3/18/2023 0 Comments Cut the rope fabric box level 11![]() ![]() In particular, there is at most one intermediate rope containing waste. Therefore: Each output rope is contained in exactly one intermediate rope again, this also applies to the waste rope. By our choice of algorithm, this is impossible. Then we can instead combine these two intermediate ropes into one and cut it into $a_0 b_10, a_1,\ldots, a_r, b_1,\ldots,b_s$ with the same number $r s$ of cuts, but with less intermediate ropes. That is: Each intermediate rope contains at most one contiguous piece from each output rope note that this applies also to the waste rope.Īssume two distinct intermediate ropes each contain a piece of the same output rope, say one intermediate rope is cut into pieces of lengths $a_0, a_1,\ldots, a_r$ by $r$ cuts and the other is cut into pieces of lengths $b_0,b_1,\ldots,b_s$ by $s$ cuts and that the pieces of length $a_0,b_0$ both end up in the same output rope. By the minimality of our algorithm, this does not happen. If two of the pieces an intermediate rope is cut into end up in the same output rope, we can rearrange the locations of the pieces on the uncut rope in such a manner that they are adjacent, which allows us to save a cut. To do so we consider all algorithms that achieve the minimal number of cuts, and among these consider the one that minimizes the number of intermediate ropes produced in step 1. Our task is to find an algorithm that minimizes the cuts performed in step 2. Combine the pieces into the $N$ output ropes and combine the waste pieces (if there are any) into a "waste rope". ![]() Combine unit length input ropes into some integer length intermediate ropes.It doesn't harm to allow the following more general form: Combine the pieces into the $N$ output ropes (plus possibly some waste).Cut some of the unit length input ropes into pieces.I tis clear that any such solution can be brought into this standard form: Remark: If the final result is correct, there is no need to go for $c$, one can directly say that $\lceil (1-\frac1q)N\rceil$ cuts are needed if $\frac CR=\frac pq$.Įdit: (Revisited this answer after a few years in order to add proof)Īllowed solutions consist of an algorithm that describes a sequence of cuts and (at no cost) concatenation operations. And if $c$is irrational, $N$ cuts are needed, I suppose. If $c=\frac13$, then $\lceil \frac23 N\rceil$ cuts suffice again.įor $c=\frac34$ or $c=\frac14$, I can do it with $\lceil \frac34 N\rceil$ cuts, for $\frac k 5$ with $1\le k\le 4$, I can do it with $\lceil \frac45N\rceil$ cuts.Ī pattern sems to emerge here, but I'm not sure if it is really optimal: $c=\frac pq$ requires $\lceil (1-\frac1q)N\rceil$ cuts. If $c=\frac23$, one can produce $\lceil \frac23 N\rceil$ pieces of $\frac23$ and combine the $\frac13$ rests for the remaining ropes, hence $\lceil \frac23 N\rceil$ cuts suffice. Indeed, $\lceil \frac N2 \rceil$ is enough if $c=\frac12$. Otherwise let $c=C-\lfloor C\rfloor$, a real number between $0$ and $1$.Įach rope must finally contain at least one cut end, thus the number of cuts is at least $\frac N2$ and it is easily solvable with $N$ cuts. Due to reordering of the boxes, the Gravity Changer Button had to be removed and replaced with another level.If $C$ is an integer, no cuts are required. Inside and out, the box has light blue stars spread everywhere on a darker blue background with two spotlights moving around the box slowly. Walkthroughs Main article: Magic Box/Walkthroughs Level Design The player has to time actions very carefully. Level 4-21 contains the second and last hidden drawing of the box. Level 4-4 includes two pairs of hats, much like the Blades, they are color-coded. This level also contains the first hidden drawing for the box. Level 4-3 introduces the concept that candy maintains it speed whilst teleporting. Level 4-1 introduces Magic Hats that teleport Candy from one hat to another with the same color.
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