(Added infobox)  | 
				 (Added highscores and api to infobox)  | 
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| type                = Component  | | type                = Component  | ||
| prerequisite1       = Saving Bytes  | | prerequisite1       = Saving Bytes  | ||
|   | | unlocks-level1      = 3 Bit Decoder  | ||
|   | | unlocks-level2      = Little Box  | ||
|   | | unlocks-component1  = 1 Bit decoder  | ||
| scored              = Yes  | |||
| highscore           = 3  | |||
| highscore-gate      = 1  | |||
| highscore-delay     = 2  | |||
| api-enum-id         = decoder1  | |||
| api-enum-number     = 24  | |||
}}  | }}  | ||
This level asks you to make a circuit that outputs On to one of two pins depending on the input signal.  | This level asks you to make a circuit that outputs On to one of two pins depending on the input signal.  | ||
Latest revision as of 21:14, 17 November 2023
| Section | Arithmetic | 
|---|---|
| Type | Component | 
| Prerequisite | Saving Bytes | 
| Unlocks | 3 Bit Decoder  Little Box  | 
| 1 Bit decoder | |
| Scored | Yes | 
| High score | 3 | 
| API | decoder1 (24) | 
This level asks you to make a circuit that outputs On to one of two pins depending on the input signal.
Solution[edit | edit source]
As the truth table reveals, one output is the inverse of the input and the other is a copy of the input. So add a Not Gate for the first output and wire things up.
