# JenPop and Genetic Algorithms on the loose

Hey there! I’ve committed a new class on JenPop, and it’s called “Individual”. It was made to represent the rules of typical genetic algorithms:

I’ve read a bit more on the subject and figured that genetic algorithms, although being able to achieve good solutions with little computing, are only suitable for problems where brute-forcing isn’t feasible. For instance, they do a nice job on combinatorial problems such as the classic knapsack problem, where you have a list of items with value v and weight w, and you must figure what’s the maximum value you can carry in a knapsack of capacity wc by combining different amounts of those items.

And that problem was chosen to serve as the first example. At this kind of problem, a naive approach can easily become unfeasible as the input list grows. With an evolutionary approach there’s a bigger probability that it will result in a better solution, given the same time limits. And even if the solution isn’t better, it will be a decent solution. The more time you let it roll, bigger is the chance of getting a great result.

For the classic scenario with a knapsack with wc = 15, and a (v,w) item list of (4, 12), (2, 2), (2, 1), (1, 1), (10, 4), I was able to achieve the following results:

30 iterations: Best result was { 0, 0, 4, 0, 2 } (v 28, w 12)
100 iterations: Best result was { 0, 3, 0, 9, 0 } (v 15, w 15)
500 iterations: Best result was { 0, 7, 1, 0, 0 } (v 16, w 15)
1000 iterations: Best result was { 0, 3, 4, 4, 0 } (v 18, w 14)
3000 iterations: Best result was { 0, 1, 1, 5, 1 } (v 19, w 12)
5000 iterations: Best result was { 0, 0, 5, 2, 2 } (v 32, w 15)

You can see that while none of those runs delivered the optimal solution, which would be { 0, 0, 3, 0, 3 } (v 36, w 15), they got close enough to provide good solutions. Also, it becomes clear that randomism takes a nice part in this, giving us the chance of achieving a great solution with only 30 iterations. These results also show that the more you let it run, bigger is the chance of getting better results.

Anyway, see you! 😀

# Code Jam 2008 Round 1A, Problem C: Numbers

Hello again! This one is one hell of a tricky problem. No wonder it’s called “Numbers”:

In this problem, you have to find the last three digits before the decimal point for the number (3 + √5)n.

For example, when n = 5, (3 + √5)5 = 3935.73982… The answer is 935.

For n = 2, (3 + √5)2 = 27.4164079… The answer is 027.

Wow, that can’t be hard, eh? Indeed! It’s really just an expression, isn’t it? Of course it is! But once you try, you get to see something beautiful happening:

```\$ python
>>> import math
>>> 3+math.sqrt(5)
5.23606797749979
>>> (3+math.sqrt(5)) ** 18
8751751364607.995
>>> (3+math.sqrt(5)) ** 19
45824765067264.016
>>> (3+math.sqrt(5)) ** 20
239941584945152.1
```

If I was clear enough, you can see where I’m going. If not, the answers for n = 18, 19 and 20 should be 607, 263 and 151. But, it gets worse as the exponent grows:

```>>> (3+math.sqrt(5)) ** 30
3.7167066517539914e+21
>>> (3+math.sqrt(5)) ** 40
5.757196418599164e+28
>>> (3+math.sqrt(5)) ** 50
8.917924847980422e+35
```

And those results should be 647, 551 and 247, respectively. As you can now clearly see, python can’t naturally handle those big floats, what a bummer! Well, the first thing to handle is the √5 value, since 2.23606797749979 does not give enough sampling to work with the needed precision. Looking a bit, I’ve found that python has a nice Decimal module that can deal with big numbers. So, let’s use it:

```>>> from decimal import *
>>> Decimal(5).sqrt()
Decimal('2.236067977499789696409173669')
```

Hmm… not as much as I was expecting. Digging a little deeper, there’s a way to increase the precision, so things might get better:

```>>> from decimal import *
>>> getcontext().prec = 256
>>> Decimal(5).sqrt()
Decimal('2.236067977499789696409173668731276235440618359611525724270897245410520925637804899414414408378782274969508176150773783504253267724447073863586360121533452708866778173191879165811276645322639856580535761350417533785003423392414064442086432539097252592627229')
```

Much better indeed! Let’s test the previous cases that way!

```>>> (Decimal(3) + Decimal(5).sqrt()) ** Decimal(18)
Decimal('8751751364607.992147917157027451941369199107402191688475662016738729585793189576921000417819488022521008035769458236805623355117964654519012995149203145651482621970904733675616603829159073699152262190237686134096053140643178581342535160175294280596644277408')
>>> (Decimal(3) + Decimal(5).sqrt()) ** Decimal(19)
Decimal('45824765067263.99400154247292878012837438073861701581979206241047287979476762412992360313819778117994498296523948378426713929520188369997784294269662668000818303924272345525577601783001726960434941938734208353272624615875651599971618368131896320524727780674')
>>> (Decimal(3) + Decimal(5).sqrt()) ** Decimal(20)
Decimal('239941584945151.9954175862094628730047694880020933281648497263958823604254329864718576171579087349895858656483590697583803423507394435817910056755829474974431677475727217968321896916634673228294874675631017566599732643899663816729269614472126021090970897308')
>>> (Decimal(3) + Decimal(5).sqrt()) ** Decimal(30)
Decimal('3716706651753989275647.999689800241818180639346264792151927705378585007996556972979408800792479361060761238505878803342888805078901553987993893381697765526226152915257781651542284398399012963211473811148141385735168384084704178517042635758936907558116236798')
>>> (Decimal(3) + Decimal(5).sqrt()) ** Decimal(40)
Decimal('57571964185991590695825047551.99997900148385229515959986036977505867324065830938072692611318337057286921280055552228787014811837140299777965403896929514221392361243371590607709492852574169582705605198217499438837728596606637811005617061688138760057438182845')
>>> (Decimal(3) + Decimal(5).sqrt()) ** Decimal(50)
Decimal('891792484798041273405092327480885247.9999985785363502863489704179083057190576105613126358868072802757567654991706904766380400266097425885364586125135675324533868405035416842283484903727056986268545269819514444772213000354546446009165968589958054059389753136')
```

Whoa, now that’s improvement! Since the small input ranges from 2 to 30, we’re covered, yay! To the large input! The first case is 910062006… much bigger! But we have faith in our solution!

```>>> (Decimal(3) + Decimal(5).sqrt()) ** Decimal(910062006)
Decimal('8.919448541446674973277872746171245933275138630248130907008894180215290388336023868239234075453030387703632363584380715738200630546564329205017138690712399728915139680036686573669248915790961561461272596637583821147468503082573913305207639264660422490421272E+654339383')
```

Aaaaaand it fails miserably. You say increase the precision? Go!

```>>> getcontext().prec = 1024
>>> (Decimal(3) + Decimal(5).sqrt()) ** Decimal(910062006)
Decimal('8.919448541446674973277872746171245933275138630248130907008894180215290388336023868239234075453030387703632363584380715738200630546564329205017138690712399728915139680036686573669248915790961561461272596637583821147468503082573913305207639264660422298653727910480913903028526252731111441440780088518055499910461897770689934875196042744680956819167861668586410285893670912335938531514972687567521690574174307478919726903450127130115796805326307689705722233029462717164687160362204672599178459426195515468743692514680427999432359495315725269307474956180961814633487158821252498531057196075527072557513060268997873150760403767129373225601570326857102899047033394123502152602095967110573983281744795256688834356089540911527223279951661861220253786982201657061127666645472375405207936633238083128248926879097462312848916323276895759197636621186062828933585137465868233553517455227951169710940268113754834340838971865855889844812381822071748466341030321954673834742153256457611174551211610695251648386140804719763671673855871318598E+654339383')
>>> getcontext().prec = 1000000000
>>> (Decimal(3) + Decimal(5).sqrt()) ** Decimal(910062006)
```

It is still computing that last command… maybe it’ll finish before Christmas! 😀 So… nothing we can do about it. Not in a natural way at least. It turns out that Code Jam organizers knew about this (that’s the whole point, indeed) and my approach, even if it’s able to solve the small input, is not the correct one. From the contest analysis:

The key in solving the problem is a mathematical concept called conjugation. In our problem, we simply note that (3 – √5) is a nice conjugate for (3 + √5). Let us define

(1)     α := 3 + √5,   β := 3 – √5,   and Xn := αn + βn.

We first note that Xn is an integer. This can be proved by using the binomial expansion. If you write everything down you’ll notice that the irrational terms of the sums cancel each other out.

(2)

Another observation is that βn < 1, so Xn is actually the first integer greater than αn. Thus we may just focus on computing the last three digits of X.

Whoa, there’s a ton of math symbols there. What it is trying to explain is that there’s another way around, and we can write a simple algorithm to calculate only what we need. Taking that math apart, this suggested approach is quite interesting:

Solution C. [the periodicity of 3 digits]

For this problem, we have another approach based on the recurrence (7). Notice that we only need to focus on the last 3 digits of Xn, which only depends on the last 3 digits of the previous two terms. The numbers eventually become periodic as soon as we have (Xi, Xi+1) and (Xj, Xj+1) with the same last 3 digits, where i < j. It is clear that we will enter a cycle no later than 106 steps. In fact, for this problem, you can write some code and find out that the cycle has the size 100 and starts at the 3rd element in the sequence. So to solve the problem we can just brute force the results for the first 103 numbers and if n is bigger than 103 return the result computed for the number (n – 3) % 100 + 3.

Hmm… so that means that 103 restarts the cycle?

```>>> getcontext().prec = 1024
>>> str((Decimal(3) + Decimal(5).sqrt()) ** Decimal(3))
'143.5541752799932702850935573994008395340997875075688231766687118531366696204097567812612610681210327990242616368247610721361045671823063636347635238890704866837369015421401333059608526503244754105771443632133610811201095485572500621467658412511120829640713240415845568781018115768290851039517558845033513247514268365293660766111722141944646319577243016028721322545978616614879992775936489738916811506453866403999620857004610560179974673677933515377954018029702474909866234538867118832726784429359876544843489516123361404699859563411214642284786110581426213620665031024800019372553994881602971764760169511058165201173143835473121683945817118282695117541643928648522370457633421861064234721885791403592429823579671023036790457862064776462707851047085181514532031422824259553334558658255398529105918381922193776593783203964365329603283979620022727823878012615898529089431909156945831424556056508947493929464193144955007377329721849300605964711369201940159834308432696751549585124838734830577613026312919174891129200931587878779'
>>> str((Decimal(3) + Decimal(5).sqrt()) ** Decimal(103))
'114167750723954075118506909152243184891636055812752354180398291472449798143.9999999999990991848908257663970808178511998207346190405521007125450976083864218743136022578058573724307353453054670910157005516822203809922099339999229730921853127935894210481666862007195197029531387123303773842565764868785959347563566969625431650825652899007545905552208934231424236695732218626707231897771500030743392334437529104775115629703470534218197873009424685334262220812905849999596760216536020750239094398005209138812883762761775627540601688798157222970582315547073108170909501918275307282209081948400177828835529550843580664303363677674984334542177398232468766365174239306558484619915875621147584977655598495728316487075831729616962367129848658475956289730636910101894486413362652617806007673400475350422005999251984236600004538944250397180445073308202368906957039259759659759171424708223454172257556499138474886346739987815229848744135302653447948550170944394443810710713254841099059181243617896686076589736820366175425243225873342031095'
>>> str((Decimal(3) + Decimal(5).sqrt()) ** Decimal(4))
'751.6594202199646689967411763468544075540238844147363216775107372289675155071512231016216206076354221948773735933299956287145489777071084090825085004176200550896187330962356998562944764142034959055300079068701456758805751299255628262705206665683384355613744512183189236100345107783526967957467183936425944549449908917791719022086541245209393177780525834150786943366387737228119962073666571129313260408882798620998009499274205440944867036809150955734258594655937993276797731329052373871815618254139351860428319959647647374674262707908876871995127080552487621508491412880200101705908473128415601764990889933055367306159005136233888840715539870984149367093630625404742444902575464770587232289900404868860256573793272870943149903775840076429216217997197202951293164969827362655006432955840842277806071505091517327117361820812917980417240893005119321075359566233467277719517523073965614978919296671974343129687014011013788730981039708828181314734688310185839130119271657945635321905403357860532468388142825668178428304890836363588'
>>> str((Decimal(3) + Decimal(5).sqrt()) ** Decimal(104))
'597790103628874365020709731601266597564608080186061135989016966589686218751.9999999999993118384917497799741823573220669410860125985943311312283719331038685561426361821548229163694601416604360587022213664498323168042300721157774333881227779113388111518769679330555901306111571209984513448987672187738925735173356845673100783866660482880393505457702913561140292994651458434392515606071335149084501581993111115430324514143366418268949755088875955878162140123915031739484539675175716477662141640774955323553027076342238845088259277403081453735076960587935422087441774083950451065986208934485511159638170169172967026214058721899896878428807562496126156550849295831738165499903388396019937912975536800221981964584521002823218293972856046263138381804404409768650023404607326059948575617153491987096854714709481244777860640664620489694413768067522712403081007060606347733537888589375255218262079999570621589681646821848243321335219275048192774559691932753194312696801857829138795089010920052048921614192197989676705158844531382420174'
```

Yay! To the git-pushed solution!

```from decimal import *                            # Improved math

getcontext().prec = 1024                         # Adjust the precision
base = (Decimal(3) + Decimal(5).sqrt())          # Calculate the base

for case in range(1,int(raw_input())+1):         # For all test cases
exp = (int(raw_input()) - 3) % 100 + 3         # Get the recurrent exponent

num = str(base ** Decimal(exp))                # Calculate
ans = num[:num.find('.')]                      # Get only the integer part

print "Case #%d: %03d" % (case, int(ans[-3:])) # Zero-lead the last 3 digits
```

One thing that’s really interesting about this is that I actually solved this (of course, not from scratch) while posting! I was already OK with the idea of only solving the small input, but reading the analysis brought some new light into this problem. Blogging FTW! See ya!

# Floating point precision, bad numbers

Hello there! While still trying to find a good way to solve the 2008 Round-1A-C problem, Numbers, I came across something that I still have to adventure a bit on: Floating Point number precision. I mean, is there a way (even at expense of time, CPU and memory usage) to fully reproduce what calculations “by hand” can offer? It might sound silly, but beside Code Jam challenges, as a programmer it has never bothered me. But since we need precise results to solve those kind of problems, I’ve found myself in need of tools, because the ones I got are clearly not OK (or I don’t know how to use them correctly).

So, I’ll try to spend some time this week to try to make an expensive but precise way to multiply floating point numbers, so that I can solve the problems without the need to use some external calculator (I’ve done before, it’s at my Code Jam github repository). Wish me luck! 🙂

# Code Jam 2008 Round 1A, Problem A: Minimum Scalar Product

Hey there, since things are getting a little more calm, I can focus once more to solve some nice Code Jam problems. And here we go, from the first official tournament round: Problem A: Minimum Scalar Product. It says:

You are given two vectors v1=(x1,x2,…,xn) and v2=(y1,y2,…,yn). The scalar product of these vectors is a single number, calculated as x1y1+x2y2+…+xnyn.

Suppose you are allowed to permute the coordinates of each vector as you wish. Choose two permutations such that the scalar product of your two new vectors is the smallest possible, and output that minimum scalar product.

There’s a catch on the text, that made me quite confused. In case you didn’t notice, it’s that damn TWO PERMUTATIONS part. Despite getting the sample results right, when the bigger test cases came, I wasn’t able to get the smallest results right! Because, guess why, I was only doing two value permutations, because to me, the text says so!

Anyway, took a look into the results and saw that participants were simply not considering this limitation and outputting the minimum possible value anyway and to hell with it. If Google thinks this would be the correct approach, who am I to judge! 😀 Just sort both arrays, invert one of them, iterate calculating the MSP and you’re done! Python code, as usual:

```for case in range(1,int(raw_input())+1): # For all test cases
size = int(raw_input())                # Since we're informed, save array size
x = map(int, raw_input().split(" "))   # Get all X parameters
y = map(int, raw_input().split(" "))   # Get all Y parameters

x.sort()                               # Sort both vectors
y.sort()                               # And sort-reverse Y
y.reverse()                            # So we can match the bigger/smaller values

msp = 0
for i in range(size):                  # For all the array's len
msp += x[i]*y[i]                     # Sum "forward" on X and "backwards" on Y

print "Case #%d: %d" % (case, msp)     # Print the sum
```

And that’s it! Quick and easy, despite the text hiccup.  Githubbin’ it. See you!

# Difficult CodeJam problem and Free Software Documentary

Well, it’s been a tough weekend. Many new things were installed at home, lots of drilling and weightlifting (oh my arms…) but feels nicer after all. 🙂 More places for the kittens to hide.

That said, I’m evolving on Africa 2010 CodeJam D problem, but as soon as I get closer to solving, another challenge appears. This time, I’m having trouble finding an heuristic to determine the right choice to make, of all the results. Perhaps I’m still getting many results and not comparing correctly to the “solved group” list. I’ll for sure find out eventually.

Also, watched InProprietario, a brazilian documentary on free software (lots of Stallman) and, beside being very crude on effects, they used some nice movie references to cleverly show some software concepts. If you can understand portuguese and would like to know more about Free Software, I highly recommend it. 🙂

# Code Jam 2010 Africa – Problem C

Whoa, that was a tough problem. Ironically it’s called Qualification Round:

You’ve just advanced from the Qualification Round of Google Code Jam Africa 2010, and you want to know how many of your fellow contestants advanced with you. To give yourself a challenge, you’ve decided only to look at how many people solved each problem.

The Qualification Round consisted of P problems; the ith problem was fully solved by Scontestants. Contestants had to solve C problems in order to advance to the next round. Your job is to figure out, using only that information, the maximum number of contestants who could have advanced.

Well, at first it seems pretty simple, but there is quite some math involved. The first case of the example input is easy to solve, but the second will most likely get you scratching your head thinking “where the #\$%* that number came from” (at least I did). 🙂 Well the secret is that you should not look at the inputs as they are, but instead, find the maximum amount of solved problem combinations there are.

I studied a bit but still could not solve it, and surrendered by checking the contest analysis. It was quite complicated to understand, until they wrote about a clever solution from the winning candidate: it consisted in lowering the possible scores to max at the division of the scores by the amount of needed problems. Just do it (don’t ask why) until there are no scores over the result of that division. That same result will be the answer. Creepy! Python code time:

```for case in range(1,int(raw_input())+1):          # For all test cases
info = map(int, raw_input().split(" "))       # Get test case information
problems = info[0]                            # Fetch number of problems
needed = info[1]                              # Fetch number of needed problems to pass
results = info[2:]                            # Fetch results for each problem

qualified = 0                                 # Initialize our results

if (problems == needed):                      # For this case, we can expect
results.sort()                            # as many contestants as the
qualified = results[len(results)-needed]  # least solved problem

else:                                         # For this case, we use a solution
out = False                               # similar to the winning one
while not out:
out = True
qualified = sum(results)/needed       # Reduce the number of possible qualifiers
for i in range(len(results)):         # And for all results
if results[i] > qualified:        # Limit their values to the current qualifiers
results[i] = qualified
out = False                   # Until they all "match" the qualifiers number

print "Case #%d: %s" % (case, str(qualified)) # Report results
```

Interestingly enough, this code is smaller than the last, and solves a more complicated problem (at least in my opinion). I don’t think I could have done a better job on this one. 🙂 That’s the beauty in learning!

# CodeJam Quickie: 2010 Africa – Problem B

Howdy! This time, problems started getting more and more complex, and that’s good, because I like the challenge! The next problem is Get to Work:

You work for a company that has E employees working in town T. There are N towns in the area where the employees live. You want to ensure that everyone will be able to make it to work. Some of the employees are drivers and can drive P passengers. A capacity of `P == 1` indicates that the driver can only transport themselves to work. You want to ensure that everyone will be able to make it to work and you would like to minimize the number of cars on the road.

You want to calculate the number of cars on the road, with these requirements:

• Every employee can get to town T.
• The only way an employee may travel between towns is in a car belonging to an employee.
• Employees can only take rides from other employees that live in the same town.
• The minimum number of cars is used.

Find whether it is possible for everyone to make it to work, and if it is, how many cars will end up driving to the office.

Whoa, that’s way more complex than the other problems, but still, not that hard! What we need to do is: check who are drivers and passengers for each town (since they can only help same town buddies), if the total capacity of cars from a town is inferior to the number of passengers, the problem cannot be solved and so, reported as “IMPOSSIBLE”. If we can carry everyone, start filling the cars from the “minivan” (can carry more passengers) to the “mini cooper”. 😀 That way we can make sure there are fewer cars on the road.

Some rough, but commented, and validated (!) python code follows:

```for case in range(1,int(raw_input())+1):                  # For all test cases
town_len, office = map(int, raw_input().split(" "))   # Get amount of towns and office

carriers = [[] for x in xrange(town_len)]             # Create new lists for every town
capacities = town_len * [0]                           # Initialize "capacity of transportation"
payload = town_len * [0]                              # Initialize "cargo" for each town

for i in range(int(raw_input())):                     # Get and go through employee list
town, capacity = map(int, raw_input().split(" ")) # Get the current employee info
if town == office:                                # If employee is from office town
continue                                      #   No need to commute, ignore

town -= 1                                         # Adjust "town" to zero-based index
payload[town] += 1                                # Employee needs to commute, count

if capacity != 0:                                 # If employee has a car
capacities[town] += capacity                  # Add to our capacities for the town
carriers[town].append(capacity)               # Append its car capacity

resultStr = ""                                        # Create our result string

for i in range(town_len):                             # For all towns
if capacities[i] < payload[i]:                    # If we can't carry everyone
resultStr = "IMPOSSIBLE "                     #   declare it impossible, give up
break;

carriers[i].sort()                                # Sort and reverse the car list
carriers[i].reverse()
for car in carriers[i]:                           # So we can start from the bigger ones
break                                     #  we're done for this town

payload[i] -= car                             # Else, fill a car, report as being used
cars += 1

resultStr += str(cars)+" "                        # Add this town filled vehicles

print "Case #%d: %s" % (case, resultStr[:-1])         # Report results, remove last space
```

As the last problem, the same can be told about this code. There are many ways to improve its performance, but this way it can already solve the large inputs in an instant. Bitbangers apart, it solves the problem and everybody gets happy. Also, I think it is quite clear to understand, even more with the comments. 🙂 Bye!

# CodeJam Quickie: 2010 Africa – Problem A

Hey there! So, another day, another problem! This time, it is called Odd Man Out, and it is quite simple to “get”:

You are hosting a party with `G` guests and notice that there is an odd number of guests! When planning the party you deliberately invited only couples and gave each couple a unique number `C` on their invitation. You would like to single out whoever came alone by asking all of the guests for their invitation numbers.

Well, just get the number that’s not duplicated! There are many ways to do this, be it counting the occurrences of every member of the list and finding the “odd man”, or the way I did, which is basically some push and pop into a new list. Python code follows:

```for case in range(1,int(raw_input())+1):        # For all test cases
raw_input()                                 # Discard the guest list size
guests = raw_input().split(" ")             # Get the list of guest IDs

guestList = []                              # Create an empty guest list

for guest in guests:                        # For all guests
if guestList.count(guest) > 0:          # If it's already on the final list
guestList.remove(guest)             #    if is a couple, remove
else:                                   # If it's not in the list yet

print "Case #%d: %s" % (case, guestList[0]) # Reports results
```

Needless to say, I bet there are many ways to improve this algorithm, but for this problem it performs decently, so no worries (unless you’re a hardcore bit banger). 🙂 See you tomorrow!

# CodeJam Quickie: 2010 Africa – Qualification C

Whoa! Keeping up with the Africa 2010 contest, enter the Qualification C problem: T9 Spelling! It was a very fun problem and says:

The Latin alphabet contains 26 characters and telephones only have ten digits on the keypad. We would like to make it easier to write a message to your friend using a sequence of keypresses to indicate the desired characters. The letters are mapped onto the digits as shown below. To insert the character `B` for instance, the program would press `22`. In order to insert two characters in sequence from the same key, the user must pause before pressing the key a second time. The space character `' '` should be printed to indicate a pause. For example, `2 2` indicates `AA` whereas `22` indicates `B`.

Reminds me of those days of texting through that old indestructible Nokias! Anyway, it’s not hard to do, but requires some data disposition. I chose to create a map to lists containing keys and the amount of keystrokes for every letter (including space, which would be 0), in order to simply match letters and print the sequence right along. Also, to know when to pause, I would check the last printed character before, so I can print a space and then the sequence.

Python code followsas usual:

```# Build a map with the number to write and its repetition count
t9 = {
"a" : ["2", 1], "b" : ["2", 2], "c": ["2", 3],
"d" : ["3", 1], "e" : ["3", 2], "f": ["3", 3],
"g" : ["4", 1], "h" : ["4", 2], "i": ["4", 3],
"j" : ["5", 1], "k" : ["5", 2], "l": ["5", 3],
"m" : ["6", 1], "n" : ["6", 2], "o": ["6", 3],
"p" : ["7", 1], "q" : ["7", 2], "r": ["7", 3], "s": ["7", 4],
"t" : ["8", 1], "u" : ["8", 2], "v": ["8", 3],
"w" : ["9", 1], "x" : ["9", 2], "y": ["9", 3], "z": ["9", 4],
" " : ["0", 1],
}

for case in range(1,int(raw_input())+1): # For all test cases
message = raw_input()                # Get the list of words

t9str = attr[0]*attr[1]              # Initialize t9str with first T9 "word"

for i in range(1,len(message)):      # From the second character on
attr = t9[message[i]]            # Get the map values
if t9str[-1] == attr[0]:         # If it's the same number as the last
t9str += " "                 # Add a space

t9str = t9str+attr[0]*attr[1]    # Append the new T9 "word"

print "Case #%d: %s" % (case, t9str) # Reports results
```

It was the hardest of the Qualification problems, but still very easy to develop. 🙂 Fun! See ya tomorrow!

# CodeJam Real Quickie: 2010 Africa – Qualification B

Hello there! Well, I can’t say anything about this problem other than it’s just super easy. Allow me to show you:

Given a list of space separated words, reverse the order of the words.

If that first sentence didn’t kill it for you, I don’t know what will. Anyway, the python code (validated, of course):

```for case in range(1,int(raw_input())+1):           # For all test cases
words = raw_input().split(" ")                 # Get the list of words
words.reverse()                                # Reverse word list
print "Case #%d: %s" % (case, " ".join(words)) # Reports results, add spaces
```

Can’t get easier than that. And I actually had to learn that reverse() does not return the reversed list… in a perfect world, it could be solved within the print line, with:

```    " ".join(raw_input().split(" ").reverse())
```

But, maybe next time… 😀 See ya, warned you it was real quick!