day 6 tcl solution

day06/sol.tcl

@@ -0,0 +1,43 @@
+#!/usr/bin/env tclsh
+package require Tcl 8.6
+set input stdin
+regexp {Time: (.*)}     [gets $input] _ time
+regexp {Distance: (.*)} [gets $input] _ dist
+
+proc solve {t d} {
+    # the distance traveled if the boat is released at time i 
+    # is equal to (t-i)*i
+    # we want to know the range of values of i for which this
+    # exceeds the best time (d)
+    #   (t-i)*i > d
+    # a little rearranging gets us the quadratic equation
+    #   i^2 - ti + d <= 0
+    # which we can determine the crossing points for
+    # using the quadratic formula (the good one, not the
+    # one you learned in school)
+    #puts "solve $t $d"
+    set h [expr {$t / 2.0}]
+    set s [expr {sqrt($h*$h - $d)}]
+    set a [expr {int($h-$s)}]
+    set b [expr {int($h+$s)}]
+    # in general these will not be at integer values,
+    # so we probe the floor and ceiling of each crossing
+    # point to determine exactly where the condition is met
+    if {($t-$a)*$a <= $d} { incr a }
+    if {($t-$b)*$b > $d}  { incr b }
+    #puts "a=$a b=$b"
+    return [expr {$b - $a}]
+}
+
+proc product {list} {
+    set p 1
+    foreach x $list {
+        set p [expr {$p * $x}]
+    }
+    return $p
+}
+
+proc smush {x} { return [join $x ""] }
+
+puts "Part 1: [product [lmap t $time d $dist {solve $t $d}]]"
+puts "Part 2: [solve [smush $time] [smush $dist]]"