The Middle Shogi Manual was compiled by the late George Hodges
This website lists some new insights into these problems

All problems were fed to the HaChu engine. This was running in tsume mode with the 'Allow repeats' setting, which puts the obligation to deviate always on the checking side. For problems where promotion played a crucial role, solution was attempted with both 'Promote on entry' switched off (which means that HaChu plays according to the rule that a piece that deferred promotion can promote on later moves that touch the zone, but not on the turn immediately following their deferral, unless they capture), as well as on (which means noncaptures can only promote when the enter the zone from without). Note that both of these rules are different from what the Middle Shogi Manual states, where the condition is that the next move with that piece cannot be a promotion.
The problems of the C series came without historic solutions, and the solutions presented in the MSM were found in recent times. Many problems so far defied solution, however. As the HaChu analysis shows, this sometimes is because the problems are flawed, so that no solution exists under the rules as we know them (or, in case of promotions, were assumed in the MSM). In many other cases HaChu did manage to find a solution, though. First we discuss these hitherto unsolved problems.
C8  Impossible
C11  The solution found by HaChu relies on the ability to promote the FL on leaving the zone with a noncapture:
1. FLx7d= K9d 2. Ln7c K10c 3. FL8e+ K10d 4. Ln9e +RCx9e 5. Ph8f mate
Under more restrictive promotion rules (zoneentry only, or after another move with the same piece, as the MSM assumes) no solution exists. This could be taken as evidence against the promoteonentry rule. As we will see later, this is really a minority vote, though. Also unsatisfactory is that the solution does not require the presence of the SE (still mate in 5), or alternatively the Ph (mate in 8).
It thus seems the design is flawed, and we conjecture that the solution the designer had in mind is the one the MSM mentions in the preface to the C problems: 1. +DK8e Gx8e {lures a piece to 8e to allow the FL to promote there} 2. FL7d= K9d 3. Ln7c K10c 4. FLx8e+ K10d 5. Ln9e +RCx9e 6. Ph8f mate, which does use all pieces in an elegant way. That 1... BT8d refutes this line must have been an oversight of the designer.
C14  A mate in 12 was found for this problem:
1. +RC10k +DKx10k (1... +VM10h 2. +DH5j K9d 3. FKx8d DEx8d 4.Ln8f K8c 5. +DH3h SM4g 6. +DHx4g P5f 7. +DHx5f DE7d 8. +DHx7d mate) 2. +DH6i K9d (2... SM8g 3. +DHx8g K9d 4. FK9g +DH9f 5. FKx9f K8c 6. +DH4c DE6c 7. +DHx6c FL7c 8. +GBx7c mate or 2... +S7h 3. Lnx7h6g K9d 4. FKx9g +VM9e 5. Ln8f K8c 6. +R2c SM5c 7. +Rx5c DE6c 8. +Rx6c FL7c 9. +GBx7c mate) 3. +DH6g K10e (3... SMx6g 4. FK9g SMx9g 5. Ln8f K8c 6. +GB9b DKx9b 7. +R2c SM5c 8. +Rx5c DE6c 9. +Rx6c FL7c 10. +Rx7c mate or 3... +DHx6g 4. FK10c DKx10c 5. Ln8f K10c 6. +GB9c K11d 7. Lnx9g9f K11c 8. +Px12c +Lx12c 9. +R11a +DK11b 10. +GB10b DKx10b 11. Ln11e mate) 4. +DH8g SMx8g 5. Lnx8g K9d 6. FK7f Px7f 7. Ln8f K 8c 8. +GB9b DHx9b 9. +R2c SM5c 10. +Rx5c DE6c 11. +Rx6c FL7c 12. +Rx7c mate
C21  There only is a solution if the FL is allowed to promote with a noncapture out of the zone.
1. DH7d+ K8b 2. +DHx!7c K9c 3. FK6f K7b 4. FL5e+ +L6d 5. +FLx6d R7c 6. +DHx!6d +DK6d 7. +DH9b mate
C22  This is a mate in 17:
1. FK1l +SMx1l (1... DH1e 2. FKx1e +VMx1e 3. +DK1d R1c 4. +DK8d3d K2b 5. +DK1d2e R2c 6. +DK3dx2c mate or 1... R1c 2. +DK3d +VMx3d 3. +DKx3d K2b 4. +VM5e Gx5e 5. +DK4d K1b 6. FK6g G5f 7. FKx5f R3c 8. +DK3d K1c 9. +DKx3c mate or 1... K2c 2. +DKx4c +VM3c 3. +DKx2e K3b 4. FK7g FL5d 5. +DH8dx5d K4a 6. +DK2ex4e +VM4b 7. +DK2c mate) 2. +VM1f DKx1f 3. DK3d K2b 4. +DK3c +VMx3c 5. +DK2d K3b 6. +DKx3c4b Kx4b 7. Ky2d+ K5c 8. Ln3d K6d 9. Lnx4c K7c 10. Ln6e and Lion mate follows
C23  No solution exists under the known rules.
C25  Mate in 12:
1. +DH11b Kx11b 2. +DK10d K10a 3. +SMx6e K9a 4. +SM2a K8b 5. +SM3b +B4b 6. +SMx4b K9a 7. +SM6d K10a 8. L12c+ K9b 9. +DK7b Ky8b 10. +DK7d K9a 11. +SM3a Ky7a 12. +SMx7a mate
When repeats are strictly forbidden, gote can be 'smoked out' four moves faster, because after 4. +SM5e K9b 5. +SM6e stepping back to 9a is no longer an option as it creates a repeat, so that the King is forced to the fatal 8th file imediately.
C26 and C27  No tsume exists under the known rules
C31  When the Silver is allowed to promote on stepping out of the zone, a mate in 9 can be achieved:
1. P6d+ +Lx12k 2. DK3d+ FKx3d 3. Sx3d= K4b 4. DK2d+ +Bx2d 5. S4e+ +SM4c 6. +Sx4c Kx4c 7. Ln4e K4b 8. B3c Kx3a 9. Ln4c mate
This mate would also work if we delete sente's Lance, Bishop and Whale, however, and is thus unlikely to be what the designer intended. A much more difficult mate, which does use all pieces, is necessary under 'promote on entry' rule:
1. P6d+ +Lx12k 2. B11j +Rx11j 3. DK3d+ FKx3d 4. Ln4e {to lure the FK to 4e, so the silver could promote there on leaving the zone} FKx4e 5. S3d= FKx3d (5... K4b 6. Sx4e+ +DK4d 7. +Sx4d +SM3d 8. +Sx3d K3d 9. DK2c+ K4b 10. DK4e mate) 6. DK2c+ K4b 7. +RC4j +P4i 8. +RCx4i +DK4d 9. +RCx4d +SM4c 10. +FL3c Kx3a 11. L1c+ +Bx1c 12. +DKx1c K2a 13. +DK1a K3b 14. +DK2b K3a 15. +DK2a mate
C33  Mate in 16:
1. +SM12b K10a 2. FL7d+ Rx7d 3. +SM11b K9c (3... K11a 4. +SM12a Kx12a 5. +DH9d FLx9d 6. FLx9d+ K11a 7. +P11b mate) 4. +SM10c Kx10c 5. FL10d+ Kx10d (5... K9b 6. FL10c Kx10c 7. Ln9e K9b 8. Lnx10e9d K10a Ln11b mate or 5... FLx10d 6. +DH10e FLx10e 7. Ln9e and Lion mate follows) 6. +DHx10e Kx10e 7. Ln9g K9d 8. Lnx11f K8e 9. Ln9g K9d 10. Ln9f K8c 11. Ln9e K9b 12. Lnx8dx7d K8a 13. Ln6c K9b 14. Ln7a K10a 15. Lnx8b8a K11a 16. Ln8a10c mate
Note that it would be trivial to force gote to make the first repetition here, by delivering a perpetual (1. +DHx9a Bx9h 2. +SM12a K10a 3. +SM11a K11a {repeat}). So that when repeats are strictly forbidden, 3. +SM11a would be a checkmate. Apparently the burden to deviate is on the checker.
C36  A straightforward mate in 9:
1. +DK8h6f +DHx6f (1... K7c 2. +DKx5e Cx5e 3. +DKx5e K8d 4. Ln10f K9c 5. Ln11d K8d 6. Ln9e mate or 1... K9c 2. +DK9i K10c 3. +P10d Kx10d 4. Ln10f K11c 5. FK11h K10c 6. Ln10e mate) 2. Ln9f +DHx9f 3. +DK6f +DHx6f (3... K9c 4. +DKx9f K10c 5. +DK11d K9c 6. FK9l +P9h 7. FKx9h DE9d 8. FKx9d mate) 4. FK12h +DH10f 5. FKx10f K9c 6. FK9e K10c 7. +P10d K11b 8. +P11c K12a 9. +P12b mate
C37  Again the promotion rule is crucial. When inzone promotion is allowed (with a oneturn delay), there is a mate in 7:
1. Phx5c= K6e 2. +RC7d BTx7d 3. Gx7e {protected by the Phoenix!} FKx7e 4. Ph4c+ +B5d 5. +Phx5d Kx5d 6. +Ky4c Kx4c 7. Ln3e mate
Note that either the Lion on 2c or on 1f (and then the +P on 6a with it) are redundant in this case, although their absence would lengthen the mate to 11 moves. When promotion is only allowed on zone entry or capture, however, a mate in 12 results where each piece has significance:
1.. +Ky4c {lures a piece to 4c so the Ph can promote there} DKx4c 2. Phx5c= K6e 3. +RC7d BTx7d 4. +G7e FKx7e 5. Phx4c+ +B5d 6. +Phx5d Kx5d 7. Ln3e K6e 8. Ln5g K5d 9. Lnx7e K4c 10. Lnx7d6c K3d 11. Ln5c K2c 12. Ln3b mate
C40  A straightforward mate in 11 with a hopping Falcon:
1. +DH4i +Px4i 2. Ln5j +Px5j 3. +DH4i K5g 4. +DH4g K6f 5. +DHx6e K5g 6. +DH4g K6f 7. +DH5f K7e 8. +P6e Kx8e 9. +DH9f K8d 10. +DH9d K7c 11. +DH6d mate
C41  This problem has a matein7 solution of which one can wonder if it is the one intended by the composer, as many of the pieces remain unused: 1. +SM7h K9g 2. +DKx4g BT8g 3. +DKx8f Gx8g 4. FK9e FL9f 5. FKx11g K8f 6. +SMx6h K7f 7. DH6g mate
C42  No solution exists under the known rules.
C45  A wonderful matein 14, based on clearing the path for the FK to 8b, and a defense to it that promotes the DE to Crown Prince, so that you have to start all over again hunting that.
1. DH9l +G8kx9l {must stay out of Lion reach} (1... K7k 2. DKx8k K6l 3. FKx6g C6h 4. FKx6h K5l 5. FK2l +RC4l 6. FKx4l mate) 2. Ln10j +Gx10j 3. +SM1e DEx1e {this offers higher reward than chasing with the +BT. Interpositions are futile, as all squares on the path are once protected, and twice attacked if you allow the checker to survive. E.g.} (3... S3g 4. +Bx3g +L4h 5. +Bx4h Cx4h 6. +SMx4h P5i+ 7. +SMx5i +DK6j 8. +SMx6j K7l 9. DH7k mate) 4. +L2f DEx2f 5. +B3g DEx3g 6. +VM4h DEx4h 7. FK8b +DK8i 8. FKx8i DE5i+ 9. FKx8l +DE6h 10. DK6k +DE7g 11. FKx10j +DE7f (11... S8h 12. FK10g +DE7f 13. DK6g +DE8f 14. DK6f mate) 12. FK6f +DE8g 13. DK6g +DE9h 14. DKx9g mate
C46  This is a mate in 11 (8 when discounting futile interpositions of +BT, C and +DH). The crux is to get the DH on the 8a1h diagonal with checking moves.
1. Ln5j +Px5j 2. +SM4k +Px4k 3. DK7j +L5j (or 3... +BT6j with same continuation, or 3... FKx7j 4. C2k K2i 5. DH2d+ Cx2d 6. +DH2f K1h 7. Lx1b+ +DH1d 8. +Lx1d C1e 9. +Lx1e mate) 4. C2k K2i 5. DH2d+ Cx2d 6. +DH2f K1h 7. DK7b+ +BT6c 8. +DKx6c FK4e 9. Lx1b+ C1e 10. +Lx1e +DH1g 11. +DKx4e mate
C50  No solution exists under the known rules.
C3  Faster (mate in 13 in stead of 16) is: 3. FLx10k +GBx10k 4. +DH12l K9k (4... K9j 5. +DH12j K9k 6. FK6k K9l 7. FK6l +R8l 8. +DH7j K10l 9. FKx8l +GB9l 10. FKx9l K11k 11. +DH11j mate) 5. FK6k +R8k 6. +RC5g K9j 7. +DH12j +BT10j 8. FK5j K8i 9. +DH12i +BT11i 10. +DHx11i K9h 11. FK7h +R8h 12. +DH8i K9g 13. +DHx8h8g mate
C6  This problem is flawed. The solution according to the MSM can be refuted through 5... Ln6d 6. +Lx5d K5b, and there is no alternative line. With a B in stead of a +FL on 2a, however, there is a mate in 7: 1. +DH6f Lnx6f 2. +Phx6f K5c 3. Bx5d+ Kx5d 4. Ln4f K5c 5. +L5i K4b 6. Ln5d K4a 7. Lnx3c mate. This does not make essential use of the Rook on 12e, however, although in its absence the mate takes two moves longer. So it is questionable whether a +FLforB copying error is the culprit.
C9  Faster (mate in 7 in stead of 9) is: 2. FK10e +RCx10e (2... K12e 3. FK12g BTx12g 4. +DHx12g K11d 5. +DKx9f mate) 3. +DK9f K12e 4. +DK12f BTx12f 5. +DH12g K11d 6. +SM12c Kx12c 7. +DHx12f12e mate
C17  After 8... K2f 9. FK2h is mate, and the best defense is 8... K2e
C19  Last move must be 5. +Lx6b, as after 5. +Bx6b gote still has 5... K6a
C29  After 2... K4d it is mate in 4: 3. C3d+ +DK4d 4. +Cx4d mate so 2... Kx3c is best defense.
C30  After 6... K9c in the main line of the MSM solution 7. +G4c K9b 8. FK8c mate is faster. Best defense is 6... K7b 7. +G4b (7. FK2b R3b {not entirely futile, as it mispositions the FK for a fast mate} 8. FKx3b K6c 9. FK3c K5d 10. FK4c K5e 11. FK4e mate) K6c 8. FK8c K5d 9. FK4c K5e 10. FK6e mate
C32  The presence of the Lion is essential here, as otherwise 2... FK2b would refute the mate. The +RC on 7h seems to be redundant, however.
C34  There seems to be a simple mate in 2 here, through 1. Ln3f???
C35  HaChu can improve the MSM solution to a mate in 11, by disposing of its DH at an earlier opportunity to move it away with check: 1. SM9d+ Kx9d (2... K9c 3. +SM10d K10b 4. +SM12b K9a 5. +SM11a K8b 6. +RCx8e mate) 2. +VMx8e K9e 3. +VM8f K9f 4. +VM8g K9g 5. DH4l +Rx4l (5... Lnx4l 6. +VM8h K9h 7. +VM8i K9i 8. C8j +DHx8j 9. +VMx8j K9j 10. +RC8l mate) 6. +VM8h K9h 7. +VM8i K9i 8. C8j +DHx8j 9. +VMx8j K9j 10. VMx1b+ +BT3c 11. +VMx3c mate
C48  In the MSM solution 11. +SM9a is not possible because of 11... DHx9a. The correct continuation is: 11. Ph11d+ DHx11d 12. +SM9a K9b 13. FM8b K10c 14. FK11b Kx11b 15. +S11c DHx11c 16. RC12a+ mate
C49  The status of 'jewel in the crown' of this problem might be jeopardized by the fact that HaChu finds a mate in 11, not using most of the pieces:
1. Bx10f +DHx10f (1... +FLx10f 2. DH10h K11f 3. FK11l +DHx11l 4. DKx11l Kx10e 5. DKx11d+ K9e 6. +SM8d K8e 7. +GBx7f mate) 2. DH10h +DHx10h (2... K11f 3. FK11l +DHx11l 4. DKx11l +DH11g 5. +P10f +FLx10f 6. DH9d+ VM10e 7. +DHx11d K12g 8. DKx12l +DH12h 9. +DHx10e mate) 3. FK11l K12g (3... +DHx11l 4. DKx11l +DH11h 5. +SM9g K11f 6. DKx11h Kx12e 7. DH12a+ +FL12b 8. +DHx12b mate) 4. +P12f +DHx12f 5. +SM9g +FL10g 6. FKx10g +DHx10g 7. DHx12l K11f 8. +P10f VMx10f 9. DH9d+ K11e 10. +DH8e K10d 11. DK12b+ mate
Of the 18 problems of the C series that were unsolved in the MSM, HaChu could prove that six have no solution under the known rules. Ten others could be solved, but for one of those the solution seems to be a 'parasitic' one not intended by the composer. For the remaining two problems it depends on the exact promotion rule. Both of those can be solved under the rule where any noncapture that touches the zone (i.e. ends and/or starts inside it) can be a promotion, after a oneturn delay since the deferral. In both these cases the solution does not look like the intended one, however, as some of the attacker's pieces can be deleted without affecting the solution. When noncaptures can only promote on entering the zone, however, these problems are proven unsolvable. One of them has a line that looks like the intended solution (by criteria of making essential use of all pieces, as well as elegance), but the line is flawed, as better defense is possible. For the other no such likely intended solution has been identified.
Of the problems 'solved' in the MSM, one problem was found to be flawed.