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 ▼ Olivier TREGER Member Posts: 120 Threads: 19 Joined: Nov 2006 06-21-2007, 12:19 PM Hello all, Despite my now decent collection of HP calcs, I must admit that, having left school quite early, I miss the necessary basis to figure out how to spell the following problem. my daughter Agathe was asked to solve the following kinda sudoku by her teacher: let's say a square 3x3 boxes. In each box a number (left to right, top to bottom) such as: a+b+c=96 d+e+f=315 g+h+i=12 but also a+d+g=72 b+e+h=40 c+f+i=126 PLEASE NOTE: e=5 I suspect there's something to do with matrices but I ain't sure. I've tried to use Mathematica (don't laugh...) to modelize (?) a formula with no success. I'm sure somebody around will tell me in three words where to dig. Thanks for help Olivier ▼ Monte Dalrymple Senior Member Posts: 297 Threads: 25 Joined: Nov 2006 06-21-2007, 12:40 PM There appears to be no solution. The first three equalities say a+b+c+d+e+f+g+h+i = 423 The last three equalities say a+b+c+d+e+f+g+h+i = 238 ... or is there a typo in your post? Monte Walter B Posting Freak Posts: 4,587 Threads: 105 Joined: Jul 2005 06-21-2007, 12:42 PM Bonjour Olivier, je pense que ... you need two more constraints. You wrote "kinda sudoku", so was there anything further required implicitely? ▼ Dave Shaffer (Arizona) Posting Freak Posts: 776 Threads: 25 Joined: Jun 2007 06-21-2007, 03:25 PM I agree with Walter: there are 9 unknowns, so you need nine equations or constraints to determine a unique solution. If this is a true sudoku, then each digit (1 through 9) should be used only once. That, plus the equations and constraint you gave provide only 8 of the necessary 9 conditions. Olivier TREGER Member Posts: 120 Threads: 19 Joined: Nov 2006 06-21-2007, 01:22 PM I'm completely sorry... Please replace "+" by "x" I thought I was unable to count. I can't type either... ▼ Massimo Gnerucci (Italy) Posting Freak Posts: 882 Threads: 23 Joined: Jan 2005 06-21-2007, 02:06 PM I believe the solution is: ```4 * 4 * 6 = 96 (2*2*2*2*2*3) * * * 9 * 5 * 7 = 315 (3*3*5*7) * * * 2 * 2 * 3 = 12 (2*2*3) = = = 72 40 126 (2*2*2*3*3) (2*2*2*5) (2*3*3*7) ``` Greetings,Massimo ▼ Olivier TREGER Member Posts: 120 Threads: 19 Joined: Nov 2006 06-21-2007, 02:41 PM That's one solution. Mine was: 8*4*3 9*5*7 1*2*6 But my question was: how to create a model that would solve the problem? ▼ Egan Ford Posting Freak Posts: 1,619 Threads: 147 Joined: May 2006 06-21-2007, 02:51 PM Was one of the constraints that you can only used each digit once (like sudoku)? If so I think you have the correct answer. ▼ Olivier TREGER Member Posts: 120 Threads: 19 Joined: Nov 2006 06-21-2007, 04:36 PM Quote: Was one of the constraints that you can only used each digit once (like sudoku)? If so I think you have the correct answer. Yes. I forgot to mention it (again). But still, I can't figure out what method to be used to solve it as a generic problem. ▼ Egan Ford Posting Freak Posts: 1,619 Threads: 147 Joined: May 2006 06-21-2007, 05:16 PM Quote: But still, I can't figure out what method to be used to solve it as a generic problem. I can think of a few. 1. Brute force. There are only 40320 possible outcomes. Try them all. If you have access to a quantum calculator you can test all possible outcomes simultaneously. 2. Random guesses. Could be faster or slower than #1. Both #1 and #2 can be easily parallelized across multiple calculators. #1 would need a unique domain for each calculator. #2 would just use different random seeds. 3. Search. Start with the column or row with the least number of possibilities (pairs) that meets the condition of the row/column with the most knowns (5 in this case). Push that on a stack, remove the 2 digits from a list of possibilities. E.g. start with def, it can only be 9,5,7 or 7,5,9, then beh, then abc, adh, etc... once you get to a point where none of the pairs will work, pop off the stack the last working pair, and try a different pair, move forward, etc... you may have to pop off multiple pairs. Eventually you'll end up with an answer. #1 and #3 will allow you to predict the worse case number of operations. ▼ Olivier TREGER Member Posts: 120 Threads: 19 Joined: Nov 2006 06-21-2007, 05:27 PM That doesn't look like an industrialized method :) What I don't understand is that when I push the 6 equations in Mathematica, it gives me back, it says: ```Equations may not give solutions for all "solve" variables. ``` although a solution can be found by simple guessing. Ain't there no way to build a standard algorithm to find out? ▼ Egan Ford Posting Freak Posts: 1,619 Threads: 147 Joined: May 2006 06-21-2007, 05:44 PM Can you also tell Mathematica a!=b!=c!=d!=e!=f!=g!=h!=i, a through i belongs to the set of positive integers, and e=5? ▼ Olivier TREGER Member Posts: 120 Threads: 19 Joined: Nov 2006 06-21-2007, 05:59 PM Quote: Can you also tell Mathematica a!=b!=c!=d!=e!=f!=g!=h!=i, a through i belongs to the set of positive integers, and e=5? I just tried: ```{a,b,c,d,f,g,h,i}<9 {a,b,c,d,f,g,h,i}>1``` with no success either ▼ Egan Ford Posting Freak Posts: 1,619 Threads: 147 Joined: May 2006 06-21-2007, 06:09 PM Needs to be >=, not just >. You also need to state somehow that each variable is unique. Karl Schneider Posting Freak Posts: 1,792 Threads: 62 Joined: Jan 2005 06-22-2007, 01:29 AM Quote: Brute force. There are only 40320 possible outcomes. Try them all. If you have access to a quantum calculator you can test all possible outcomes simultaneously. An arithmetical problem solved by brute force (i.e., testing all possible arrangements of input values for solutions) and by more-intelligent means was discussed in this recent thread: As Walter pointed out, the problem cannot be solved directly, because there are eight "degrees of freedom" for the nine inputs, but only six equations. Standard linear-algebra techniques can't be used to simplify the problem, because the problem isn't linear (whereas a standard sudoku puzzle is). -- KS Egan Ford Posting Freak Posts: 1,619 Threads: 147 Joined: May 2006 06-21-2007, 02:01 PM Is this what the problem looks like? ```.---.---.---. | a | b | c | = 96 |---+---+---| | d | e | f | = 315 |---+---+---| | g | h | i | = 12 '---'---'---' \ \ \ \ \ \-> = 126 \ \----> = 40 \-------> = 72 ``` ▼ Olivier TREGER Member Posts: 120 Threads: 19 Joined: Nov 2006 06-21-2007, 02:39 PM Yes it does look like this ▼ Allen Senior Member Posts: 562 Threads: 27 Joined: Oct 2006 06-21-2007, 08:39 PM Using this 48/49/50 program you can find the Divisors of each number: ``` << { } OVER SQRT 2 SWAP FOR N ' SETS UP RANGE TO RUN 2 TO SQRT(N) IF OVER N / FP 0 == THEN N+ END ' APPEND TO LIST IF DIVISIBLE NEXT DUP2 / + ' APPEND LIST ALL DIVISORS > SQRT(N) SORT DUP 10 < * ' CRUDE MASK TO HIDE ALL VALUES <=9 >> YIELDS THE POSSIBLE DIVISORS FOR EACH ROW: .---.---.---. step1 step 2 | a | b | c6| = 96 = [2 3 4 6 8] [2 8] |---+---+---| | d9| e5| f7| = 315 = [3 5 7 9] [solved] |---+---+---| | g | h | i3| = 12 = [2 3 4 6] [1 4] '---'---'---' \ \ \ \ \ \-> = 126 = [2 3 6 7 9] [3 6] \ \----> = 40 = [2 4 5 8] [1 2 4 8] \-------> = 72 = [2 3 4 6 8 9] [1 2 4 8] ``` So you can conclude that: ```Step 1 1. e must be 5 (given) 2. f must be 7 (no other common divisors except 126 and 315) 3. d must be 9 (=315/(5*7)) Step 2 4. i must be 3 (only CD with 126 and 12 (step 2)) 5. eliminate [1 2 4 8] as divisors of 126 b/c must be for 40 and 72 6. c must be 6 (=126/(7*3) 7. either a or b must be 8. Try a: g must be 1 , leaving 96=8*2*6 works Try b: g must be 4, leaving 2*8*6 ALSO WORKS. ``` Is there more than one possible solution? perhaps I am missing something Edited: 21 June 2007, 8:47 p.m. ▼ Dave Shaffer (Arizona) Posting Freak Posts: 776 Threads: 25 Joined: Jun 2007 06-21-2007, 11:09 PM "4. i must be 3 (only CD with 126 and 12 (step 2))" No - a 6 works, too (to give Olivier's solution)! (I had already prepared the following): Actually, if we read between the lines, there ARE other constraints. If this is a sudoku of the normal type, perhaps the most compelling constraint is that the values must be (positive) integers. And, as noted by others, they must lie between 1 and 9 and in fact include ALL the integers from 1 to 9. So, let’s try some educated guess work. First, note that abc=96, def=315, and cfi=126 tell us that none of a, b, c, d, e, f, i can be a 1. (If one of the variables in a triple product was one, the maximum product would be 8x9=72.) Thus, we know already that either g or h must be 1. ghi=12 tells us that the other two (the pair g and i or the pair h and i) must be either the pair 2 and 6 or the pair 3 and 4. (Turns out, both work.) Now, since beh = 40 and e=5, we know that bh = 8. If b and h are positive integers (between 1 and 9!), then b and h must be the pair 1 and 8 or the pair 2 and 4. Similarly, def = 315 leads to the conclusion that df = 63 and thus d and f are the pair 7 and 9. Since adg = 72, d must be a 9 (and f a 7), because if d was a 7, it would not divide evenly into 72. We now have for sure: e = 5 (given), d = 9, and f = 7. Also, from adg = 72, and d = 9, we get ag = 8. We already ascertained that g must be 1, or 2 or 6, or 3 or 4. Since all numbers must be integers, ag = 8 rules out g = 6 or 3, so g must be 1 or 2 or 4. Note, too, that cfi = 126 gives ci = 18 (because f = 7). So, c and i are either the pair 2 and 9 or the pair 3 and 6. Because d = 9, c and i must be the pair 3 and 6. At this point, try b = 1 or 2 or 4 or 8, using abc = 96. Look for inconsistencies with the known values. Do the same for other possible pairs. When you are done, I think you will find that there are at least two solutions consistent with all the triple product conditions: a b c d e f g h i = 8 4 3 9 5 7 1 2 6 (as reported by Olivier) a b c d e f g h i = 8 2 6 9 5 7 1 4 3 (my alternate solution) Edited: 21 June 2007, 11:09 p.m. ▼ Olivier TREGER Member Posts: 120 Threads: 19 Joined: Nov 2006 06-22-2007, 02:36 AM So... If I read well all the answers, there seem to be no way to automate the search for a solution. Right? Oh, by the way, many thanks to all of you for these explanations. It happens that I found a solution but it seemed to be a matter of chance.

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