This is my first attempt at the Kenyan part of a Kenyan Killer, so I thought that my steps might be a bit more trial and error than those by more experienced solvers; as it happened, there wasn’t any need for T & E.
I’ve started with HATMAN’s max-min table, which was given in small text, since it’s easy to work out from minimum totals for 2,3 and 5 cell cages and maximum totals for 2,3,4,5 cell cages, plus the specified order of the totals. Thanks HATMAN for doing that routine task for us.
1. H = 1/3 G, G = 3/4 A, min A = 17, max A = 23, the only multiple of 4 is 20 -> A = 20 -> G = 15 -> H = 5
1a. A = 2/3 B, A = 20 -> B = 30
1b. The minimum values of all the other cage totals are adjusted upward to keep them in the same order, maximum values of K,C,P,E,R and T,D,M reduced.
2. F = 3/5 L, min F = 21, max F = 24, min L = 32, max L = 35, only way to get 3/5 ratio is F = 21, L = 35
3. Q = 3/4 T, min T = 22, max T = 27, the only multiple of 4 is 24 -> T = 24, Q = 18
3a. G = 15, Q = 18, A = 20 -> U = 16, J = 17, N = 19 to maintain the correct order of cage totals
3b. Minimum values of M,B adjusted upward.
4. Q = 2/3 D, Q = 18 -> D = 27
4a. Minimum value of M adjusted upward.
5. C = 4/5 P, min C = 7, max C = 11, only multiple of 4 is 8 -> C = 8, P = 10
5a. Minimum values of E,R adjusted upward, maximum value of K reduced.
6. M = K + F, min K = 6, max K = 7, F = 21, min M = 28, max M = 29 -> K = 7, M = 28
7. S = E + R + 6, sum of cage totals already calculated (with repeats for H, C, P and Q) = 349 -> total of E + R + S = 56, min S = 31, max S = 34, only way to get E + R + S = 56 with S = E + R + 6 is S = 31, E + R = 25
7a. Min E = 11, max E = 13, min R = 12, max R = 14 -> E + R = 25 can be either 11 + 14 or 12 + 13. At this stage I cannot see any relationship in the diagram to determine which of these options is correct.
7b. Innies for N23 don’t immediately determine which value is correct for E, innies for N9 don’t immediately determine which value is correct for R, although they do give the first placements in R7C67.
7c. Similarly E = 11 or 12 give 2 outies for R1234 = 16 or 17 so don’t determine which value for E is correct.
8. Just realised that, since HATMAN said that this is based on Para’s puzzle, E should be 11 and R should be 14. However I’ll be happier if this can be proved without this clue.
[I checked by PM with HATMAN and he confirmed that it’s necessary to start solving the puzzle to determine which pair of values for E and R is correct.
Even though HATMAN commented that Para’s puzzle is paper solvable, after the cage pattern has been determined, I’ve just used normal elimination solving even though this may be overkill.]
Prelims
a) R34C3 = {14/23}
b) R3C89 = {16/25/34}, no 7,8,9
c) R5C56 = {19/28/37/46}, no 5
d) R7C34 = {17/26/35}, no 4,8,9
e) R7C67 = {17/26/35}, no 4,8,9
f) R8C89 = {14/23}
g) R67C8 = 13 or 14, no 1,2,3
h) 8(3) cage at R1C6 = {125/134}
i) 21(3) cage at R2C6 = {489/579/678}, no 1,2,3
j) 19(3) cage at R5C3 = {289/379/469/478/568}, no 1
k) 10(3) cage at R6C5 = {127/136/145/235}
l) 27(4) cage at R1C8 = {3789/4689/5679}, no 1,2
m) 18(5) cage at R5C7 = {12348/12357/12456}, no 9
n) 35(5) cage at R4C5 = {56789}
Steps resulting from Prelims
9a. 27(4) cage at R1C8 = {3789/4689/5679}, 9 locked for N3
9b. 35(5) cage at R4C5 = {56789}, locked for R4
10. 45 rule on N9 3 innies R7C789 = 24 -> R7C7 = 7, R7C9 = 8, R7C8 = 9, R7C6 = 1, clean-up: no 9 in R5C5
10a. 13(2) or 14(2) cage at R6C8, R7C8 = 9 -> R6C8 = {45}
11. 18(5) cage at R5C7 = {12348} (only remaining combination), 1,2,3,4 locked for N6 -> R6C8 = 5, clean-up: no 2 in R3C9
11a. R67C8 = [59] = 14 -> R = 14 -> E = 11 (step 7a)
11b. 11(3) cage at R2C4 = {128/137/146/236/245}, no 9
12. R7C7 = 6 -> R7C56 = 4 = [13], clean-up: no 7,9 in R5C56
[There’s the obvious 2,4 killer pair, but quicker is …]
13. R4C56 = {56} (hidden pair in R4), locked for N5, clean-up: no 4 in R5C56
13a. Naked pair {28} in R5C56, locked for R5 and N5 -> R4C4 = 4, clean-up: no 1 in R3C3
13b. R4C4 = 4 -> R35C4 = 13 = [67], R6C4 = 9, no 1 in R3C89, no 2 in R7C3
14. Naked triple {123} in R4C123, locked for N4
14a. R6C4 = 9 -> R56C3 = 10 = [64], R6C9 = 2, clean-up: no 1 in R4C3, no 2 in R7C4, no 3 in R8C8
14b. Naked pair {23} in R34C3, locked for C3 -> R7C34 = [53]
15. 45 rule on N78 using R7C6 = 1, 3 outies R5C1 + R6C12 = 20, R6C12 = {78} = 15 -> R5C1 = 5, R5C2 = 9
15a. R5C2 = 9 -> R34C2 = 6 = [42/51]
16. 45 rule on N78 using R7C6 = 1, 2 innies R7C12 = 8 = {26}, locked for R7 and N7 -> R7C5 = 4
17. 45 rule on N8 1 remaining outie R8C3 = 1, clean-up: no 4 in R8C89
17a. R8C89 = [23], clean-up: no 4 in R3C8, no 5 in R3C9
17b. R3C89 = [34] -> R34C3 = [23], R3C2 = 5, R4C2 = 1 (step 15a), R4C1 = 2, R3C7 = 8, R3C56 = [79], R2C6 = 4 (cage sum), R3C1 = 1, R4C789 = [987], R5C789 = [341], R7C12 = [62]
18. R3C5 = 7 -> R2C45 = 4 = [13]
19. R9C8 = 1 (hidden single in C8), R9C9 = 6 (hidden single in N9)
20. Naked triple {258} in R1C456, locked for R1 -> R12C9 = [95], R12C7 = [12], R1C6 = 5, R1C3 = 7, R12C8 = [67]
21. Naked pair {28} in R1C45, locked for 30(5) cage at R1C2 -> R2C3 = 9, R1C2 = 4 (cage sum)
and the rest is naked singles.
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