As with so many of our Ruby Q. problems, this is another pathfinding challenge. We're probably pretty use to seeing that pruning of the search space is usually a big win in these cases and this problem is no exception. Let's talk a little about exactly why that is the case. When you think about the operations allowed by this quiz, one thing that becomes obvious is that n * 2 and n / 2 are opposites. If you do one and then the other, you end up right back at the number you had before you applied any operations. That kind of busy work isn't helpful. In fact, visiting any number a second time is pointless since we've already seen an equal or faster path for the same thing. There's a lot more duplication in the problem than the opposite operations too. Let's start working 2 to 9 by hand, so we can see that: 2 2, 4 (double) 2, 4, 8 (double) 2, 4, 8, 16 (double) 2, 4, 8, 4 (halve) 2, 4, 8, 10 (add two) 2, 4, 2 (halve) 2, 4, 2, 4 (double) 2, 4, 2, 1 (halve) 2, 4, 2, 4 (add two) 2, 4, 6 (add two) 2, 4, 6, 12 (double) 2, 4, 6, 3 (halve) 2, 4, 6, 8 (add two) 2, 1 (halve) 2, 1, 2 (double) 2, 4, 2, 4 (double) 2, 4, 2, 1 (halve) 2, 4, 2, 4 (add two) 2, 1, 3 (add two) 2, 4, 3, 6 (double) 2, 4, 3, 5 (add two) 2, 4 (add two) 2, 4, 8 (double) 2, 4, 8, 16 (double) 2, 4, 8, 4 (halve) 2, 4, 8, 10 (add two) 2, 4, 2 (halve) 2, 4, 2, 4 (double) 2, 4, 2, 1 (halve) 2, 4, 2, 4 (add two) 2, 4, 6 (add two) 2, 4, 6, 12 (double) 2, 4, 6, 3 (halve) 2, 4, 6, 8 (add two) There's a lot of paths already and we're not there yet. Let's look at the exact same tree, but with one simple rule of pruning applied: We can toss out any operation that results in a number we have already seen. Watch how that changes things: 2 2, 4 (double) 2, 4, 8 (double) 2, 4, 8, 16 (double) 2, 4, 8, 10 (add two) 2, 4, 6 (add two) 2, 4, 6, 12 (double) 2, 4, 6, 3 (halve) 2, 1 (halve) 2, 1, 3 (add two) 2, 4, 3, 5 (add two) Those two trees go to the same depth and both represent the same set of numbers. However, the second one is over three times less work. Imagine how much we can save as the numbers keep growing and growing. Another important optimization involves limits. Even with our simple 2 to 9 example, we can be up to 64 after only five operations (2, 4, 8, 16, 32, 64). 64 is a long way from 9 and probably not helping us get there. We can limit upper and lower bounds for the numbers, or even by limiting the steps the path can take. (Florian Pflug made a great post in the quiz thread about the latter.) The only thing to be careful of with an optimization like this is that you make sure you don't impose a limit low enough to prevent an optimal solution. Many other optimizations were used. Some, like storing instance data in Integer objects, have that questionable code smell and are probably best avoided. Other solutions, while super fast on huge inputs, did not produce the shortest path in all cases. Finally, many optimizations involve timing various elements of Ruby syntax for minor increases here and there, but that's more detail than we need to go into in this summary. Given that, let's examine a nice solution, by Tristan Allwood, using the two optimizations described above: require 'set' class MazeSolver def solve start, finish visited = Set.new tul, tll = if start > finish [(start << 1) + 4, nil] else [(finish << 1) + 4, nil] end solve_it [[start]], finish, visited, tul, tll end def solve_it lpos, target, visited, tul, tll n = [] lpos.each do |vs| v = vs.last next if tul and v > tul next if tll and v < tll return vs if v == target d = v << 1 # double h = v >> 1 unless (v & 1) == 1 # half p2 = v + 2 # plus 2 n << (vs.clone << d) if visited.add? d n << (vs.clone << h) if h and visited.add? h n << (vs.clone << p2) if visited.add? p2 end return solve_it(n, target, visited,tul, tll) end end if __FILE__ == $0 puts MazeSolver.new.solve(ARGV[0].to_i, ARGV[1].to_i).join(" ") end Tristan's solution makes use of bit operations, because they tend to be faster than multiplication and division. All you need to know about these is that n << 1 == n * 2 and n >> 1 == n / 2. The primary interface method for the code above is solve(). It takes the start and finish numbers. As you can see, it sets up a visited Set object to keep track of the numbers we've seen, assigns upper and lower limits, then hands off to solve_it(). In solve_it(), each path of numbers is walked and expanded by the three operations. Note the calls to visited.add?() before new paths are added. This is the optimization keeping us from revisiting numbers. The next if tul and v > tul line skips to the next iteration if we've passed the upper limit. That's the other big optimization. After another level of paths have been added, solve_it() just recurses to find the next set of operations. This only ever goes as deep as there are steps in the solution, so there's not much danger of overrunning the stack for problems we can reasonably solve. The final if statement of the program triggers the process from the two parameters passed to the program. My thanks to the many, many participants that generated great solutions and discussion, especially all you new guys! I also need to thank the quiz creator who was very involved in the discussion and gave me a bunch of tips for this summary. Tomorrow, we have a dice rolling challenge for all you RPG players out there...

on 2006-01-05 18:44

on 2006-01-05 18:44

> Tristan's solution makes use of bit operations, because they tend to be faster > than multiplication and division. All you need to know about these is that n << > 1 == n * 2 and n >> 1 == n / 2. Are they really faster? Ruby bits are not directly in CPU registers. My rule of thumb is that every method call in Ruby takes a huge amount of time, whether it is a bitshift or a multiplication (or even a regexp check). For the record, I did a quick test: kero@pc67140460:~/tmp$ time ruby -e '1_000_000.times { |i| i << 1 }' real 0m2.683s user 0m1.970s sys 0m0.710s kero@pc67140460:~/tmp$ time ruby -e '1_000_000.times { |i| i << 1 }' real 0m2.687s user 0m1.880s sys 0m0.810s kero@pc67140460:~/tmp$ time ruby -e '1_000_000.times { |i| i * 2 }' real 0m2.684s user 0m2.080s sys 0m0.600s kero@pc67140460:~/tmp$ time ruby -e '1_000_000.times { |i| i * 2 }' real 0m2.689s user 0m2.060s sys 0m0.640s No significant differences whatsoever. Bye, Kero. PS: fun quiz, but that should be clear with all my postings :)

on 2006-01-05 18:44

On Jan 5, 2006, at 9:37 AM, Kero wrote: >> Tristan's solution makes use of bit operations, because they tend >> to be faster >> than multiplication and division. All you need to know about >> these is that n << >> 1 == n * 2 and n >> 1 == n / 2. > > Are they really faster? Good point. Thanks for showing us the real story. James Edward G. II

on 2006-01-05 18:44

On Jan 5, 2006, at 10:37 AM, Kero wrote: > time, whether it is a bitshift or a multiplication (or even a > regexp check). > For the record, I did a quick test: As I suspected. Stupid C idioms. :) ~ ryan ~

on 2006-01-05 18:44

The shift operator comes in handy and can save a lot of time. These are equivalent operations: >time ruby -e '1_000_000.times { || 2**31 }' real 0m33.225s user 0m32.452s sys 0m0.015s >time ruby -e '1_000_000.times { || 1<<31 }' real 0m3.469s user 0m3.421s sys 0m0.000s >time ruby -e '1_000_000.times { || 2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2 }' real 0m13.839s user 0m13.656s sys 0m0.015s

on 2006-01-05 20:57

On Jan 5, 2006, at 5:37 AM, Ruby Q. wrote: > My thanks to the many, many participants that generated great > solutions and > discussion, especially all you new guys! I also need to thank the > quiz creator > who was very involved in the discussion and gave me a bunch of tips > for this > summary. I think that the purely binary pattern/bit-twiddling based solutions should at least get an honorable mention. Great quiz! --Steve

on 2006-01-05 22:54

Hey guys, I've just come back from a 2 day holiday and discover my code on rubyquiz.com.. Wow! For my first entry on ruby-quiz (and i'm fairly new to ruby to boot) i'm quite happy :) Cheers all! I've noticed a few comments regarding my use of bitshift, I didn't expect them to be faster (nor intend them to be) but I just tend to think in terms of the binary idioms when it comes to +ve odd numbers, mult/div 2. The fact that I felt the compulsion to write comments next to them indicating what I was doing should probably be an indication that I should have been using more standard notation... Anyways thanks again for making my day :) Cheers, Tristan

on 2006-01-05 23:18

Lou V. wrote: > real 0m3.469s > Kero wrote: This is an implementation detail, the ** operator of Fixnum calls rb_big_pow converting self to a Bignum regardless of the value. '<<' checks the size of the result and uses real bitshifting if possible. While this was interresting to me it's not the explanation of the effect above because the value 2**31 is always a bignum, but the implementation of the ** operator in Bignum isn't very fast (while shifting the internal representation of the bignum is faster) To prove my point that this isn't realy because bitshifts are faster try this: 1_000_000.times { || 2.0**30 } At least on my machine this is faster than any of your three examples above. (if you choose smaller values '<<' will be the fastest because it doesn't convert to bignums as stated above) cheers Simon

on 2006-01-06 01:46

On 2006-01-05, Simon Kröger <removed_email_address@domain.invalid> wrote: > Lou V. wrote: > >> The shift operator comes in handy and can save a lot of time. >> These are equivalent operations: [snip 3 examples with considerable speed differences] > > This is an implementation detail, perhaps. I'd say ** takes *more* time; not that bitshifts take *less* time. > the ** operator of Fixnum calls rb_big_pow converting self to a Bignum > regardless of the value. '<<' checks the size of the result and uses > real bitshifting if possible. worse: kero@chmeee:~$ time ruby -e '100_000.times { || 200**10 }' real 0m1.740s user 0m1.540s sys 0m0.072s kero@chmeee:~$ time ruby -e '100_000.times { || 2**10 }' real 0m2.122s user 0m1.928s sys 0m0.064s The float value ranks first with about 0.5 seconds runtime. If Fixnum#** is slower than Bignum#**, that's not a detail, that's plain weird :) Something to keep in mind... Bye, Kero.

on 2006-01-06 04:37

Then I guess I just love implementation details. ;) >ruby -e "puts (2**29).class" Fixnum >time ruby -e '1_000_000.times { 2**29 }' real 0m25.788s user 0m25.686s sys 0m0.031s >time ruby -e '1_000_000.times { 2.0**29 }' real 0m1.233s user 0m1.234s sys 0m0.000s >ruby -e "puts (2.0**29).class" Float >time ruby -e '1_000_000.times { 1<<29 }' real 0m0.559s user 0m0.577s sys 0m0.000s >ruby -e "puts (1<<29).class" Fixnum >time ruby -e '1_000_000.times { 2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2 }' real 0m5.970s user 0m5.983s sys 0m0.015s >ruby -e "puts (2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2).class" Fixnum

on 2006-01-06 15:26

Kero <removed_email_address@domain.invalid> writes: >> Tristan's solution makes use of bit operations, because they tend to be faster >> than multiplication and division. All you need to know about these is that n << >> 1 == n * 2 and n >> 1 == n / 2. > > Are they really faster? Ruby bits are not directly in CPU registers. IIRC, a bitshift on modern Intel processors is extremely inefficient because it has to throw-away all kind of assumptions. I even think n << 1 gets compiled to n+n by good compiler. (Can't test at the moment, I'm on PPC only here.) > user 0m1.880s > No significant differences whatsoever. These microbenchmarks are totally pointless; you'd need to write them in assembler since you don't know which code your compiler generated for <<. > Bye, > Kero. > > PS: fun quiz, but that should be clear with all my postings :) Absolutely, I enjoyed it, even if I didn't sent in my solution--a breadth-first search like so many others. (I used the cute trick of int.send(X, 2) for X being :/, :* or :+, which I hadn't seen yet anywhere...)