more part 2 changes, still unfinished

day10/input

@@ -0,0 +1,98 @@
+35
+111
+135
+32
+150
+5
+106
+154
+41
+7
+27
+117
+109
+63
+64
+21
+138
+98
+40
+71
+144
+13
+66
+48
+12
+55
+119
+103
+54
+78
+65
+112
+39
+128
+53
+140
+77
+34
+28
+81
+151
+125
+85
+124
+2
+99
+131
+59
+60
+6
+94
+33
+42
+93
+14
+141
+92
+38
+104
+9
+29
+100
+52
+19
+147
+49
+74
+70
+84
+113
+120
+91
+97
+17
+45
+139
+90
+116
+149
+129
+87
+69
+20
+24
+148
+18
+58
+123
+76
+118
+130
+132
+75
+110
+105
+1
+8
+86

day10/main.go

@@ -0,0 +1,47 @@
+package main
+
+import (
+    "fmt"
+    "sort"
+    "math"
+    "strconv"
+    "strings"
+    "io/ioutil"
+)
+
+func main() {
+    d, _ := ioutil.ReadFile("input")
+    strings := strings.Split(string(d),"\n")
+    ints := []int{0} // start with the connector
+    for _, s := range strings {
+        i, err := strconv.Atoi(s)
+        if err == nil {
+            ints = append(ints, i)
+        }
+    }
+    sort.Ints(ints)
+    counts := make(map[int]int)
+    ints = append(ints, ints[len(ints)-1]+3) // add the phone
+    // Part 1: Find the number of 1-gaps and 3-gaps
+    prev := 0
+    var gaps []int
+    for _, c := range ints[1:] {
+        d := c - prev
+        prev = c
+        counts[d] += 1
+        gaps = append(gaps, d)
+    }
+    fmt.Println("Part 1:", counts[1] * counts[3])
+    // Part 2: Find number of skippable adapters
+    skippable := 0
+    for i, _ := range ints {
+        if !(i == 0 || i+1 == len(ints)) { // The first and last cannot be removed
+            if ints[i+1] - ints[i-1] <= 3 {
+                skippable += 1
+            }
+        }
+    }
+
+    fmt.Println(gaps)
+    fmt.Printf("Part 2: %f\n", math.Pow(2, float64(skippable))-1) // All possible combinations for n, according to wikipedia
+}

day10/notes

@@ -0,0 +1,9 @@
+if the one before and the one after an adapter can connect, that adapter can be dropped
+find all these possible drops
+number of possible combinations of these drops
+https://en.wikipedia.org/wiki/Binomial_coefficient maybe?
+https://en.wikipedia.org/wiki/Combination "Number of k-combinations for all k"
+(2^n)-1
+Applying this blindly doesn't work because some combinations are invalid (multiple adapters next to each other cannot all be removed, because that leads to greater than 3 overall). I don't want to have to check every possible combination, that would be slow.
+Some adapters can only be skipped in the case others are not. Could I check for these and count them as one?
+3s cannot be changed. Runs of 1s can be changed.