Detailed mathematical approach

Problem Summary

For each carrying capacity \(C\), let \(M(C,R)\) be the minimum number of security cards needed to get through \(R\) rooms. The final task is

$\sum_{C=3}^{40} M(C,30).$

The key is that the rooms form a one-dimensional transport problem: to solve room \(r\), we must first move enough cards one room deeper to solve the remaining \(r-1\) rooms.

Mathematical Approach

1) Base case.

With one room, we need one card to enter it and one more card to leave it, so

$M(C,1)=2.$

The code stores the shifted helper value

$F_C(r)=M(C,r)-1,$

so the base becomes

$F_C(1)=1.$

2) What must be transported one room inward.

Suppose we already know \(M(C,r-1)\). Then, to clear \(r\) rooms, we must first arrange that many cards inside the next room, because once we are there for the final time, the rest of the problem is exactly an \((r-1)\)-room instance.

So the transport target is

$M(C,r-1)=F_C(r-1)+1.$

3) Why one shuttle trip contributes only \(C-2\) net cards.

A shuttle trip means: move forward one room carrying up to \(C\) cards, leave some cards there, and come back to fetch more. A non-final round trip consumes

1) one card to go forward through the door,

2) one card to return through the same door.

Therefore a round trip can deposit at most

$C-2$

cards net in the deeper room.

4) Why the last trip is special.

On the last trip we do not come back, so that trip can deliver one extra card. If we make \(t\) forward trips in total, then the maximum number of cards accumulated one room deeper is

$ (t-1)(C-2)+(C-1)=t(C-2)+1. $

We need this to be at least \(M(C,r-1)\), hence

$t(C-2)+1\ge M(C,r-1).$

So the minimum number of forward trips is

$t=\left\lceil\frac{M(C,r-1)-1}{C-2}\right\rceil=\left\lceil\frac{F_C(r-1)}{C-2}\right\rceil.$

5) Total card cost of those trips.

Each of the first \(t-1\) trips is a full out-and-back shuttle, so it costs 2 door usages. The last trip is one-way, so it costs 1 door usage. Therefore transporting the needed stock one room deeper adds

$2(t-1)+1=2t-1$

cards to the requirement.

This gives the recurrence

$M(C,r)=M(C,r-1)+2t-1,$

with

$t=\left\lceil\frac{M(C,r-1)-1}{C-2}\right\rceil.$

6) Helper-form recurrence used in the code.

Since \(F_C(r)=M(C,r)-1\), the recurrence becomes

$F_C(1)=1,$

$t_r=\left\lceil\frac{F_C(r-1)}{C-2}\right\rceil,$

$F_C(r)=F_C(r-1)+2t_r-1,$

and finally

$M(C,r)=F_C(r)+1.$

The implementation writes the ceiling in the equivalent integer form

$t_r=\left\lceil\frac{F_C(r-1)+C-3}{C-2}\right\rceil,$

which avoids fractions.

7) Worked checkpoints.

The code verifies

$M(3,6)=123,\qquad M(4,6)=23.$

Hence

$M(3,6)+M(4,6)=146,$

and another checkpoint is

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

Algorithm

1) For a fixed \(C\), initialize \(F_C(1)=1\).

2) For \(r=2,3,\dots,R\), compute \(t_r\) and update \(F_C(r)\).

3) Recover \(M(C,R)=F_C(R)+1\).

4) Sum over all capacities \(C\) in the required range.

Complexity Analysis

For each \(C\), the recurrence has length \(R\), so the total cost is

$O\bigl((C_{\max}-C_{\min}+1)R\bigr)$

with only

$O(1)$

extra memory.

Checks And Final Result

The checkpoints are

$M(3,6)=123,\qquad M(4,6)=23,\qquad \sum_{C=3}^{4}M(C,6)=146,$

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

The final answer is

$\sum_{C=3}^{40}M(C,30)=34315549139516.$

Further Reading

1. Problem page: https://projecteuler.net/problem=327
2. Recurrence relations: https://en.wikipedia.org/wiki/Recurrence_relation
3. Integer transport intuition: https://en.wikipedia.org/wiki/Integer_programming