# Transport Networks - An Introduction

This is a new type of puzzle and I don't know yet, whether or not they are fun to solve. So what better to do with them, than throw them at the best puzzlers out there and see what happens? Feedback welcomed.

Here are the first few "Transport Networks" puzzles for "warm-up" and testing the "fun-factor" of this type of puzzles. More might be coming in future.

Each puzzle consists of a network of nodes and links. The goal of the puzzle is to move all number-tokens from the starting place via the links and nodes to the bottom collection space. The following rules apply:

• All tokens must be moved from top to bottom to solve a puzzle.

• The order in which the tokens are moved, or how many moves they make, is of no concern.

• No two number tokens can be "in the game" at the same time, i.e. at any time, only one number-token can be on the links or nodes.

• A number token can only move along the links in the direction given by the arrow.

• A number token can only move over links which are labelled by a number, which is a integer-division factor of the token's number.
( example: A token with label $12$ can only move across links labelled $1$, $2$, $3$, $4$, $6$ or $12$. )

• Whenever a token passes a link, the label of the link is incremented by +1.
( example: A token with label $12$ may pass a link labelled $6$. After it has passed, that link is now labelled $7$. )

• It is not necessary to use all links or all nodes.

### A valid solution consists of:

• A solution for each of the four networks
• The description of all token-movements from start to finish for each network
• Some feedback about the puzzle - was it fun? How was the solution found? etc.

## Network #1

Move all the grey # tokens from the yellow box into the green destination area. ## Network #2

The same, just a bit more challenging ## Network #3

A network with some twist. ## Network #4

The final network with some variance. Number tokens may "rest" in the blue rectangles while other tokens are 'in the game'. The number of tokens in the area is not limited. Final note: I know of solutions for all of the four, but these solutions are not necessarily unique. Indeed, it is more likely that there are multiple solutions per network.

• Looks interesting! Perhaps you should post an example with a solution? – Beastly Gerbil May 1 '17 at 21:11
• Must all nodes be used? – Beastly Gerbil May 1 '17 at 21:18
• @BeastlyGerbil I had the solution for the first network prepared, but then realized it is actually simple enough to find an easy solution. I think it is more rewarding for the solver to "learn" the process by himself. If I'm proven wrong, I'll post the example-answer in a couple of days. – BmyGuest May 1 '17 at 21:19
• Network 4 has an arrow marked 0. Intentional or not? I ask because division by zero is no good as far as I know. – Hakdo May 2 '17 at 2:26
• If you labeled the nodes with A, B, C, ... etc., it would be much simpler to write down a solution. – Gerhard May 4 '17 at 11:08

## (No spoilers, so proceed with caution)

There are 9 solutions for the first, around 17 for the second, and many for the other two.

## Recurring Heuristics

(o) Check the future "timeline" of path numbers: By "timeline", I mean "the numbers that the path can turn into". In the second puzzle, the two exit's maximum timeline is $2 \rightarrow3\rightarrow4\rightarrow5\rightarrow6\rightarrow7\rightarrow8\rightarrow9$. But as $26$ must go at second place, and take a new exit, the max timeline becomes $2 \rightarrow3\rightarrow4\rightarrow5\rightarrow6\rightarrow7\rightarrow8$.

(i) Check the order of the first and the last numbers.: For example, in the first puzzle, the beginning and the end is $(3,1)$. So in future, there would be $(3,1), (4,2), (5,3), (6,4)$. Now, $(5,3)$ must be for the $15$, and $(6,4)$ must be for the $12$. $(3,1)$ has now only one choice, so it must be $18$, so and $(4,2)$ must be $8$. This makes the order unique. Also, for example, in the second puzzle, $26$ must go for a exit for the first time.

(ii) Find upper bounds on link numbers: After a certain point, you can ignore any prime factors since they aren't possible to "reach" by th tokens. In #3, the max number is $\text{maximum node number}+\text{number of tokens} = 8+8 = 16$. For the second, the useful prime factors are $9 \rightarrow 1,3,9; \; \; 12 \rightarrow 1,2,3,4,6,12 \; \; 20 \rightarrow 1,2,4,5,10; \; \; 30 \rightarrow 1,2,3,5,6,10,15; ...$ but the key ones are $26 \rightarrow 1,2,13$, $105 \rightarrow 1,3,5,7,15$ and $56 \rightarrow 1,2,4,7,8,16$.

(iii) Timeline "death by missing primes": If a link ever reaches a prime number not in any tokens, it is never usable again. For example, in the second puzzle, there is a node with $12$. The max timeline of it is $12 \rightarrow 13 \rightarrow 14 \rightarrow 15 \rightarrow 16 \rightarrow 17 \rightarrow 18 \rightarrow 19$. But as there are no numbers with factor of $17$, the timeline 'dies' at $17$, and the modified timeline becomes $12 \rightarrow 13 \rightarrow 14 \rightarrow 15 \rightarrow 16$, and it can be passed over max $4$ times.

## Solutions

First puzzle: By following the heuristics, the order is unique $18, 8,15,12$. Now, There is only three choice for each: Go through Right/Central/Left path. I brute forced the $3^4$ possibilities in head(most dies off instantly, can be done under 5 minutes), so the $9$ possible solutions are $LRCL, LRCC, LRCR, RLCL, RLCC, RLCR, RRCL, RRRL, RRRR$

Second puzzle: It begins to get annoying from here. I found $17$ solutions. Firstly, after doing the prime factorization, divide into two subcases: Using 10node, or without using it. If you don't use it, then there's only one entry: $1$, and the timeline of it is $1 \rightarrow 2 \rightarrow3\rightarrow4\rightarrow5\rightarrow6\rightarrow7\rightarrow8$. After following some obvious deduction, the order how the number goes are $1 \rightarrow One \; \; from \; \; the \; \; set \{12,20,30,60\}, 2 \rightarrow 26, 3 \rightarrow 9, 4,5,6 \rightarrow One \; \; from \; \; the \; \; set \{12,20,30,60\}, 7 \rightarrow 105, 8 \rightarrow 56$. Now, there are $11$ ways to take them, and all of them produces a solution !. Doing analogous for the case of using $10$, you first see you can use it only once because there are no numbers with a divisor of $11$. Out of all the possibilites in this case, I checked $6$ to be working, giving a total of $6+11 =17$ possible solutions.

Third puzzle: Analogous to two. I was lazy and annoyed with the lack of near primes, so I didn't check all, but there are $>12$ solutions.

Fourth puzzle: There are many more than 6 solutions. Compute the timeline-image first. The bridge can be passed over only two times, because of no multiple of $7$, the line with $10$ can passed only once, and the bottom right road to exit can be passed on max three times, because of the same reason. Now subcases: Using only $1$ node entry, and using both $3$ and $1$ node entry. Using only $3$ node entry is impossible, because it can be used only $4$ times, and there are $5$ tokens. There are $2^5-2=30$ possibilities for only the entry for this, so I am very lazy for that. For the sub-case Using only $1$ node entry, see the timeline-image (Now you have no choice till the orange dot): For the order of the tokens, The number has to simultaneously be divided by $1 \rightarrow lcm(1,2), 2 \rightarrow lcm(2,3), 3 \rightarrow lcm(3,4), 4 \rightarrow lcm(4,5), 5 \rightarrow lcm(5,6)$, which gives after some deduction, two possibilities $100/20 , 6, 12, 100/20, 30$. Now notice two must go over the bridge to the left, since the right road to exit can only be passed over $3$ times. There are $4$ possibilities of who goes to the left of the bridge $100/20, 6/12$. Each possibility works, with some sub-cases (of $12$ to rest or not rest on the left side rest), giving a total of $6$ possibilities.

• Welcome to Puzzling. Great first answer. :) – Alconja May 2 '17 at 11:33
• Welcome here and thanks for the answers. You sort of overdid it by trying to find all solutions - which was not required. The frist valid would have been enough, and finding that would hopefully not be frustrating ;c) This puzzle is similar to graphical mazes - there you often have also more ways to go from A to B, but you "solve" it by just drawing one (the simplest, if possible). So street-fighting-maths is just fine. I think your "Recurring Heuristics" section is exactly the excellent reasoning I was hoping somebody put up!! +1 for that right away. – BmyGuest May 2 '17 at 12:56
• As for getting the "accepted" answer (which you deserve), could I ask you to just outline one solution path for each grid. Purpose beeing, that (non-methematical) people would like to just "see" an easy answer for comparison. Also why are you not using spoiler tags? Not that it matters too much. I would maybe just spoiler-tag the given "easy" solution and put it on top - long (and good) explainations bottom. – BmyGuest May 2 '17 at 12:58
• As for the (welcomed) suggestions: I tried building up a net with prime/near prime numbers only first, but then solutions stood out to be too obvious even without your (clever) elimination steps on start. So, in a sense, the - mutliple-possiblities - were kind of meant to add distraction & 'fun', not frustration :c) But then again, this was not meant to lead to mathematically complete solution with "all possibilities" like you did. Oh, and I do have various other ideas on these networks, but I did want to test-pilot the 'easy' ones first, to see if this tis type of puzzle has its audience. – BmyGuest May 2 '17 at 13:09

First, sorry, but I haven't done 2 and 3. I started with 1 to check the concept then jumped to 4. 2 and 3 looked tedious with too many tokens and I haven't attempted them yet. Though for 4 I took one guess and found solution so these might be reasonably fast.

Solution for 1:

18,8,15,12 is the only order because of A and H. 18 needs to go through C, G to enable 8 to get through, the rest is mostly arbitrary. There are in total 8 options I believe. 8 left (15 mid, 12 anywhere) or right (15 mid (12 anywhere) or right (12 left or right)).

Solution for 4:

You quickly see you are limited by 7 and 11. So H and F can be traveled twice and I once. L can be traveled 3x. M is going to be available to any, then 6, 12 and 30, then 12, 20, 100, then 20, 30, 100, then 6, 12, 30. One possible solution would be to have 6 going ACE. 20 goes ACEEHKM and 6 goes EHKM. 100 goes BDJLM. So does 30. And finally 12. But I assume there are tons of other solutions I haven't found.

Puzzle seems an interesting concept, but there are too many solutions IMO, leading to guesswork instead of thinking and deduction. There could be some requirement to do leaps in reasoning to quickly eliminate less probable paths to find maybe not unique solution but one of few. But these puzzles have too many of solutions and it smells like guessing instead of solving. I argue against further complications. Keep it simple in concept and hard to solve.