Computer Science Canada World Puzzle Championship Question |
| Author: | funzone36 [ Wed Nov 26, 2008 4:04 pm ] |
| Post subject: | World Puzzle Championship Question |
Can anyone help me solve this puzzle from WPC? This has a unique solution. The following is a list of 14 names of German musical groups. All 26 letters of the alphabet have a different integer value from 1 to 26. Next to each word the sum of its letters is given. Punctuation signs separate words, as well as spaces. Find the value of all letters. AND ONE 49, 41 APOPTYGMA BERZERK 134, 81 BEBORN BETON 67, 77 CAMOUFLAGE 113 DEPECHE MODE 82, 52 DE/VISION 27, 65 EDGE OF DAWN 52, 10, 57 FROZEN PLASMA 68, 96 HAUJOBB 61 IRIS 36 LOWE 45 MESH 60 MIND.IN.A.BOX 61, 27, 10, 39 SEABOUND 111 I think one could potentially set this puzzle up as a system of 25 equations with 26 variables. Any help is appreciated. |
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| Author: | cavetroll [ Wed Nov 26, 2008 5:26 pm ] |
| Post subject: | Re: World Puzzle Championship Question |
Not sure what I do for the whole question, but I know where you can start. If you look at the second last one, there is the word "A", therefore you know that A=10, this is a good start. Then I would look at "IRIS", since it has too "I"'s and letters can't have negative values, you know that I <=16 ((36-1-2)-1)/2 (since 1 and 2 are the smallest possible values for "R" and "S",minus one because it must be an integer, divided by two because there are two "I"'s, ). Since I<=16, N must be >=11(because of the second last one). Now we can look at the first one with the word "AND", since A=10 and D<=26, we can further narrow down N. 10+11=21, 49-21>26. Therefore N must be >=13, and thus I must be <=14. Basically I would just continue along like this until, I have narrowed down each number to a smaller and smaller range, working with small words and words with repeated letters first. I believe this method would lead you to the right answer eventually. |
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| Author: | jbking [ Wed Nov 26, 2008 5:35 pm ] |
| Post subject: | RE:World Puzzle Championship Question |
Well, you start with A = 10 since there is that one line. W = 52-49 = 3 (From AND = 49 and that being 3 of the 4 letters in DAWN.) I'd imagine there are ways to continue from there, e.g. note MODE and ONE share some letters giving the following line: MD = 34 (Taking the in out of mind) MD-N = 11 N = 23 49 = 33+D => D = 16 E = 11 M = 22 (From MD = 34) 22+O+16+11 = 52 O = 3 F = 7 (Using OF) Do you see how this keeps going? |
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| Author: | md [ Wed Nov 26, 2008 6:06 pm ] |
| Post subject: | RE:World Puzzle Championship Question |
26 variables, 25 equations. Make a big friggin huge matrix and reduce. Easy as pie |
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| Author: | funzone36 [ Thu Nov 27, 2008 3:27 pm ] |
| Post subject: | Re: World Puzzle Championship Question |
Ok, here's what I have: A=10 W=8 L=19 N=23 I=4 D=16 E=11 M=18 O=7 F=3 G=14 But I'm still stuck with 14 other equations. |
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| Author: | jbking [ Thu Nov 27, 2008 8:56 pm ] | ||
| Post subject: | Re: World Puzzle Championship Question | ||
Notice that if you use what you have that the BEBORN simplifies to this: B+B+R = 26 That means B < 13 as R is at least 1, really at least 2 as it has to be even. However, BETON after substituting reduces to: B+T = 36 That means B >=10 as T is at most 26. Now, given that you have 10,11 already accounted for this means B = 12 Which leads to this set of next values(sorry about the alignment not quite working out):
Which leaves us with the following: Q and Y Using what we have for APOPTYGMA, we have 10+13+7+13+24+Y+14+18+10 = 134 Y = 134 - 109 = 25 Q, by process of elimination then is the missing value of 17. |
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| Author: | funzone36 [ Fri Nov 28, 2008 4:08 pm ] |
| Post subject: | RE:World Puzzle Championship Question |
I think J = 9 and H = 5 since MESH=60 18+11+26+5=60 Thank you for helping. |
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| Author: | jbking [ Sun Nov 30, 2008 12:47 am ] |
| Post subject: | RE:World Puzzle Championship Question |
Yes, I did get them backwards, sorry. Minor brain fart there. |
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