Project Euler Problem 327

Solution

Answer: 34315549139516

Let $M(C,R)$ be the minimum number of cards needed to pass through $R$ rooms when we may carry at most $C$ cards.

The key is to derive a recurrence for moving cards through one door.

Step 1: Transporting cards through one door

Suppose we need $x$ cards available in the next room.

A round trip through a door:

start with $C$ cards,
use 1 card to enter,
leave as many as possible in the next room,
use 1 card to return.

This deposits at most $C-2$ cards into the next room.

A final one-way trip:

start with $y+1$ cards,
use one to enter,
end in the next room holding $y$ cards.

So if

$$x=t(C-2)+y,\qquad 1\le y\le C-1,$$

then:

$t$ round trips deposit $t(C-2)$,
one final trip deposits $y$.

Total cards required in the current room:

$$g_C(x)=tC+(y+1).$$

We choose the decomposition minimizing this quantity.

Step 2: Recursive structure

To cross $R$ rooms, there are $R+1$ doors (including the exit door).

Let $f_k$ be the minimum cards needed before $k$ remaining doors.

Base case:

$$f_0=0.$$

Recurrence:

$$f_{k+1}=g_C(f_k).$$

Then:

$$M(C,R)=f_{R+1}.$$

This reproduces the examples:

$M(3,6)=123$,
$M(4,6)=23$,
$\sum_{C=3}^{10}M(C,10)=10382$.

Python implementation

def transport_cost(C, x):
    """
    Minimum cards needed in the current room
    to end up in the next room with x cards.
    """
    # One trip is enough
    if x <= C - 1:
        return x + 1

    best = float("inf")

    # Try all possible sizes of the final trip
    for y in range(1, C):
        remaining = x - y

        # Must be exactly covered by round trips
        if remaining % (C - 2) == 0:
            t = remaining // (C - 2)
            cost = t * C + (y + 1)
            best = min(best, cost)

    return best

def M(C, R):
    """
    Minimum cards needed to traverse R rooms.
    """
    f = 0  # no doors remaining

    # R rooms => R+1 doors (including exit)
    for _ in range(R + 1):
        f = transport_cost(C, f)

    return f

# Compute the required sum
answer = sum(M(C, 30) for C in range(3, 41))

print(answer)

Code walkthrough

transport_cost(C, x) computes the minimum cards required to move $x$ cards through a single door.
If $x \le C-1$, one trip suffices, so cost is x + 1.
Otherwise, try every possible final-trip size $y\in[1,C-1]$.
The remaining cards must come from full round trips depositing exactly $C-2$ cards each.
M(C, R) repeatedly applies this transport rule.
There are R + 1 doors total.
Start from f = 0 (nothing needed after the final exit).

Verification against examples

The code gives:

M(3,6) = 123 ✓
M(4,6) = 23 ✓
sum(M(C,10) for C in range(3,11)) = 10382 ✓

So the recurrence is consistent with all stated checks.

The computed value is:

Answer: 34315549139516