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Just Enough: Difference between revisions
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Text replacement - "Slide Delay" to "slide delay"
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Begin by moving the first block R 2U 5R D, then the second R 2U 6R 2D onto the [[ice]], and lock it into place with the other block. Collect a third block, move it R 2U 5R D as well, and lock a second block into the ice the same way. When Chip slides at the right time, he can pick up two of the blocks without having to slide all the way back to the start. | Begin by moving the first block R 2U 5R D, then the second R 2U 6R 2D onto the [[ice]], and lock it into place with the other block. Collect a third block, move it R 2U 5R D as well, and lock a second block into the ice the same way. When Chip slides at the right time, he can pick up two of the blocks without having to slide all the way back to the start. | ||
When cloning block 4, use a [[spring step]]: [1/2] RL, which alters the [[ | When cloning block 4, use a [[spring step]]: [1/2] RL, which alters the [[slide delay]] slightly, and also move ''3D'' 2R to avoid releasing the [[teeth]], which would land on the [[force floor]] and contribute to the slide delay negatively. Chip must wait a total of [4 1/2] for the sliding blocks to line up properly, so this [2] loss can be spared. Carry the block out onto the ice strip underneath and wait the remaining [2 1/2] before following it; the clock should tick down to 471 on the second ice space. | ||
Continue sliding R, which will pick up the first block, then jump DR back to the original space, and R to collect the second sliding block. Go back to the start to collect the fourth block, then move it onto the suction boots as normal. In addition to two [1/2] waits already exorcised from the route, a full [1] is saved when Chip can follow the block onto the boot with absolutely zero [[block echo]], compared to the 345.4 route which lets the teeth fully escape. (The [1/2] spent waiting for the block would keep Chip from being able to use a [[spring slide]], which will cost another [1/2].) | Continue sliding R, which will pick up the first block, then jump DR back to the original space, and R to collect the second sliding block. Go back to the start to collect the fourth block, then move it onto the suction boots as normal. In addition to two [1/2] waits already exorcised from the route, a full [1] is saved when Chip can follow the block onto the boot with absolutely zero [[block echo]], compared to the 345.4 route which lets the teeth fully escape. (The [1/2] spent waiting for the block would keep Chip from being able to use a [[spring slide]], which will cost another [1/2].) | ||