# ShareMyScreen/AdventOfCode/2022/10/CathodeRayTube

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Another simulation problem. Well, I won't complain. Again I'll just write a verb for each action. First to read in the data. One adjustment is needed: J requires each verb to have an argument, so we need to change the 'noop' instruction to have one.

```   require 'strings'
]lines =. <;._2 ('noop';'noop 0') stringreplace LF ,~ wd 'clippaste'
+-------+--------+------+-------+------+-------+-------+-------+------+------+-------
|addx 15|addx -11|addx 6|addx -3|addx 5|addx -1|addx -8|addx 13|addx 4|noop 0|addx -1...
+-------+--------+------+-------+------+-------+-------+-------+------+------+-------
```

This seems so simple that I won't use a script. The whole history can be contained in the sequence of X values at the end of each cycle. We can conveniently initialize X to 1.

```X =: 1
noop =: {{ i. 0 0 [ X =: X , {: X }}   NB. X value is unchanged
addx =: {{ i. 0 0 [ X =: X , (0,y) + {: X }}   NB. X value changes at end of second cycle
".@> lines
```

The (i. 0 0) was to give each execution a non-printing result. Now I check the results, remembering that what the problem calls cycle 1 is the value recorded in 0{X.

```   (_1 + 20 + 40 * i. 6) { X
21 19 18 21 16 18
(*   X {~ <:) (20 + 40 * i. 6)
420 1140 1800 2940 2880 3960
```

For part 2 we have to see, for each clock cycle, whether a cursor is within 1 position of the value in the X register. I'll create an array of the positions, which are repeated for each line, and the X values, with 1 added to get to the middle of the sprite. Then compare.

```   ' *' {~ (6 40 \$ 1 + i. 40) (2 > |@-) (6 40 \$>:X)
**** *  * ***  *  * **** ***  *  * ****
*    * *  *  * *  * *    *  * *  *    *
***  **   *  * **** ***  *  * *  *   *
*    * *  ***  *  * *    ***  *  *  *
*    * *  * *  *  * *    *    *  * *
**** *  * *  * *  * **** *     **  ****
```