![]() 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. The amount of starts required in mobile versions to unlock this box is 30, but on desktop and browser versions, it is 20.If $C$ is an integer, no cuts are required. 1.4): Changed location of a star at the top right, removed a rope and an automatic rope. 1.4): Changed the locations of the stars, bubble and Om Nom. 1.4): Changed the locations of the stars and devices. 1.4): Changed the locations of the stars and Om Nom, and removed one automatic rope. 1.4): A star is relocated and an automatic rope is added. 1.4): Stars, air cushions and Om Nom are relocated. 1.4): Ropes and a row of spikes behind the air cushion are relocated. Exclusive to the background itself are pockets in the upper right and lower left corners that may indicate the box is made of the same fabric as clothes. Walkthroughs Main article: Fabric Box/Walkthroughs Level Designīoth the background and the box itself is made of a green fabric that has a darker green polka dot pattern. ![]() It is located at the bottom of the level, left to Om Nom, right to the third star. Level 2-17 introduces timed stars that exist for a set period of time. The only way to get rid of the spider is to cut the rope before he reaches the candy. Level 2-9 introduces the well-known spider that climbs up or down the rope trying to get the candy. These spinning spikes are similar to normal spikes but moving. Level 2-8 is the first appearance of spinning spikes forcing the player to time their cuts and swings, bringing in a new level of precision. The faster you tap the air cushion, the more air that comes out. It introduces air cushions that the player must use to maneuver the candy over obstacles, get stars, etc. Level 2-1 is the first level of the Fabric Box the player sees. ![]()
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