Suppose it is an Army Enigma I. Choose reflector UKW-B or UKW-C. Choose three rotors from I, II, III, IV and V. For each of the 120 reflector/rotors permutations, there are 17,576 starting indicator positions to test.
Menu
Ciphertext $\tt RRHKJ \ KNJXE \ UWNFO \ T$ has 16 characters. Crib $\tt WETTE \ RVORH \ ERSAG \ E$ has 16 characters too. Align crib to ciphertext at position 1 and there is no clash.
1 1 1
1 5 0 5 6
RRHKJ KNJXE UWNFO T <- ciphertext
WETTE RVORH ERSAG E <- crib
Assume turnover (t/o) position is at 13, 14, 15 or 16. We use letters at positions 1-12 and 16 to build a menu. It has 10 letters, 2 loops, 11 links of which one is a double link. There are two subsidiary letters and one subsidiary link.

Default ring setting is 26 26 26 (Z Z Z). Drum setting for character positions 1, 2, ..., 12 are Z Z A, Z Z B, ..., Z Z L and position 16 is Z A P. Bombe returns 32 stops. It is noted that Gordon Welchman called them "drops". Many Bletchley Park veterans called them stories. I like the term story. It is not to be trusted until verified.
Fri 26 May 2023 13:59:35 BST
UKW-B, I II IV, offsets 2 3 15 (B C O), Cable E, wire o is a stop, drums C D Q, golden drums W V I : O
UKW-B, I II IV, offsets 1 3 23 (A C W), Cable E, wire q is a stop, drums B D Y, golden drums X V A : Q
UKW-B, I III II, offsets 14 6 5 (N F E), Cable E, wire j is a stop, drums O G F, golden drums K S T : J
UKW-B, I III II, offsets 19 11 10 (S K J), Cable E, wire s is a stop, drums T L K, golden drums F N O : S
UKW-B, I III IV, offsets 14 9 10 (N I J), Cable E, wire b is a stop, drums O J L, golden drums K P N : B
UKW-B, I III IV, offsets 0 19 17 (Z S Q), Cable E, wire j is a stop, drums A T S, golden drums Y F G : J
UKW-B, I IV V, offsets 22 6 8 (V F H), Cable E, wire s is a stop, drums W H K, golden drums C R O : S
UKW-B, I V II, offsets 13 21 24 (M U X), Cable E, wire y is a stop, drums N X Y, golden drums L B A : Y
UKW-B, II III IV, offsets 15 4 14 (O D N), Cable E, wire z is a stop, drums P E P, golden drums J U J : Z
UKW-B, II V IV, offsets 8 14 23 (H N W), Cable E, wire i is a stop, drums I Q Y, golden drums Q I A : I
UKW-B, III IV I, offsets 5 20 25 (E T Y), Cable E, wire y is a stop, drums F V Z, golden drums T D Z : Y
UKW-B, III IV V, offsets 25 24 1 (Y X A), Cable E, wire y is a stop, drums Z Z D, golden drums Z Z V : Y
UKW-B, IV I III, offsets 16 11 7 (P K G), Cable E, wire l is a stop, drums R L H, golden drums H N R : L <-- ture stop
UKW-B, IV III II, offsets 2 9 19 (B I S), Cable E, wire s is a stop, drums D J T, golden drums V P F : S
UKW-B, V II III, offsets 8 7 19 (H G S), Cable E, wire l is a stop, drums K H T, golden drums O R F : L
UKW-B, V II III, offsets 1 23 21 (A W U), Cable E, wire n is a stop, drums D X V, golden drums V B D : N
UKW-B, V II IV, offsets 11 8 23 (K H W), Cable E, wire c is a stop, drums N I Y, golden drums L Q A : C
Fri 26 May 2023 14:20:26 BST (17 stops, 21 mins)
Fri 26 May 2023 14:08:06 BST
UKW-C, I III II, offsets 18 0 13 (R Z M), Cable E, wire q is a stop, drums S A N, golden drums G Y L : Q
UKW-C, I IV V, offsets 4 18 23 (D R W), Cable E, wire d is a stop, drums E T Z, golden drums U F Z : D
UKW-C, I V IV, offsets 23 16 4 (W P D), Cable E, wire q is a stop, drums X S F, golden drums B G T : Q
UKW-C, II I III, offsets 2 21 12 (B U L), Cable E, wire q is a stop, drums C V M, golden drums W D M : Q
UKW-C, II IV V, offsets 2 7 21 (B G U), Cable E, wire a is a stop, drums C I X, golden drums W Q B : A
UKW-C, III II I, offsets 16 5 17 (P E Q), Cable E, wire q is a stop, drums Q F R, golden drums I T H : Q
UKW-C, III II I, offsets 21 19 19 (U S S), Cable E, wire v is a stop, drums V T T, golden drums D F F : V
UKW-C, IV II I, offsets 1 16 11 (A P K), Cable E, wire e is a stop, drums C Q L, golden drums W I N : E
UKW-C, IV II III, offsets 8 12 3 (H L C), Cable E, wire g is a stop, drums J M D, golden drums P M V : G
UKW-C, IV III II, offsets 3 24 25 (C X Y), Cable E, wire d is a stop, drums E Y Z, golden drums U A Z : D
UKW-C, IV III V, offsets 17 20 7 (Q T G), Cable E, wire y is a stop, drums S U J, golden drums G E P : Y
UKW-C, IV V I, offsets 11 10 25 (K J Y), Cable E, wire f is a stop, drums M M Z, golden drums M M Z : F
UKW-C, IV V II, offsets 24 19 9 (X S I), Cable E, wire d is a stop, drums Z V J, golden drums Z D P : D
UKW-C, V I III, offsets 18 23 13 (R W M), Cable E, wire f is a stop, drums U X N, golden drums E B L : F
UKW-C, V II III, offsets 10 16 8 (J P H), Cable E, wire c is a stop, drums M Q I, golden drums M I Q : C
Fri 26 May 2023 14:28:36 BST (15 stops, 20 mins)
Stop condition
Suppose test register is connected to Cable $\tt E$. "In order to detect a 'drop,' the bombe needs to look for one of the two situations: a case in which current does not get back to any test register terminal other than [its steckered letter, $\tt l$, in our case], or one in which there is vast majority of [in fact, all of] the register will fill up." (Welchman 2021, p.241)
Gordon Welchman (2021) The Hut Six Story: Breaking the Enigma Code. Cleobury Mortimer, Shropshire, UK: M & M Bladwin.
On one side of the diagonal board, there are twenty-six 26-way cables. Each cable is equivalent to a letter in the menu. Each double-ended letchworth scrambler is equivalent to a link in the menu. Let's say, Cable I has 26 wires a, b, ..., i, j, ..., z and Cable J has a, b, ..., i, j, ..., z. The diagonal board connects wire Ij to wire Ji. Therefore, the are 351 distinct connections. It is the upper triangle of an 26x26 matrix.
Suppose each wire in the test register is a circuit on its own. Each wire has four possible states. Firstly, it connects to one circuit which is represented by an ${\tt 1}$. Secondly, it connects to 25 circuits which is represented by a ${\tt \sim}$. These two cases are the conditions we want for a stop. Thirdly, it connects to all 26 circuits, which means there are contradictions, and is represented by an ${\tt =}$. Fourthly, a wire is open circuit which is represented by a ${\tt 0}$. The last case happens when a letter is not in the menu and the diagonal board did not propagate to all its wires.
UKW-B, IV I III, offsets 16 11 7 (P K G), Cable E, wire l is a stop, drums R L H, golden drums H N R : L
wire a b c d e f g h i j k l m n o p q r s t u v w x y z wire
Cable A - 0 0 0 0 ~ 0 0 ~ 0 ~ ~ 0 0 ~ ~ 0 0 ~ 0 ~ ~ ~ ~ ~ 0 0 - A Cable
Cable B - 0 0 0 ~ 0 0 ~ 0 ~ ~ 0 0 ~ ~ 0 0 ~ 0 ~ ~ ~ ~ ~ 0 0 - B Cable
Cable C - 0 0 ~ 0 0 ~ 0 ~ ~ 0 0 ~ ~ 0 0 ~ 0 ~ ~ ~ ~ ~ 0 0 - C Cable
Cable D - 0 ~ 0 0 ~ 0 ~ ~ 0 0 ~ ~ 0 0 ~ 0 ~ ~ ~ ~ ~ 0 0 - D Cable
Cable E - ~ ~ ~ ~ ~ ~ ~ 1 ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ - E Cable
Cable F - 0 0 ~ 0 ~ ~ 0 0 ~ ~ 0 0 ~ 0 ~ ~ ~ ~ ~ 0 0 - F Cable
Cable G - 0 ~ 0 ~ 1 0 0 0 ~ 0 0 ~ 0 ~ ~ 0 ~ ~ 0 0 - G Cable
Cable H - 1 ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ - H Cable
Cable I - 0 ~ ~ 0 0 0 ~ 0 0 ~ 0 ~ ~ 0 1 ~ 0 0 - I Cable
Cable J - 1 ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ - J Cable
Cable K - ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ - K Cable
Cable L - 0 0 0 ~ 0 0 ~ 0 ~ ~ 0 ~ ~ 0 0 - L Cable
Cable M - 0 0 ~ 0 0 ~ 0 ~ ~ 0 ~ ~ 0 0 - M Cable
Cable N - 0 ~ ~ ~ ~ 0 ~ ~ 0 ~ ~ ~ 0 - N Cable
Cable O - 1 ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ - O Cable
Cable P - 0 0 ~ 0 ~ ~ ~ ~ ~ 0 0 - P Cable
Cable Q - 0 ~ 0 ~ ~ ~ ~ ~ 0 0 - Q Cable
Cable R - 1 ~ ~ ~ ~ ~ ~ ~ ~ - R Cable
Cable S - 0 ~ 1 0 ~ ~ 0 0 - S Cable
Cable T - 1 ~ ~ ~ ~ ~ ~ - T Cable
Cable U - ~ ~ ~ ~ ~ ~ - U Cable
Cable V - 0 ~ ~ ~ 0 - V Cable
Cable W - ~ ~ ~ ~ - W Cable
Cable X - ~ ~ 1 - X Cable
Cable Y - 0 0 - Y Cable
Cable Z - 0 - Z Cable
wire a b c d e f g h i j k l m n o p q r s t u v w x y z wire
XZ EL KG WI US maybe steckered
steckered pair - EL KG US WI XZ -
self-steckered - H J O R T -
unknown - A B C D F M N P Q V Y
Now, apply current to connection Cable $\tt E$ wire $\tt l$. It energises itself but no other wires in the test register. We know that $\tt E$ is steckered to $\tt L$. Using the checking machine, by the wiring alone, it can be found that $\tt GK, IW, SU, ZX$ are also steckered and $\tt H, J, O, R, T$ are self-steckered.
When it is not a stop, e.g. UKW-B, IV I III, 16 11 8, apply current to any one wire of Cable $\tt E$ energises all 26 wires in the test register.
UKW-B, IV I III, offsets 16 11 8 (P K H) is not a stop
wire a b c d e f g h i j k l m n o p q r s t u v w x y z wire
Cable A - 0 0 0 0 = 0 0 = 0 = = 0 0 = = 0 0 = 0 = = = = = 0 0 - A Cable
Cable B - 0 0 0 = 0 0 = 0 = = 0 0 0 = 0 0 = 0 = = 0 = = 0 0 - B Cable
Cable C - 0 0 = 0 0 = 0 = = 0 0 = = 0 0 = 0 = = 0 = = 0 0 - C Cable
Cable D - 0 = 0 0 = 0 = = 0 0 0 = 0 0 = 0 = = 0 = = 0 0 - D Cable
Cable E - = = = = = = = = = = = = = = = = = = = = = = - E Cable
Cable F - 0 0 = 0 = = 0 0 0 = 0 0 = 0 = = 0 = = 0 0 - F Cable
Cable G - 0 = 0 = = 0 0 0 = 0 0 = 0 = = 0 = = 0 0 - G Cable
Cable H - = = = = = = = = = = = = = = = = = = = - H Cable
Cable I - 0 = = 0 0 0 = 0 0 = 0 = = 0 = = 0 0 - I Cable
Cable J - = = = = = = = = = = = = = = = = = - J Cable
Cable K - = = = = = = = = = = = = = = = = - K Cable
Cable L - 0 0 0 = 0 0 = 0 = = 0 = = 0 0 - L Cable
Cable M - 0 = = 0 0 = 0 = = = = = 0 0 - M Cable
Cable N - 0 = 0 = = = = = = = = = 0 - N Cable
Cable O - = = = = = = = = = = = = - O Cable
Cable P - 0 0 = 0 = = 0 = = 0 0 - P Cable
Cable Q - 0 = 0 = = = = = 0 0 - Q Cable
Cable R - = = = = = = = = = - R Cable
Cable S - 0 = = = = = 0 0 - S Cable
Cable T - = = = = = = = - T Cable
Cable U - = = = = = = - U Cable
Cable V - = = = = 0 - V Cable
Cable W - = = = = - W Cable
Cable X - = = = - X Cable
Cable Y - 0 0 - Y Cable
Cable Z - 0 - Z Cable
wire a b c d e f g h i j k l m n o p q r s t u v w x y z wire
Turnover at position 16
At t=0, initial ring setting is 26 26 26 (Z Z Z) making the middle rotor to turnover at t=16. It suggests that DG is a steckered pair. Since GK is a pair, there is a contradiction. So, the possibility of middle rotor turnover at position 16 is eliminated.
Turnover at position 15
Change ring setting to 26 26 01 (Z Z A) making middle rotor to turnover at t=15. And hence at t=0, starting indicator is P K H (16 11 08). At t=15, middle rotor turns over and yet again indicator is P L W.
steckered pair - EL GK IW SU ZX -
self-steckered - H J O R T -
unknown - A B C D F M N P Q V Y
X - RRHKJ KNJXE UWNFO T
X1 - RRHGJ G-JZL SI--O T
X8 - ILTTL R-ORH LR--K L
Y - WETTE R-ORH ER--G E
crib WETTE RVORH ERSAG E
diff ^ ^^
t/o ~ ~
At t=7, scrambler $S_7 = {\tt (AJ) (BX) (CU) (DK) (ET) (FR) (GM) (HY) (IL) (NV) (OQ) (PW) (SZ)}$. It is clear that NV is a steckered pair making N > V > N > V. At t=13, scrambler $S_{13} = {\tt (AG) (BR) (CZ) (DI) (EJ) (FQ) (HT) (KS) (LN) (MX) (OY) (PW) (UV)}$, it follows that N > V > U > S. Let's make two wild guesses that FD is a pair and A is self-steckered. We have,
steckered pair - EL GK IW SU ZX - NV FD
self-steckered - H J O R T - A
unknown - B C M P Q Y
X - RRHKJ KNJXE UWNFO T
X1 - RRHGJ GVJZL SIVDO T
X8 - ILTTL RNORH LRUAK L
Y - WETTE RVORH ERSAG E
crib WETTE RVORH ERSAG E
diff
t/o ~ ~
Therefore, we can conclude that ring setting is ?? ?? 01 (? ? A), starting indicator is ? ? H (?? ?? 08), EL GK IW SU ZX NV FD are steckered pairs and H J O R T A are self-steckered. Also, indicator/ring offset for the left and the middle rotors are 16 11. Any settings satisfy the above conditions will yield the correct ciphertext/plaintext pair.
Alternative solutions for turnover at position 15
There is no dispute that NV is a steckered pair. However, it is equally valid to make two alternative wild guesses by saying that F is self-steckered and QA is a pair. Given that ring setting is ?? ?? 01 (? ? A), starting indicator is ? ? H (?? ?? 08), any settings that satisfy that EL GK IW SU ZX NV QA are steckered pairs, H J O R T F are self-steckered and indicator/ring offset for the left and the middle rotors are 16 11 will return the correct ciphertext/plaintext pair too.
steckered pair - EL GK IW SU ZX - NV QA
self-steckered - H J O R T - F
unknown - B C D M P Y
X - RRHKJ KNJXE UWNFO T
X1 - RRHGJ GVJZL SIVFO T
X8 - ILTTL RNORH LRUQK L
Y - WETTE RVORH ERSAG E
crib WETTE RVORH ERSAG E
diff
t/o ~ ~
Turnover at position 14
There is at least one contradiction. Therefore, the possibility of middle rotor turnover at position 14 is eliminated.
Turnover at position 13
There is at least one contradiction. Therefore, the possibility of middle rotor turnover at position 13 is eliminated.
Stop for UKW-C, I IV V, 4 18 23, Cable E, wire d
There is at least one contradiction at each of turnover positions at 13, 14, 15 and 16. Therefore, the possibility this stop can be rejected altogether.
Insufficient information
First, the ciphertext is too short, we cannot determine the rest of steckered pairs and self-steckered letters. Second, there is no message preamble, we are unable to determine ring setting for the left and the middle rotors.
End note
(1) The Turing-Welchman Bombe actually spins the left drums (at the first row of each bank) fastest, then the middle drums (at the second row) and the right drums (at the third row) slowest. It doesn't affect the stops themselves but only the order that they are coming out from the Bombe. (2) For an unknown reason, the markings on drums I, II, III, IV and V are shifted by 1, 1, 1, 2 and 3 places respectively comparing to the Enigma machines. Magnus Ekhall and Fredrik Hallenberg describe this oddity in the Turing Bombe Tutorial for their very comprehensive Bombe Simulator. I presume it is the same in the Checking Machines.
Frank Carte (2010) The Turing Bombe. The Rutherford Journal 3. http://rutherfordjournal.org/article030108.html
Magnus Ekhall and Fredrik Hallenberg (2020) Turing Bombe Tutorial. https://www.lysator.liu.se/~koma/turingbombe/TuringBombeTutorial.pdf