Problem 327: Rooms of Doom

Detailed mathematical approach

Problem Summary

A row of \(R\) rooms is linked by automatic doors: one into room \(1\), one between each pair of neighbours, and one out of room \(R\). Every passage needs a security card, which the door uses up. A machine in the starting room dispenses unlimited cards, but scanners in every room — including the starting one — lock every door forever if we hold more than \(C\) cards or leave one on the floor; only each room's box stores cards safely. With \(M(C,R)\) the least number of cards we must take from the machine to get through \(R\) rooms and out, the task is to evaluate

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

The examples in the statement fix the reading of these rules: three rooms can be cleared with six cards while carrying at most three; \(M(3,6)=123\) and \(M(4,6)=23\), so \(\sum M(C,6)=146\) for \(3\le C\le4\); and \(\sum M(C,10)=10382\) for \(3\le C\le10\). ^[1]

The whole solution in one sentence: an optimal route ferries into the next room just enough cards to finish the remaining \(r-1\) rooms, and since a round trip through a door burns two cards while the last one-way trip burns one, the minimum cost satisfies a two-line recurrence.

Mathematical Approach

Cards are only ever destroyed by doors, so counting door crossings instead of cards turns the puzzle into a one-dimensional transport problem, where a single inequality per door pins the answer down.

1. Rooms, doors and what a journey is

Number the starting room \(0\) and the others \(1,\dots,R\); let door \(j\) join rooms \(j-1\) and \(j\) for \(1\le j\le R\), and let door \(R+1\) lead out of room \(R\). There are \(R+1\) doors; the statement confirms the count, since walking straight into three rooms uses three cards and leaves us trapped.

At every instant a card is in the machine, in our hand, or in a box; nothing may be dropped. Doors are the only consumers and each crossing destroys one card. In a journey with \(T\) crossings, delete every card that never opens a door, together with its dispensing, storing and collecting: each crossing still has its card, and holding fewer cards cannot break the capacity rule, so the trimmed journey is legal and dispenses exactly \(T\). Hence \(M(C,R)\) is the least number of crossings a journey can get away with.

2. Why three rooms cost six cards

Take \(C=3\) and \(R=3\). Walking straight through fails: four doors need four cards and we may never hold more than three. The statement's walk spends one of its first three cards to enter room \(1\), leaves one in the box there and spends the third to return; three fresh cards then reopen door \(1\), the stored card restores a hand of three, and doors \(2\), \(3\) and \(4\) follow. Six cards, six crossings: door \(1\) three times, the others once.

3. Each door is crossed an odd number of times

Fix a door \(j\) and let \(u_j\) be its number of forward crossings and \(d_j\) its number of backward ones. The journey starts on the near side of door \(j\) and ends on its far side, and the sides alternate, so

$$u_j=d_j+1,\qquad t_j=u_j+d_j=2u_j-1,$$

where \(t_j\) is the total for door \(j\). Every door is crossed an odd number of times, and the cost is \(\sum_{j=1}^{R+1}t_j\). In the walk above \(u_1=2\) and \(u_2=u_3=u_4=1\), giving \(3+1+1+1=6\).

4. What the far side of a door can afford

Take any successful journey, stopped when it first leaves room \(R\). For \(1\le j\le R+1\) let \(T_j\) count its crossings of doors \(j,j+1,\dots,R+1\), so that \(T_{R+1}=1\) and \(T_1\) is its total. For \(1\le j\le R\) call rooms \(j,j+1,\dots,R\) the far side of door \(j\). Everything rests on one inequality.

Claim. For \(1\le j\le R\),

$$T_{j+1}\le u_j(C-2)+1.$$

Proof. A card is spent where the person using the door stands. Any crossing of a door beyond \(j\) is made from room \(j\) or deeper, and so is a backward crossing of door \(j\); only its forward crossings are paid from room \(j-1\). The far side must therefore supply \(T_{j+1}+d_j\) cards. But cards appear only in the machine, on the near side, so each reaches the far side carried forward through door \(j\), and just before such a crossing we hold at most \(C\) cards, one of which the door destroys. Each of the \(u_j\) forward crossings thus delivers at most \(C-1\) cards, and cards carried back out only lower the supply, so \(T_{j+1}+d_j\le u_j(C-1)\); with \(d_j=u_j-1\) this is the claim. ∎

The claim is tight in Section 2, where \(T_2=3\) and \(u_1(C-2)+1=3\). Note where \(C-2\) comes from: a round trip spends one card on the door inward and one outward, so only \(C-2\) of the \(C\) cards we hold are really delivered. The last, one-way trip needs no return card, and that is the \(+1\).

5. The recurrence, as a lower bound

Since \(u_j\) is an integer and \(u_j\ge1\), the claim rearranges to ^[2]

$$u_j\ge\max\left(1,\left\lceil\frac{T_{j+1}-1}{C-2}\right\rceil\right).$$

Combining it with \(T_j=T_{j+1}+2u_j-1\) from Section 3 gives \(T_j\ge g(T_{j+1})\), where

$$g(x)=x+2\max\left(1,\left\lceil\frac{x-1}{C-2}\right\rceil\right)-1.$$

Both terms of \(g\) are non-decreasing, so \(g\) is, and that lets the inequalities be chained: \(T_{R+1}=1\) gives \(T_R\ge g(1)=2\), then \(T_{R-1}\ge g(T_R)\ge g(2)=3\), and so on down to \(T_1\). Writing \(m_k\) for \(g\) applied \(k\) times to \(1\), every journey has \(T_1\ge m_R\), so \(M(C,R)\ge m_R\). For now \(m_k\) is only a lower bound on the supply a room needs when \(k\) rooms and the last door remain: one card opens that door, so \(m_0=1\); one gets us into a single room and one out, so \(m_1=2\). From there \(m_k\ge2\), so for \(C\ge3\) the ceiling is at least \(1\) and the \(\max\) never matters again.

6. The shuttle strategy attains the bound

That bound is attained, by induction on \(r\) starting at \(r=0\), where the single card of \(m_0\) opens the last door. Suppose \(m_r\) cards are available in the room we stand in — in its box, to be collected a handful at a time, or straight from the machine when that room is room \(0\) — and suppose inductively that \(m_{r-1}\) cards in the next room suffice for the rest. Put \(t=\max(1,\lceil (m_{r-1}-1)/(C-2)\rceil)\) and make \(t\) forward trips through the next door:

• on each of the first \(t-1\) trips, carry \(C\) cards, spend one on the door, store \(C-2\) in the box of the next room, and keep one card to come back;
• on the last trip, carry the shortfall \(s=m_{r-1}-(t-1)(C-2)\) plus one card for the door, cross, and do not return.

For \(t=1\) the definition of \(t\) gives \(s=m_{r-1}\le C-1\); for \(t\ge2\) the ceiling gives \((t-1)(C-2)\lt m_{r-1}-1\) and \((t-1)(C-2)\ge m_{r-1}-1-(C-2)\), so again \(1\le s\le C-1\). The last load \(s+1\) is therefore legal, and

$$m_r=m_{r-1}+2t-1=(t-1)C+s+1,$$

so the stock pays for the \(t-1\) full loads and the last one exactly, leaves nothing behind, and deposits precisely \(m_{r-1}\) cards next door. With Section 5 this proves

$$M(C,r)=M(C,r-1)+2t_r-1,\qquad M(C,1)=2,$$

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

7. The form the programs use

The subtraction inside the ceiling disappears if we track the shifted value \(F_C(r)=M(C,r)-1\). Then \(F_C(1)=1\) and

$$\begin{aligned}t_r&=\left\lceil\frac{F_C(r-1)}{C-2}\right\rceil,\\ F_C(r)&=F_C(r-1)+2t_r-1.\end{aligned}$$

Only integers occur, thanks to the identity \(\lceil n/m\rceil=\lfloor(n+m-1)/m\rfloor\) for \(m\ge1\): ^[2] the ceiling is \(F_C(r-1)+C-3\) divided by \(C-2\) and rounded down, which is what integer division does.

8. A worked example: \(M(4,6)=23\)

With \(C=4\) the divisor is \(C-2=2\). Starting from \(F_4(1)=1\), each step halves the previous value upwards and adds twice that, minus one:

• \(r=2\): \(t=\lceil 1/2\rceil=1\), \(F=1+1=2\), so \(M(4,2)=3\);
• \(r=3\): \(t=\lceil 2/2\rceil=1\), \(F=2+1=3\), so \(M(4,3)=4\);
• \(r=4\): \(t=\lceil 3/2\rceil=2\), \(F=3+3=6\), so \(M(4,4)=7\);
• \(r=5\): \(t=\lceil 6/2\rceil=3\), \(F=6+5=11\), so \(M(4,5)=12\);
• \(r=6\): \(t=\lceil 11/2\rceil=6\), \(F=11+11=22\), so \(M(4,6)=23\).

For \(C=3\) the divisor is \(1\), so \(t_r=F_3(r-1)\) exactly and \(F_3(r)=3F_3(r-1)-1\). From \(F_3(1)=1\) this gives \(1,2,5,14,41,122\) and hence \(M(3,r)=2,3,6,15,42,123\) for \(r=1,\dots,6\). The third entry is the six-card walk of Section 2 and the sixth is the \(123\) of the statement; adding \(M(4,6)=23\) reproduces the checkpoint \(146\).

9. How fast the cost grows

The recurrence \(F_3(r)=3F_3(r-1)-1\) is first-order linear, solved by shifting to its fixed point \(x=3x-1\), that is \(x=\tfrac12\): the quantity \(F_3(r)-\tfrac12\) triples at every step. ^[3] Since \(F_3(1)-\tfrac12=\tfrac12\), this gives \(F_3(r)=(3^{r-1}+1)/2\), OEIS A007051, ^[4] and therefore

$$M(3,r)=\frac{3^{r-1}+3}{2}.$$

For a general fixed capacity, put \(a=C/(C-2)\) and write the ceiling as \(F/(C-2)+\theta\) with \(0\le\theta\lt1\). The recurrence becomes \(F_C(r)=aF_C(r-1)+2\theta-1\), so one step is multiplication by \(a\) up to an error in \([-1,1)\). Each step raises \(F_C\) by at least \(1\), so \(F_C(r)\) tends to infinity and successive ratios tend to \(a\): the cost is geometric with ratio \(C/(C-2)\) — near \(1\) for large \(C\), but \(3\) when \(C=3\). A larger capacity only shrinks the ceilings in \(g\), so \(M(C,R)\) is non-increasing in \(C\) and \(M(3,R)\) is the largest term. The opposite extreme is just as easy: with \(C\ge R+1\) we hold one card per door and walk straight out, so \(M(C,R)=R+1\), which settles \(C=31,\dots,40\) at once — each contributes \(31\), for a total of \(310\).

10. The sum that is asked for

Nothing more is needed: for each of the \(38\) capacities run the recurrence of Section 7 for \(29\) steps to reach \(r=30\), add \(1\), and total the results. All of this is the discrete cousin of the jeep problem, where a vehicle crosses a desert by caching fuel in dumps and coming back for more. ^[5] The boxes are the dumps, and the two cards of a round trip are the fuel it wastes.

How the Code Works

The C++ program ^[6] transcribes Section 7. M(c, rooms) keeps the shifted value in one variable f, initialised to \(1\) for one room, and loops r from \(2\) to rooms. Each iteration sets trips to (f + c - 3) / (c - 2), the integer-division form of \(\lceil F/(C-2)\rceil\), then f = f + 2 * trips - 1; it returns f + 1. For \(C=3\) the divisor is \(1\) and the added constant \(0\), so no special case is needed, and since \(F\ge1\), trips is never \(0\) — hence no counterpart of the \(\max(1,\cdot)\) of Section 5. solve adds M over the capacities and main prints the total.

The rest is scaffolding. Options holds the defaults \(3\), \(40\) and \(30\); parse_arguments, with parse_int_after_prefix, overrides them from --c-min=, --c-max= and --rooms=, accepts --skip-checkpoints, rejects unknown options and empty or non-digit values, and demands \(c_{\min}\ge3\) — which keeps \(C-2\) positive — plus \(c_{\max}\ge c_{\min}\) and at least one room. One gap: the digits go into an int unchecked, so an argument above \(2^{31}-1\) overflows before those tests run, as a sanitiser build confirms; the defaults are unaffected. run_checkpoints then verifies the four given values, exiting with code \(2\) on failure. Arithmetic uses u128 for unsigned __int128, ^[7] which the standard library cannot print, so to_string_u128 builds the decimal digits by repeated division.

Python and Java carry the same two functions. Python's solve takes c_min, c_max and rooms as optional arguments, returns the total as a decimal string printed under the module's entry guard, and uses // for floor division on unbounded integers. ^[8] Java's zero-argument solve() fixes the three parameters internally and is printed by main; its M returns a long, so shifted value, trip count and running sum are all signed 64-bit. Neither file parses a command line or carries checkpoints, so those checks come from C++ alone.

Complexity and Verification

For one capacity the recurrence performs \(R-1\) updates, each an integer division plus a fixed number of additions, subtractions and multiplications: over \(K\) capacities that is \(O(KR)\) operations and, in fixed-width arithmetic, \(O(1)\) memory. Here it is \(38\cdot29=1102\) updates, plus \(92\) inside the checkpoints, measured at about \(7\) microseconds, so the run time of all three programs is process start-up. Python keeps the same few variables, but its integers grow to \(O(R)\) bits, making its cost \(O(R^2)\) bit operations per capacity as \(R\) rises.

The size of the numbers is the only real limit. By Section 9 every term is at most \(M(3,30)\), itself below \(3^{29}\), so the sum stays under \(38\cdot3^{29}\), below \(2.7\cdot10^{15}\) and far inside the signed 64-bit range \(-2^{63}\) to \(2^{63}-1\). ^[8] Java's long is therefore correct here, and we measured the margin: with \(C=3\) the exact values stay inside a signed 64-bit integer up to \(R=41\) and inside u128 up to \(R=82\), beyond which those two programs would wrap silently, as neither uses checked arithmetic. Python is unaffected.

The claims above were checked by computation. Two independent shortest-path searches over the true state space — position, hand and the contents of every box, with free moves for dispensing, storing and collecting and one unit of cost per crossing — ran on \(28\) pairs \((C,R)\) with \(3\le C\le12\), among them \(M(3,5)=42\), \(M(4,6)=23\) and \(M(7,7)=10\). A journey of at most \(B\) crossings needs at most \(B\) cards, so states holding more may be dropped, and a search with \(B\) the predicted value settles both bounds: it returned that value every time, and with \(B\) one smaller found no journey at all, confirming Sections 4 and 5 independently of the proof.

The strategy of Section 6 was also simulated literally, checking after each action that the hand holds at most \(C\) cards and that every unused card is in the hand or a box, and at the exit that hand and boxes are empty; its dispensed count equals \(M(C,R)\) for \(3\le C\le25\), \(1\le R\le14\). Also checked over finite ranges: the ceiling identity of Section 7, the shortfall bounds of Section 6, the closed form and the error bound of Section 9, and that \(M\) is non-increasing in \(C\) and equals \(R+1\) exactly when \(C\ge R+1\) (the last for all \(3\le C\le79\), \(R\le59\)). The three programs produce byte-identical output at the required parameters and the C++ checkpoints pass.

Footnotes and References

The references below supply the background facts used above. The per-door inequality of Section 4, the monotone chaining that turns it into a recurrence and the matching shuttle strategy are derived here.

1. Project Euler 327 — Rooms of Doom. The official statement describes doors that use up their cards, scanners that lock everything if too many cards are held or one is dropped, and a storage box in every room; C is the maximum carried, R the number of rooms, M(C, R) the minimum taken from the dispenser. Its values: six cards for three rooms, M(3, 6) = 123, M(4, 6) = 23 with sum 146, and 10382 for the sum over 3 ≤ C ≤ 10 at R = 10.
2. Wikipedia — Floor and ceiling functions. The ceiling of x is the least integer greater than or equal to x, and the section on quotients gives, for a positive integer m, the identity ⌈n/m⌉ = ⌊(n + m − 1)/m⌋. Section 5 rounds the trip count up with the definition, Section 7 evaluates that ceiling with the identity.
3. Wikipedia — Linear difference equation. For an equation with a constant term the article defines the steady state as the value that reproduces itself, computes it in the first-order case as b/(1 − a₁), and substitutes the deviations from it to get a homogeneous equation with a geometric solution: the fixed-point shift of Section 9.
4. OEIS A007051 — (3ⁿ + 1)/2. The entry lists the terms 1, 2, 5, 14, 41, 122, 365, 1094, 3281, 9842 and records both the closed form (3ⁿ + 1)/2 and the recurrence a(n) = 3a(n − 1) − 1 with a(0) = 1 — the shifted values of Section 9 for C = 3, with r = n + 1.
5. Wikipedia — Jeep problem. Also called the desert crossing problem, and traced to a ninth-century collection in which a camel carries grain: a vehicle holding a fixed amount of fuel pushes fuel into dumps, keeping just enough to return, and later trips refuel from them. Its range is a partial sum of the reciprocals of the integers when it returns every time, and of the reciprocals of the odd integers for a one-way crossing, so fuel and dumps grow exponentially with distance — the comparison drawn in Section 10.
6. C++ — Euler327.cpp. The linked immutable revision contains Options, parse_int_after_prefix, parse_arguments, M, solve, to_string_u128 and run_checkpoints, and is the implementation record for this article. The searches and the simulation above are additional checks, not tests packaged in that file.
7. GCC manual — 128-bit integers. The page states that __int128 and unsigned __int128 are a GCC extension giving 128-bit integers on targets with a wide enough integer mode. This is the u128 alias, whose values the program prints by hand.
8. Java — Long gives the range of long as −2⁶³ to 2⁶³ − 1, and Python — built-in types states that integers have unlimited precision. The bound above proves that the computation fits in the Java range with room to spare.