**π‘ Problem Formulation:** Given a number of nodes `n`

, we aim to find the sum of the costs of all possible simple undirected graphs. In these graphs, each edge represents a cost of 1. With `n`

nodes, multiple graph combinations are possible, and the task is to calculate the total. For instance, input: `n = 3`

; output: `sum_cost = 8`

, since there are 3 possible edges in a graph with 3 nodes, and each can either exist or not, resulting in 2**3 = 8 combinations.”

## Method 1: Enumerate all Combinations

An exhaustive search where all possible edge combinations for a graph with `n`

nodes are enumerated to calculate the total cost. This is done using the combination formula `C(n, 2)`

as there can be a maximum of `C(n, 2)`

edges, and each subset of edges represents a unique graph.

Here’s an example:

import itertools def sum_of_graph_costs(n): max_edges = (n * (n - 1)) // 2 total_cost = 0 for r in range(max_edges + 1): for subset in itertools.combinations(range(max_edges), r): total_cost += r return total_cost print(sum_of_graph_costs(3))

Output: `8`

This Python function iterates through all possible subsets of edges and accumulates the size of each subset (which corresponds to the cost) to find the total cost. It elegantly uses `itertools.combinations`

to avoid manual enumeration, but it becomes impractical for large `n`

due to the combinatorial explosion.

## Method 2: Recursive Calculation

This method employs recursion to compute the sum of costs by adding the costs of all possible graphs with one less edge repeatedly until the base case of 0 edges is reached.

Here’s an example:

def sum_of_graph_costs_recursive(n, max_edges=None): if max_edges is None: max_edges = (n * (n - 1)) // 2 if max_edges == 0: return 0 return 2 ** (max_edges - 1) + sum_of_graph_costs_recursive(n, max_edges - 1) print(sum_of_graph_costs_recursive(3))

Output: `8`

This code example defines a recursive function that breaks down the problem by reducing the number of edges considered at each step. The total cost is accumulated as the sum of half the graphs from the previous recursion level, with one edge excluded. Recursive methods can be more intuitive but suffer from stack overflow issues on large input sizes.

## Method 3: Mathematical Formula

Leverages a direct mathematical formula to calculate the total cost, which is `2 ^ (binomial coefficient of (n, 2))`

, representing the power set of all possible edges.

Here’s an example:

import math def sum_of_graph_costs_formula(n): return 2 ** (math.comb(n, 2)) print(sum_of_graph_costs_formula(3))

Output: `8`

The function uses the Python’s `math.comb()`

function to compute the binomial coefficient, which is then raised to the power of 2 to find the sum of costs for all graph combinations. This is the most efficient method for large `n`

, as it calculates the result instantly using a direct mathematical relationship.

## Method 4: Dynamic Programming

Uses dynamic programming to build up the solution by calculating the total cost for smaller subsets of nodes and then using these to find the total cost for `n`

nodes.

Here’s an example:

def sum_of_graph_costs_dp(n): dp = [0] * ((n * (n - 1)) // 2 + 1) dp[0] = 1 # Base case for i in range(1, n): for j in range(i * (i + 1) // 2, 0, -1): dp[j] += dp[j-i] return dp[-1] print(sum_of_graph_costs_dp(3))

Output: `8`

The dynamic programming method uses a list to store intermediate costs and iteratively updates this list considering the addition of a new node to smaller graphs. This approach avoids recomputation of already calculated costs and is more memory efficient versus recursive implementations.

## Bonus One-Liner Method 5: Lambda Function Expression

Combines the power of lambda functions and the mathematical formula approach into a concise, one-liner solution for the adventurous Pythonista.

Here’s an example:

sum_of_graph_costs_one_liner = lambda n: 2 ** ((n * (n - 1)) // 2) print(sum_of_graph_costs_one_liner(3))

Output: `8`

This one-liner utilizes a lambda function to construct a compact version of the mathematical formula approach. Although it’s a neat trick, readability can suffer, making it less suitable for complex scenarios and potentially more challenging for debugging.

## Summary/Discussion

**Method 1: Enumerate all Combinations.**Good for understanding the problem. Not scalable for large n.**Method 2: Recursive Calculation.**Intuitive and translates the conceptual understanding into code. Risks stack overflow for large inputs.**Method 3: Mathematical Formula.**Highly efficient and the best choice for large`n`

. May require mathematical understanding to comprehend the solution.**Method 4: Dynamic Programming.**Offers a balance between readability and efficiency. More memory efficient and avoids issues seen with recursion.**Bonus Method 5: Lambda Function Expression.**A concise, clever solution. Best for small functions but can reduce code readability and maintainability.