):
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
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.
I'll now use first-step analysis focusing specifically on the parity of the number of rolls.
Let's define:
- P₀: Probability of rolling an odd number of times when starting in State 0
- P₁: Probability of rolling an odd number of times when starting in State 1
- P₂: Probability of rolling an odd number of times when starting in State 2
Using first-step analysis, we can write equations for each state:
For State 0:
- 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.
- 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.
Now I have a system of 3 equations with 3 unknowns:
P₀ = 1 - 5P₀/6 - P₁/6
P₁ = 1 - P₁/6 - P₂/6 - 4P₀/6
P₂ = 1 - P₁/6 - 4P₀/6
Simplifying:
- 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)
(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
The third condition is always true when X ≤ 0.5 (which is our assumption for Case 1).
Further simplifying:
11P₀/6 + P₁/6 = 1
4P₀/6 + 7P₁/6 + P₂/6 = 1
4P₀/6 + P₁/6 + 6P₂/6 = 1
Let's work on the first condition:
X + Y(1-X) > (1-Y)(1-X)
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
This is a linear system of equations that I can solve.
Similarly for the second condition:
X + (1-Y)(1-X) > Y(1-X)
X + (1-X) - Y(1-X) > Y(1-X)
X + (1-X) > 2Y(1-X)
1 > 2Y(1-X)
1/(2(1-X)) > Y
So for Case 1, we need:
- 2X + 2Y - 2YX > 1
- Y <1/(2(1-X))
Next thought needed: True
Total thoughts estimate: 8
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.
Thought 4: I need to continue analyzing Case 1 and then move on to Case 2.
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.
For Case 1 (X ≤ 0.5), we have the constraints:
- 2X + 2Y - 2YX > 1
- Y <1/(2(1-X))
Taking the system of equations from the end of Thought 4:
11P₀/6 + P₁/6 = 1
4P₀/6 + 7P₁/6 + P₂/6 = 1
4P₀/6 + P₁/6 + 6P₂/6 = 1
Let's rewrite the first constraint:
2X + 2Y - 2YX > 1
2Y(1-X) > 1 - 2X
Y > (1 - 2X)/(2(1-X))
Let's multiply all equations by 6 to eliminate fractions:
11P₀ + P₁ = 6
4P₀ + 7P₁ + P₂ = 6
4P₀ + P₁ + 6P₂ = 6
So for Case 1, we need:
- Y > (1 - 2X)/(2(1-X))
- Y <1/(2(1-X))
Subtracting the third equation from the second:
(4P₀ + 7P₁ + P₂) - (4P₀ + P₁ + 6P₂) = 6 - 6
6P₁ - 5P₂ = 0
6P₁ = 5P₂
P₁ = 5P₂/6
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:
(1 - 2X)/(2(1-X)) <1/(2(1-X))
1 - 2X <1
-2X <0
X > 0
From the first equation:
11P₀ + P₁ = 6
11P₀ + 5P₂/6 = 6
66P₀ + 5P₂ = 36
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 let's analyze Case 2 (X > 0.5): We have pieces of lengths (1-X), YX, and (1-Y)X
Now solving for P₀ in terms of P₂:
24P₀ + 41P₂ = 36
24P₀ = 36 - 41P₂
P₀ = (36 - 41P₂)/24 = 3/2 - 41P₂/24
The triangle inequality conditions are:
- (1-X) + YX > (1-Y)X
- (1-X) + (1-Y)X > YX
- YX + (1-Y)X > (1-X)
Substituting back into 66P₀ + 5P₂ = 36:
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
Simplifying:
- (1-X) + YX > (1-Y)X
- (1-X) + (1-Y)X > YX
- X > (1-X) (since YX + (1-Y)X = X)
Now we can find P₁:
P₁ = 5P₂/6 = 5(54/431)/6 = 270/2586 = 45/431
The third condition is always true when X > 0.5 (which is our assumption for Case 2).
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.
Thought 5: Continuing with Case 2 (X > 0.5), let's simplify the remaining triangle inequality conditions:
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.
First condition:
(1-X) + YX > (1-Y)X
(1-X) + YX > X - YX
(1-X) + 2YX > X
1 - X + 2YX > X
1 + 2YX > 2X
2YX > 2X - 1
Y > (2X - 1)/(2X)
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.
Second condition:
(1-X) + (1-Y)X > YX
(1-X) + X - YX > YX
(1-X) + X > 2YX
1 > 2YX
1/(2X) > Y
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.
So for Case 2, we need:
- Y > (2X - 1)/(2X)
- Y <1/(2X)
Let me verify this with the equations we derived:
- Let Q₀, Q₁, Q₂ be the probabilities of rolling an even number of times from states 0, 1, and 2 respectively.
- Since the events "odd number of rolls" and "even number of rolls" are complementary: Q₀ = 1-P₀, Q₁ = 1-P₁, Q₂ = 1-P₂
This means Y must be in the range ((2X - 1)/(2X), 1/(2X)). For this range to be valid, we need:
(2X - 1)/(2X) <1/(2X)
2X - 1 <1
2X <2
X <1
Now, I can derive the equations for Q₀, Q₁, Q₂ directly:
Which is true for our domain. So we have a valid range for Y when X > 0.5.
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.
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.
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] 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.
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 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):
- Y must be between (1 - 2X)/(2(1-X)) and 1/(2(1-X))
Now I'll provide a conclusion that summarizes the problem, our approach, and the final answer.
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.
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.
Area = ∫₀^(0.5) [1/(2(1-X)) - (1 - 2X)/(2(1-X))] dX
= ∫₀^(0.5) [(1 - (1 - 2X))/(2(1-X))] dX
= ∫₀^(0.5) [2X/(2(1-X))] dX
= ∫₀^(0.5) [X/(1-X)] dX
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
This integral equals -X - ln(1-X) evaluated from 0 to 0.5, which is:
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.
So the probability for Case 1 is 0.193/0.5 = 0.386.
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
For Case 2 (X > 0.5):
- Y must be between (2X - 1)/(2X) and 1/(2X)
Since we start from State 0 (no part of the sequence rolled yet), the answer to our original question is P₀ = 216/431.
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.
To verify this result, we solved the system multiple times and checked that our values satisfied the original equations.
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.
So the probability for Case 2 is also 0.193/0.5 = 0.386.
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.
The total probability is the sum of the probabilities for both cases: 0.386 + 0.386 = 0.772.
Wait, I should double-check my calculations. Let me reconsider the integrals.
So the probability for Case 1 is 0.193/0.5 = 0.386.
For Case 2 (X > 0.5), the constraints are:
- Y > (2X - 1)/(2X)
- Y <1/(2X)
The area for which the triangle can be formed is:
∫₀.₅^1 [1/(2X) - (2X - 1)/(2X)] dX
Simplifying the integrand:
[1 - (2X - 1)]/(2X) = (2 - 2X)/(2X) = (1 - X)/X
So we have:
∫₀.₅^1 (1 - X)/X dX = ∫₀.₅^1 (1/X - 1) dX
= ln|X|₀.₅^1 - X₀.₅^1
= ln(1) - ln(0.5) - (1 - 0.5)
= 0 - (-0.693) - 0.5
= 0.693 - 0.5
= 0.193
So the probability for Case 2 is also 0.193/0.5 = 0.386.
The total probability is the sum of both: 0.386 + 0.386 = 0.772.
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.
Wait, that's not right. I need to be more careful about the sample space...
- [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 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] 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 values satisfy all original equations.
- [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.
=== 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:
- Case 1 (X ≤ 0.5): Area = 0.193
- Case 2 (X > 0.5): Area = 0.193
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.
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.
Create a self-looping Chain of Thought node that can:
- Generate thoughts to solve a problem step by step
- Revise previous thoughts when necessary
- Branch to explore alternative approaches
- Track thought numbers and adjust total thoughts dynamically
- Generate and verify hypotheses
- Provide a final solution when reasoning is complete
- Solve a problem step-by-step by maintaining and executing a structured plan.
- Critically evaluate the previous step's reasoning and results before proceeding.
- Refine the plan by breaking down complex steps into nested sub-steps.
- Update the status of plan steps (`Pending`, `Done`, `Verification Needed`) and record concise results.
- Handle potential errors identified during evaluation by adjusting the plan.
- 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
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
flowchart LR
@ -19,35 +20,65 @@ flowchart LR
## 3. Utilities
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
### Shared Store Design
```python
shared = {
"problem": "The problem statement goes here",
"thoughts": [], # List of thought objects
"current_thought_number": 0,
"total_thoughts_estimate": 5, # Initial estimate, can change
"solution": None # Final solution when complete
"problem": str, # The problem statement.
"thoughts": list[dict], # List of thought dictionaries generated so far.
"current_thought_number": int, # Counter for the current thought being generated.
"solution": str | None # Stores the final conclusion text when finished.
}
```
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
{
"content": "The actual thought text",
"thought_number": 1,
"is_revision": False,
"revises_thought": None,
"branch_from_thought": None,
"branch_id": None,
"next_thought_needed": True
"thought_number": int, # The sequence number of this thought.
"current_thinking": str, # Detailed text of the evaluation and thinking for this step.
"planning": list[dict], # The updated plan structure (list of dictionaries).
"next_thought_needed": bool # Flag indicating if the loop should continue.
}
```
### Chain of Thought Node
- `type`: Regular (self-looping)
- `prep`: Read the problem and all thoughts so far from shared store
- `exec`: Call LLM to generate next thought or solution
- `post`: Update shared store with the new thought and decide whether to continue or finish
The `planning` list contains dictionaries representing steps, which can be nested:
"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.
@ -3,7 +3,7 @@ from flow import create_chain_of_thought_flow
defmain():
# 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 --
5.**RefinePlan(Sub-steps):**Ifa"Pending"stepiscomplex,adda`sub_steps`keytoitsdictionarycontainingalistofnewstepdictionaries(status:"Pending")breakingitdown.Keeptheparentstep's status "Pending" until all sub-steps are "Done".
zachary62