The three rules of Ruby Quiz: 1. Please do not post any solutions or spoiler discussion for this quiz until 48 hours have passed from the time on this message. 2. Support Ruby Quiz by submitting ideas as often as you can: http://www.rubyquiz.com/ 3. Enjoy! Suggestion: A [QUIZ] in the subject of emails about the problem helps everyone on Ruby Talk follow the discussion. -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= by Stephen Waits Like many homeowners, I have a residential monitored alarm system installed in my house. It consists of a few keypads and several remote sensors. My alarm has three modes of operation: 1. Off - completely disarmed 2. Away - everything (perimeter and interior) armed 3. Stay - perimeter armed The keypad consists of 12 buttons, which includes the digits '0' through '9', plus '*' and '#'. The '*' and '#' are only used for special programming. The digits '1', '2', and '3' are also labeled "Off", "Away", and "Stay". This corresponds to the three system modes mentioned above. The security code is 4 digits long. To set or change the system's mode, you simply enter your 4 digit security code, followed by the digit ('1', '2', or '3') which corresponds to the mode you want to set. For example, if the security code was 1234: 12341 - sets system to "Off" 12342 - sets system to "Away" 12343 - sets system to "Stay" What if you make a mistake when you're entering your code? Yes, this is where an interesting observation comes into play. To answer the question... if you make a mistake you just start over. In other words, the keypad seems to only care that the last 5 keypresses match a valid code+command sequence. For example, if you entered the first two digits of your code incorrectly, you could just simply start over with the correct code, without any kind of reset: 8912341 ++ => first 2 keypresses erroneous, oops! +++++ => last 5 keypresses == valid code+command sequence The thought occurs to me, and I'm sure to the reader, that perhaps this can be exploited. For example, if you entered the sequence 1234123, you are actually testing 3 different codes: 1234123 => what you entered 12341 => 1234 + 1/off .23412 => 2341 + 2/away ..34123 => 3412 + 3/stay Problems: 1. What is the shortest sequence of digits you can come up with that tests every code in this alarm system? The worst case is 5*10**4, or 50000 keypresses. 2. What if the security code was 5 digits instead of 4? Worst case here is 6*10**5, or 600000 keypresses. 3. What if the "stop digits" changed from [1, 2, 3], to just [1]? For instance, maybe the system is armed (mode 2 or 3) and only actually beeps when switching to mode 1. (See Assumptions below) 4. Can you also minimize the number of duplicate code tests? Assumptions: * We assume the keypad will actually let you go on entering digits for this long. I haven't tested it, but it seems silly that it might actually allow this. However, as a long time comp.risks reader, almost nothing surprises me. * We assume that the keypad will always signify a valid code+command sequence, regardless of mode. In reality, if you set to Away when it's already in Away, it might not emit it's "success" signal. #!/usr/bin/env ruby -w # # Stephen Waits <steve@waits.net> # class AlarmKeypad # init a keypad, with length of security code, and the code's # stop digits def initialize(code_length = 4, stop_digits = [1,2,3]) # remember the length of the security code @code_length = code_length # and which digits cause a code to be checked @stop_digits = stop_digits # and reset our data structures to 0 clear end # reset keypad to initial state def clear # an array of each code and how many times it's been entered @codes = Array.new(10**@code_length,0) # last N+1 keypad button presses @key_history = [] # total number of keypad button presses @key_presses = 0 end # press a single digit def press(digit) # add digit to key history @key_history.shift while @key_history.size > @code_length @key_history << digit @key_presses += 1 # see if we just tested a code if @stop_digits.include?(@key_history.last) and @key_history.length > @code_length @codes[@key_history[0,@code_length].join.to_i] += 1 end end # find out if every code had been tested def fully_tested? not @codes.include?(0) end # find out if an individual code has been tested # NOTE: an actual keypad obviously doesn't offer this functionality; # but, it's useful and convenient (and might save duplication) def tested?(code) @codes[code] > 0 end # output a summary def summarize tested = @codes.select { |c| c > 0 }.size tested_multiple = @codes.select { |c| c > 1 }.size puts "Search space exhausted." if fully_tested? puts "Tested #{tested} of #{@codes.size} codes " + "in #{@key_presses} keystrokes." puts "#{tested_multiple} codes were tested more than once." end end if $0 == __FILE__ # a random button presser, 3 digit codes a = AlarmKeypad.new(3) a.press( rand(10) ) until a.fully_tested? a.summarize # sequential code entry, 4 digit codes a = AlarmKeypad.new(4) ("0000".."9999").each do |c| next if a.tested?(c.to_i) c.split(//).each { |d| a.press(d.to_i) } a.press(rand(3)+1) end a.summarize # sequential code entry, 5 digit codes, only [1] as a stop digit a = AlarmKeypad.new(5,[1]) ("00000".."99999").each do |c| next if a.tested?(c.to_i) c.split(//).each { |d| a.press(d.to_i) } a.press(1) end a.summarize end

on 2006-03-24 14:56

on 2006-03-24 15:57

```
On Mar 24, 2006, at 7:56 AM, Ruby Quiz wrote:
> by Stephen Waits
Just by a neat little quirk of serendipity this quiz was launched on
the quiz creator's birthday. Happy birthday Stephen!
James Edward Gray II
```

on 2006-03-24 17:28

```
On Mar 24, 2006, at 6:57 AM, James Edward Gray II wrote:
> Happy birthday Stephen!
Thanks! I'm looking forward to those solutions...
--Steve
```

on 2006-03-24 19:24

On Mar 24, 2006, at 8:28 AM, Stephen Waits wrote: > > On Mar 24, 2006, at 6:57 AM, James Edward Gray II wrote: > >> Happy birthday Stephen! > > Thanks! I'm looking forward to those solutions... > > --Steve > Happy birthday Steve. In case you're curious, my solution is... not doing so well. I thought I'd built up a decent model by working with code lengths I could fit in my head (e.g. 1 or 2 digits), but it actually does much worse than the naive solution for 4 and 5 digit codes (it does better for 1-3 digit codes). So, back to the drawing board for me. Tim

on 2006-03-24 20:24

Timothy Bennett wrote: > > Happy birthday Steve. Thanks Tim! > In case you're curious, my solution is... not doing so well. Well, keep at it. :) --Steve

on 2006-03-27 17:49

I think that there are more key presses then you think. There are (10**4)*3 different combinations which totals 30,000. But that takes 30,000*5 key presses which is 150,000 key presses. So if the code was 5 digits, it is 1,500,000 key presses to test them all by brute force. Anyway, here is my first solution. It is unrefined, but I haven't really had time to work on it much. <code> # Shane Emmons # # Quiz 72 B&E # # This code is very ugly and inefficient, but I wanted to get # something out there to look at. Sorry I didn't have more # time to work on this quiz. all_possible = "" 10.times do |key1| 10.times do |key2| 10.times do |key3| 10.times do |key4| 3.times do |key5| current_code = Array.new current_code << key1 << key2 << key3 << key4 << key5 + 1 next if all_possible.include?( current_code.to_s ) left, left_over = Array.new( current_code ), Array.new right, right_over = Array.new( current_code ), Array.new while true do left_over << left.shift right_over << right.pop if left.length == 0 all_possible += current_code.to_s break elsif all_possible =~ /^#{left.to_s}/ all_possible = left_over.to_s + all_possible break elsif all_possible =~ /#{right.to_s}$/ all_possible += right_over.to_s break end end end end end end end print "string: ", all_possible, "\n" print "string length: ", all_possible.length, "\n" </code> Shane

on 2006-03-27 18:18

Hmm, this is a tricky one :) This solution is pretty ugly (I've not seriously tried to clean it up, either) and slow. There are actually two solutions in here, one that's fast and 'good enough' for the smaller digit counts, and one that's *very* slow but gives shorter results with less retesting of codes. Neither of these are all that much better than simply running through and checking each code against AlarmKeypad#tested? but then that's to be expected I guess, perhaps this is why real alarms don't tend to duplicate that functionality :P I ran benchmarked tests to get an idea of how it did. The tests are doing simple sequential tests (seq), sequential using tested? (seq/chk), the simple/quick(er) way (simple) and the slow, better way (best). Check out the 'best' time for 4 digit codes - there has to be room for improvement there :/ ### 2 digits, [1,2,3] stops ### user system total real seq. 0.010000 0.000000 0.010000 ( 0.005530) Search space exhausted. Tested 100 of 100 codes in 300 keystrokes. 39 codes were tested more than once. seq/chk 0.010000 0.000000 0.010000 ( 0.008893) Search space exhausted. Tested 100 of 100 codes in 234 keystrokes. 11 codes were tested more than once. simple 0.000000 0.000000 0.000000 ( 0.006074) Search space exhausted. Tested 100 of 100 codes in 243 keystrokes. 15 codes were tested more than once. best 0.050000 0.000000 0.050000 ( 0.054995) Search space exhausted. Tested 100 of 100 codes in 223 keystrokes. 2 codes were tested more than once. ### 3 digits, [1,2,3] stops ### user system total real seq. 0.070000 0.000000 0.070000 ( 0.064838) Search space exhausted. Tested 1000 of 1000 codes in 4000 keystrokes. 513 codes were tested more than once. seq/chk 0.060000 0.000000 0.060000 ( 0.062613) Search space exhausted. Tested 1000 of 1000 codes in 2944 keystrokes. 224 codes were tested more than once. simple 0.090000 0.000000 0.090000 ( 0.086938) Search space exhausted. Tested 1000 of 1000 codes in 2960 keystrokes. 223 codes were tested more than once. best 6.430000 0.050000 6.480000 ( 6.741495) Search space exhausted. Tested 1000 of 1000 codes in 2623 keystrokes. 50 codes were tested more than once. ### 4 digits, [1,2,3] stops ### user system total real seq. 0.820000 0.020000 0.840000 ( 0.871923) Search space exhausted. Tested 10000 of 10000 codes in 50000 keystrokes. 6108 codes were tested more than once. seq/chk 0.690000 0.010000 0.700000 ( 0.704795) Search space exhausted. Tested 10000 of 10000 codes in 33395 keystrokes. 2569 codes were tested more than once. simple 1.830000 0.030000 1.860000 ( 1.869937) Search space exhausted. Tested 10000 of 10000 codes in 33555 keystrokes. 2620 codes were tested more than once. best 781.350000 5.520000 786.870000 (796.666829) Search space exhausted. Tested 10000 of 10000 codes in 28614 keystrokes. 435 codes were tested more than once. ### 5 digits, [1,2,3] stops ### user system total real seq. 11.010000 0.140000 11.150000 ( 11.248501) Search space exhausted. Tested 100000 of 100000 codes in 600000 keystrokes. 69494 codes were tested more than once. seq/chk 7.450000 0.060000 7.510000 ( 7.563379) Search space exhausted. Tested 100000 of 100000 codes in 371028 keystrokes. 30902 codes were tested more than once. simple 136.960000 6.250000 143.210000 (145.916674) Search space exhausted. Tested 100000 of 100000 codes in 371394 keystrokes. 31105 codes were tested more than once. best (not tested)

on 2006-03-27 23:19

semmons99@gmail.com wrote: > I think that there are more key presses then you think. There are > (10**4)*3 different combinations which totals 30,000. But that takes > 30,000*5 key presses which is 150,000 key presses. So if the code was 5 > digits, it is 1,500,000 key presses to test them all by brute force. > Anyway, here is my first solution. It is unrefined, but I haven't > really had time to work on it much. Hi Shane, I'm not sure I follow you.. With 4 digit codes, there are 10,000 unique codes, or 10**4. Each code takes a maximum of 5 keypresses to test, therefore 5*10**4 == 50k keypresses. Regardless, thanks for your solution! --Steve

on 2006-03-27 23:21

```
Ross Bamford wrote:
> Hmm, this is a tricky one :)
Indeed.. thanks for the solutions Ross!
--Steve
```

on 2006-03-28 15:39

You are right that there are 10,000 unique codes, but each of those codes can have any of the three unique modifier keys attached to it so 10,000 * 3 = 30,000 unique combinations you must press. And, since there are 5 keys in the 4 key code + 1 modifier key, it is 30,000 * 5 = 150,000. At first I didn't realize this until I wrote the loop to generate all of the unique code/modifier key combinations. The one key to this I believe, is that you have to treat the modifier key (1,2,3) as part of the code not a seperate entity. Here is some code to prove the solution. <code> output, num_codes = '', 0 10.times do |x1| 10.times do |x2| 10.times do |x3| 10.times do |x4| 3.times do |x5| output += x1.to_s + x2.to_s + x3.to_s + x4.to_s + x5.to_s num_codes += 1 end end end end end print "number codes: ", num_codes.to_s, "\n" print "output length: ", output.length, "\n" </code>

on 2006-03-28 16:53

I think part of the problem was that it didn't matter which modifier key you hit, the system would confirm the code. So while 150,000 is right if you need to test every combo with every modifier, 50,000 is what the problem was looking for (i.e., every combo with _any_ modifier).

on 2006-03-28 18:16

On Mar 28, 2006, at 6:53 AM, Matthew Moss wrote: > I think part of the problem was that it didn't matter which modifier > key you hit, the system would confirm the code. So while 150,000 is > right if you need to test every combo with every modifier, 50,000 is > what the problem was looking for (i.e., every combo with _any_ > modifier). That's right. Sorry for the confusion. --Steve

on 2006-03-28 19:24

I understand now, here is my modified solution, it does work a bit better. <code> all_possible = "" 10.times do |key1| 10.times do |key2| 10.times do |key3| 10.times do |key4| current_code = Array.new current_code << key1 << key2 << key3 << key4 next if all_possible =~ /#{current_code.to_s}(1|2|3)/ left, left_over = Array.new( current_code ), Array.new right, right_over = Array.new( current_code ), Array.new while true do left_over << left.shift right_over.insert( 0, right.pop ) if left.length == 0 if key1 == 0 all_possible += current_code.to_s + "1" elsif key1 == 1 all_possible += current_code.to_s + "2" else all_possible += current_code.to_s + "3" end break elsif all_possible =~ /^#{left.to_s}(1|2|3)/ all_possible = left_over.to_s + all_possible break elsif all_possible =~ /#{right.to_s}$/ if key1 == 0 all_possible += right_over.to_s + "1" elsif key1 == 1 all_possible += right_over.to_s + "2" else all_possible += right_over.to_s + "3" end break end end end end end end print "string length: ", all_possible.length, "\n" </code>

on 2006-03-29 14:18

On Tue, 2006-03-28 at 01:18 +0900, Ross Bamford wrote: > Hmm, this is a tricky one :) This solution is pretty ugly (I've not > seriously tried to clean it up, either) and slow. Probably a bit late on (sorry) but this is a cleaned up version that does much the same and is marginally quicker.

on 2006-03-29 19:02

On Mar 29, 2006, at 3:44 AM, Ross Bamford wrote: > Probably a bit late on (sorry) but this is a cleaned up version that > does much the same and is marginally quicker. Thanks Ross!

on 2006-03-30 05:18

I apologize for the 11th hour solution; this week was busy. It took me a while to come up with an algorithm that would actually produce fewer keystrokes than the naive just write-out-every-code-so-long-as- it-isn't-already-tested solution in every case. After trying a few (I believe 3 is the number) poor solutions, I was getting frustrated, so I figured I'd just try to work through it backwards. This worked out rather well, I think (it beats the naive solution by over 70,000 strokes on the 5-digit, 3-stop case). It was pretty fun, too: I actually had a reason to use lambda and shift (which I've never used before), as well as inject (which I've only used once before). Of course, it takes a while to run (the 5 digit, 3 stop code case took nearly 40 minutes on a dual 2.7GHz PowerMac G5), so I'm providing the output as well as the code. So, here're my results: user system total real One stop digit: 2 digits 0.010000 0.000000 0.010000 ( 0.006755) Search space exhausted. Tested 100 of 100 codes in 271 keystrokes. 0 codes were tested more than once. ============================================================ 3 digits 0.110000 0.000000 0.110000 ( 0.132552) Search space exhausted. Tested 1000 of 1000 codes in 3439 keystrokes. 0 codes were tested more than once. ============================================================ 4 digits 9.580000 0.260000 9.840000 ( 11.005426) Search space exhausted. Tested 10000 of 10000 codes in 40951 keystrokes. 0 codes were tested more than once. ============================================================ 5 digits 1208.900000 23.850000 1232.750000 (1439.590285) Search space exhausted. Tested 100000 of 100000 codes in 468682 keystrokes. 46 codes were tested more than once. ============================================================ Three stop digits: 2 digits 0.010000 0.000000 0.010000 ( 0.005834) Search space exhausted. Tested 100 of 100 codes in 219 keystrokes. 0 codes were tested more than once. ============================================================ 3 digits 0.170000 0.000000 0.170000 ( 0.196276) Search space exhausted. Tested 1000 of 1000 codes in 2545 keystrokes. 4 codes were tested more than once. ============================================================ 4 digits 16.580000 0.300000 16.880000 ( 19.149404) Search space exhausted. Tested 10000 of 10000 codes in 28105 keystrokes. 96 codes were tested more than once. ============================================================ 5 digits 1910.240000 27.430000 1937.670000 (2240.603938) Search space exhausted. Tested 100000 of 100000 codes in 299559 keystrokes. 1517 codes were tested more than once. ============================================================ My code is attached, along with alarm_keypad.rb (which contains the AlarmKeypad class). Tim

on 2006-03-30 09:20

Hi, I wanted to discuss this problem from a more mathematical perspective. Anyone that's interested, please comment on this, I'm curious what others' thoughts are. It strikes me that there should be a mathematical way to determine the minimum number of presses necessary to cover all key codes. This would of course be based on the code length (n), and the number of stop codes (m). I also believe that attaining the minimum number of strokes requires that there be no duplicated codes, but that having that nice "0 codes were tested more than once" message doesn't guarantee that you have the best solution. For examples, I'm going to talk about the easiest case to think about: n = m = 1. That is, each code is just 1 digit in length, and there is only one stop code (we'll choose 1, but it really doesn't matter). Now, the general formula for the worst case (the brute force just-enter-every-code method, without the tested? check), is (n + 1) * 10 ** n. In this example, the worst case is 20: 01112131415161718191 It is easy to see that the best case here is 19: 0112131415161718191 Now let's switch the example to n = 1, m = 3, where the stop codes are 1, 2, and 3. This is now the best case: 01231415161718191 (length 17) A little bit more experimentation should be enough to convince anyone that the length of the best solution will be 20 - m when n = 1, and m is in the range [1,9]. The formula breaks if m = 10, so I'm just going to throw that out to keep the model simple right now. The 20 in this case is the length of the worst case scenario, so I believe that this can be generalized to worst-case-length - amount-of-overlap The worst-case-length is of course (n + 1) * 10 ** n; I don't think that needs further study. The part that I haven't been able to generalize is the "amount of overlap." For n = 1, we're good, but that's not very interesting, and the formula breaks when n = 2. Take this case: n = 2, m = 1. I worked out a solution by hand, and I don't believe it's possible to go below 271 keystrokes. I haven't been able to figure out a general formula that works for n = 1 and for that n = 2, m = 1 data point. I think it may be based on the number of occurrences of the stop codes in the code set (let's call this number k), but I'm not sure. To be clear on what I mean, for the n = 2 case, the digit 1 appears 20 times in the 00..99 range. This is the same for 2 and 3, so if m = 3, stop codes appear 60 times in the code set. In general, k = m * n * (10 ** (n - 1)) But I can't find a way to work k in, and I've only got those data points to work with. There are a couple of other potential data points, but I don't know if they're valid. For example, my solution found the n = 2, m = 3 case solution with 219 strokes and no overlap. However, I don't know for sure, yet, whether 219 is the correct minimum for that case. I think it probably is, but I don't know. To explain why I'm not sure, take this example for n = 1, m = 1: 000000112131415161718191 Running that through the keypad will produce no duplicate tests, but it's obviously not a minimal solution. So, I can't trust that the program's solution is the true minimum, even if no codes were tested more than once. Anyway, some other possible data points that my program generated (but have not been verified) are: n = 3, m = 1 => 3,439 n = 4, m = 1 => 40,951 For all other cases, my solution didn't manage to avoid duplicate testing. So, that's an outline of what I've figured out so far. If anyone has ideas about how one might predict the length of the ideal solution, speak up :) Tim

on 2006-03-30 09:39

On Thu, 2006-03-30 at 16:17 +0900, Timothy Bennett wrote: > Hi, I wanted to discuss this problem from a more mathematical > perspective. Anyone that's interested, please comment on this, I'm > curious what others' thoughts are. > (I'm not a mathematician (by any stretch of the word) so I may be utterly wrong in all of this). I think this problem is basically the shortest common superstring problem: http://www2.toki.or.id/book/AlgDesignManual/BOOK/B... Which is basically how my solution (and I think yours too) approached it. The upshot is that it's easy to find a common superstring, and even a reasonably short common superstring, but very difficult to find the _shortest_ or determine how long it would be (it's NP complete I think). That said, while working on it my limited mathematical knowledge kept taunting me that there was a much easier way given that we were dealing with numbers only, but I couldn't find it and gave up after a bit - I'd love to see something like that done.

on 2006-03-30 17:21

On Mar 29, 2006, at 11:37 PM, Ross Bamford wrote: > _shortest_ or determine how long it would be (it's NP complete I > think). I definitely need to study more math and algorithms. Yes, this does seem to be a variation on the shortest common superstring, and the sources I'm finding say that it is NP-complete. Unfortunately, the existence of the stop digits seems to complicate matters somewhat. The most thorough discussion I've found online (so far, haven't looked very long yet) is a PDF discussing, of all things, the Pokemon trading card game: http://home.earthlink.net/~mstamp1/papers/poke.pdf Nothing I've found so far discusses how one might predict the length of the shortest common superstring, though. I feel, that for our rather narrow problem domain, that it should be possible. Especially since the substrings in the more traditional superstring problems are random to some degree, while we know what all of our strings are. Hm, I'll have to think on this more. Tim