slightly more idiomatic tcl?

lmul is a more idiomatic name than product.
using join+eval is probably slower than foreach but the lists are small
so i don't care.

day06/input

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+Time:        48     87     69     81
+Distance:   255   1288   1117   1623

day06/sample1.in

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+Time:      7  15   30
+Distance:  9  40  200

day06/sol.py

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+from math import sqrt
+sample = {
+    "time":     [7, 15,  30],
+    "distance": [9, 40, 200],
+}
+
+input = {
+    "time":     [   48,     87,     69,     81,],
+    "distance": [  255,   1288,   1117,   1623,],
+}
+
+
+def solve(data):
+    time = data["time"]
+    dist = data["distance"]
+
+    part1 = 1
+    for t, best in zip(time, dist):
+        ways = 0
+        for i in range(t):
+            v = i
+            d = (t-i)*v
+            if d > best:
+                ways += 1
+
+        part1 *= ways
+
+    return part1
+
+
+def fancy_solve(data):
+    time = data["time"]
+    dist = data["distance"]
+
+    for t, best in zip(time, dist):
+        # 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
+        #   (t-i)*i > best
+        # a little rearranging gets us the quadratic equation
+        #   i^2 - ti + best <= 0
+        # which we can determine the crossing points for
+        # using the quadratic formula (the good one, not the
+        # one you learned in school)
+        h = t/2
+        s = sqrt(h*h - best)
+        #print(h-s, 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
+        a = int(h-s)
+        b = int(h+s)
+        if (t-a)*a <= best:
+            a += 1
+        if (t-b)*b > best:
+            b += 1
+
+        ways = b - a
+        return ways
+
+
+print(solve(sample))
+print(solve(input))
+
+def part2(data):
+    return {"time":     [int("".join(str(x) for x in data["time"]))],
+            "distance": [int("".join(str(x) for x in data["distance"]))]}
+
+print(solve(part2(sample)))
+print(fancy_solve(part2(sample)))
+print(fancy_solve(part2(input)))
+                
+

day06/sol.tcl

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+#!/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 lmul {list} { return [expr [join $list *]] }
+proc smush {list} { return [join $list ""] }
+
+puts "Part 1: [lmul [lmap t $time d $dist {solve $t $d}]]"
+puts "Part 2: [solve [smush $time] [smush $dist]]"