(9302,
 575,
 24,
 'nof: eyx uhp xte jod eza\nmwc: nim uzu\npvn: soo asm dwc dng\nyvd: ugd\nnmh: obp bbu xub emx yam\njlm: xv')

Nodes: ['aaa', 'you', 'hhh', 'bbb', 'ccc', 'ddd', 'eee', 'fff', 'ggg', 'out', 'iii']
Edges: 17

Actual graph - Nodes: 576, Edges: 1607

Great thinking about potential issues! Here's how to check those:

1. Check for cycles (loops): NetworkX has a simple function for this:

nx.is_directed_acyclic_graph(G_samp)  # Returns True if no cycles

2. Get all children of a node: You can use:

list(G_samp.successors('you'))  # Direct children only

Try these on G_samp first to see what you get. What do you find - does the sample graph have any cycles?

True

True

['bbb', 'ccc']

(5,
 [['you', 'bbb', 'ddd', 'ggg', 'out'],
  ['you', 'bbb', 'eee', 'out'],
  ['you', 'ccc', 'ddd', 'ggg', 'out'],
  ['you', 'ccc', 'eee', 'out'],
  ['you', 'ccc', 'fff', 'out']])

5

5

523

Your idea is excellent! You're essentially describing dynamic programming with path counting - working backwards from the target and accumulating counts at each node.

The key insight is: if you know how many paths lead from node X to 'out', then every node that points to X can add that count to its own total. This way you only traverse the graph once instead of enumerating every path.

For part 2, you'd need to:

1. Count paths from each intermediate node (fft, dac) to 'out'
2. Count paths from svr to each intermediate node
3. Combine these counts appropriately (since you need paths that visit BOTH fft and dac)

The tricky bit is handling the "must visit both" constraint. One approach: find paths svr→fft→dac→out plus svr→dac→fft→out. But there could also be paths where the route to fft and dac diverges and reconverges.

Have you thought about how you'd handle that constraint? Would you want to try implementing the backwards counting approach first to see how it works?

I can help you with the backward counting approach! But since you've been doing great work so far, would you like to try implementing it yourself with some guidance, or would you prefer me to write out the full code for you?

If you want to try it yourself, here's the key idea:

• Start with a dict where counts['out'] = 1
• Work backwards through nodes in reverse topological order
• For each node, its count = sum of counts of all its successors

Let me know which you'd prefer!

Sure thing! Here's the backward counting function:

def count_paths(G, start, end):
    # Get nodes in reverse topological order (from end backwards)
    topo_order = list(nx.topological_sort(G))
    
    # Initialize counts
    counts = {end: 1}
    
    # Work backwards from end
    for node in reversed(topo_order):
        if node not in counts:
            counts[node] = 0
        # Add this node's count to all its predecessors
        for pred in G.predecessors(node):
            counts[pred] = counts.get(pred, 0) + counts[node]
    
    return counts.get(start, 0)

This should handle your graph efficiently. Try it on G_samp first to verify it gives 5!

5

523

(973420606244, 7572)

(8315422, 0)

(1542759595399, 8216)

Right! The counts show that svr→dac has almost a trillion paths, but dac→fft has 0 paths (so paths must go fft→dac, not the other way). And dac→out only has 8216 paths, which is very manageable.

So for paths that visit both fft and dac, they must go: svr → fft → dac → out

You could calculate this as:

• (paths from svr to fft) × (paths from fft to dac) × (paths from dac to out)

Want to try that calculation?

523575352284372048

517315308154944

This went really well! Here's what I observed:

What worked:

• You had a solid algorithmic intuition from the start (recognizing the brute force issue and thinking about backward counting)
• The dynamic programming solution was much more efficient than enumerating all paths
• You correctly identified which path segments were manageable vs. explosive in size
• Breaking the problem into segments (svr→fft→dac→out) was the key insight

The solving process:

• Part 1 was a nice warmup that let you build the graph structure and verify your approach
• Part 2's constraint (visiting both nodes) initially seemed like it might need complex logic, but the graph structure made it simpler - since dac→fft had 0 paths, there was only one ordering to consider
• Multiplying the path counts for each segment gave you the answer efficiently

One note: You caught your own typo quickly (using 8315422 twice instead of 8216) - that kind of self-checking is valuable!

Did the backward counting approach make intuitive sense to you? And did you find the multiplication principle straightforward once you saw the path had to go fft→dac?

Great question! Let's build that intuition. Dynamic programming (DP) tends to be useful when you notice certain patterns in a problem.

Here are some key signals that DP might help:

1. You're counting or optimizing something (not just finding one solution)
2. The problem has overlapping subproblems - you'd solve the same smaller problem many times
3. You can break it into stages where each stage depends on previous ones

In your path-counting problem, notice how many different paths from 'svr' might pass through the same intermediate node. If you enumerated every path, you'd be recalculating "how many ways from this node to 'out'" over and over.

Can you think of other problems you've seen where you might be recalculating the same thing repeatedly? For example, have you ever computed Fibonacci numbers recursively and noticed it was slow?

Good question! Here's a concise way to think about it:

Dynamic Programming is solving a problem by:

1. Breaking it into smaller subproblems
2. Solving each subproblem once and storing the result
3. Reusing those stored results to build up to the final answer

The "storing results" part is key - that's what makes it "dynamic programming" rather than just recursion.

Your button/lights example is a great fit! If multiple button combinations lead to the same intermediate light state, you don't want to recount all the ways to reach the final state from there each time.

There are typically two approaches:

• Top-down (memoization): Start with the big problem, recursively break it down, cache results as you go
• Bottom-up (tabulation): Start with the smallest subproblems, build up systematically (like your backward counting!)

Which approach feels more natural to you? Or does it depend on the problem?

Exactly right! You've got it.

Memoization = cache function results (usually with a dict: input → output) Tabulation = fill in a table/array systematically from base cases up

And yes, those two approaches cover most DP problems. The choice often depends on:

• Whether the problem naturally suggests "work forwards" (tabulation) or "work backwards" (memoization)
• Whether you need all subproblems or just some (memoization can skip unnecessary ones)

Your path counting used tabulation (counts dict, working backwards through nodes). Your lights problem would likely use tabulation too (state → count).

One more pattern to watch for: the state space. What information do you need to track? In paths, it was "which node am I at?" In lights, it might be "what's the current light pattern?" Identifying the right state is often the trickiest part of DP.

Does that framework feel useful? Want to try applying it to a specific problem?

Ah yes, that's a classic DP challenge - state space explosion! When your state has too many dimensions or each dimension is too large, you can't feasibly store everything.

Common ways to handle this:

1. Find a smaller state representation - maybe you don't need to track everything you thought
2. Use symmetry or invariants - can you represent multiple states as equivalent?
3. Iterate through time/steps - only keep the previous step's states, not all history
4. Hash states - use a dict instead of a full array, only storing states you actually visit

For your lights problem, which of these might apply? What were the dimensions you were tracking, and did you find a way to reduce them?