=====Walkthrough===== Following will be a walkthrough (the documentation in a form acceptable for submission, provided as an example of both the form I'd like to see submissions in, but also providing illumination in the process of solving a letter division puzzle) of the following puzzle: XL +--------- ILNXLN | XYXCILJY -JIJNLJN ======= LJCCYJY -YMTJCTN ======= CLXNCY base: 9 letters: CIJLMNTXY difficulty: easy The intent of the documentation (and therefore the project you are performing) is NOT merely to SOLVE the problem, but to document the step-by-step process of solving the problem. That is demonstrated here. As this is a puzzle encoded in a base OTHER THAN 10, a few considerations have to be kept in mind. First, in base 9 the available counting digits are ONLY 0, 1, 2, 3, 4, 5, 6, 7, and 8. The digit symbol of 9 does NOT exist in base 9. Similarly, 8+1 = 10. Because 8 is the last counting digit available, when we exhaust the digits, we cycle to 0 and carry 1. However, 10 base 9 should be pronounced and thought of as "one zero base 9", and NOT "ten". Ten is a decimal (base 10) construct, and 10 base 10 does NOT equal 10 base 9. Lacking 9 means any memorized mathematical facts, especially with respect to carried/multi-place values, are no longer valid. The same properties and concepts are there, but the results will be different. Being on odd base, that also means that all even and odd numbers do not end with an even or odd digit (some certainly do, in alternating fashion of runs of counts of 9). ====Getting underway==== Looking for anything to get us started, let's look at a couple layers of the left-most subtractions: * X-J=zero * J << X, but also, J < X * we are being taken from * next to it: Y-I=L * Y << I, we need to borrow * L-Y=zero * Y << L, but also: Y < L * we are being taken from * next to it: J-M=C * J << M, we need to borrow What about C? If J << M, is J also less than C? Let's check. Starting with smallest value of J, and because J << M, the smallest value of M that is greater than this current J possibility: J+10 0+10 - M - 1 ==== --> ==== C 8 To be sure, we don't yet know if this is being taken from, but at the moment, we're only looking to see if some pattern emerges suggesting the relationship of J and C. Let's tick J and M up by one (J of 1, M of 2): J+10 1+10 - M - 2 ==== --> ==== C 8 Interesting, same value of C in both cases (but only if J is exactly one less than M). So let us offset that a bit, with J of 2, M of 4: J+10 2+10 - M - 4 ==== --> ==== C 7 And how about the maximum spread: J of 0, M of 8: J+10 0+10 - M - 8 ==== --> ==== C 1 In all these cases, J remained smaller than C. We have no confirmation on C vs. M, as they flip, but that doesn't matter, we can now confidently say that J << C as well. Our initial relational chains: J < X << C J < X << M Y < L << I Our chart: C = { 2, 3, 4, 5, 6, 7, 8 } I = { 2, 3, 4, 5, 6, 7, 8 } J = { 0, 1, 2, 3, 4, 5 } L = { 1, 2, 3, 4, 5, 6, 7 } M = { 2, 3, 4, 5, 6, 7, 8 } N = { 0, 1, 2, 3, 4, 5, 6, 7, 8 } T = { 0, 1, 2, 3, 4, 5, 6, 7, 8 } X = { 1, 2, 3, 4, 5, 6 } Y = { 0, 1, 2, 3, 4, 5, 6 } ====Discovering N (our zero value)==== Looking at the right-most subtractions, we find a clear zero value: * J-N=J * Y-N=Y For if there was a borrow taking place, how could we subtract from it and get the same starting value (for J or Y). If J was 0, 0+10=10, 10-N=10? N cannot be 10. So, N is less than J, and N is less than Y. The only value N can be that allows J to be subtracted from and results in J (same case with Y), is for N to be zero. And so it is. Also: the right-most digit of the divisor is N, and we see that no matter what we multiply it by, X or L, we end up with N. Exactly as multiplying by zero would do, because we are. Our updated relational chains: N << J < X << C N << J < X << M N << Y < L << I Our updated chart: C = { 3, 4, 5, 6, 7, 8 } I = { 3, 4, 5, 6, 7, 8 } J = { 1, 2, 3, 4, 5 } L = { 2, 3, 4, 5, 6, 7 } M = { 3, 4, 5, 6, 7, 8 } N = { 0 } T = { 1, 2, 3, 4, 5, 6, 7, 8 } X = { 2, 3, 4, 5, 6 } Y = { 1, 2, 3, 4, 5, 6 } Our current key: 0 N 1 2 3 4 5 6 7 8 Looking for other subtractions involving N, we have a couple candidates: * top row: C - N = C * this means we are now borrowing * AND we are not being taken from * bottom row: Y - C = N * this indicates C << Y, because if it were the other way, what could fit? * it also indicates that C < Y, because we are being taken from. Our updated relational chains: N << J < X << C < Y < L << I N << J < X << C < Y < L << M Our updated chart: C = { 3, 4 } I = { 6, 7, 8 } J = { 1, 2 } L = { 5, 6 } M = { 6, 7, 8 } N = { 0 } T = { 1, 2, 3, 4, 5, 6, 7, 8 } X = { 2, 3 } Y = { 4, 5 } A further optimization can take place: * Because J < X, and the exceptionally small range of possibilities for them, one of them will be 2. So: nothing else can be 2 (impacting T). * Same with C < Y < L, in whatever combination proves to be true, 4 and 5 will be used, so nothing else can be 4 or 5 (ie, once again, T). Our updated chart, from this recent optimization: C = { 3, 4 } I = { 6, 7, 8 } J = { 1, 2 } L = { 5, 6 } M = { 6, 7, 8 } N = { 0 } T = { 1 3 6, 7, 8 } X = { 2, 3 } Y = { 4, 5 } ====refinement of T yields J, X, C, Y, and L through cascade==== T remains our largest unknown, so perhaps we can whittle it down a bit? If we're lucky, doing so may even reveal some other letters (for example, if we could determine it is not one). Let's see, by looking for subtractions we might be able to work from: * bottom row: C-T=L * we know that C << L (by exactly 2) * are we borrowing? Yes, because L is larger than C. * are we being taken from? No: because in C-J=X, C is larger than J. * bottom row: J-T=C * we know that J << C (by as little a difference as 2) * are we borrowing? Yes, because in Y-C=N, N is 0. * are we being taken from? No, because Y-N=Y does not need to. So, from these cases, we can make the determination that: * C << T * J << T in both cases due to the borrowings we have confirmed. How does that impact things? Well, that eliminates T as being one, leaving only one value that could be one (J), and since we know J is one less than X, X is now confirmed as 2. It also reveals C, as if T is greater than C, T also cannot be three. With X being confirmed as 2, C is the only value remaining as 3. And we know it is in a tight coupling with Y and L. Our updated chart, from this recent realization: C = { 3 } I = { 6, 7, 8 } J = { 1 } L = { 5 } M = { 6, 7, 8 } N = { 0 } T = { 6, 7, 8 } X = { 2 } Y = { 4 } And our updated relational chains: N << J < X << C < Y < L << I N << J < X << C < Y < L << M N << J < X << C < Y < L << T Our current key: 0 N 1 J 2 X 3 C 4 Y 5 L 6 7 8 Only I, M, and T remain at this point, so we simply focus on trying to whittle them down, to the point of further identification. Ideally there will be some subtractions now where two of the three letters are known, allowing for very easy determinations. Let's see: ===Finding I==== For I, we have a couple candidates, in the top row. One does not require borrowing, nor is it being taken from: * I-L=C This will do nicely. I I -L -5 == --> == C 3 I = 3 + 5 = 8. Our updated chart: C = { 3 } I = { 8 } J = { 1 } L = { 5 } M = { 6, 7 } N = { 0 } T = { 6, 7 } X = { 2 } Y = { 4 } And our updated relational chains: N << J < X << C < Y < L << M << I N << J < X << C < Y < L << T << I Our current key: 0 N 1 J 2 X 3 C 4 Y 5 L 6 7 8 I ====Determining M==== There's only one instance where M appears in the puzzled, and that is in the subtraction J-M=C in the bottom-most row of subtractions. In that particular subtraction, we know: * J << M, so it has to borrow. * In the next subtraction to the right, C << T, so it is taking from us as we evaluate this problem (so long as we are mindful of that) J+10-1 1+10-1 10 - M - M - M ====== --> ====== --> ====== C 3 3 So, solving for M, we find that it equates to 6. Which then, through process of elimination, gives us T as 7. The final relational chain: N < J < X < C < Y < L < M < T < I The chart: C = { 3 } I = { 8 } J = { 1 } L = { 5 } M = { 6 } N = { 0 } T = { 7 } X = { 2 } Y = { 4 } The puzzle key: 0 N 1 J 2 X 3 C 4 Y 5 L 6 M 7 T 8 I And there we go, puzzle solved. Now to verify: ====Verifying our key==== The best way to verify the puzzle with our key is to convert the dividend and divisor to its numeric equivalent, perform the division, and compare the resulting quotient and remainder against those found in the letterified puzzle: divisor: ILNXLN --> 850250 dividend: XYXCILJY --> 24238514 And let's do some long division! +--------- 850250 | 24238514 Let's get some factors generated so that we can have an easier time: 850250 * 0 = 0 850250 * 1 = 850250 850250 * 2 = 1810510 850250 * 3 = 2760760 850250 * 4 = 3721120 850250 * 5 = 4671370 850250 * 6 = 5631630 850250 * 7 = 6581880 850250 * 8 = 7542240 It would appear that for our first subtraction, 850250 best fits approximately twice (1810510): 2 +--------- 850250 | 24238514 -1810510 ======= 5133414 And for 5133414? 850250 best fits five times (4671370): 25 <-- quotient +--------- 850250 | 24238514 -1810510 ======= 5133414 -4671370 ======= 352034 <-- remainder Converting our quotient and remainder back to letters: * quotient: 25 --> XL * remainder: 352034 --> CLXNCY And comparing against the problem we were given: * quotient: XL <-> XL * remainder: CLXNCY <-> CLXNCY We have a match on both counts! Verification successful! This puzzle is now complete.