« November 2005 | Main | January 2006 »
December 28, 2005
Prizegiving
There will be a prizegiving at Bletchley Park on the afternoon of Friday 31st March. As always tickets are very limited, but if you would like to come email us and we will put your name on the list for the ticket lottery. The afternoon will include a tour of the museum.
Hope you are having a good holiday,
Posted by Harry at 09:30 AM | Comments (16)
December 20, 2005
Rotor 2 wiring
Here it is:
18, 23, 35, 16, 28, 3, 13, 26, 14, 22, 19, 32, 15, 0, 2, 20, 10, 4, 27, 7, 21, 24, 9, 5, 12, 29, 33, 31, 25, 1, 17, 8, 11, 34, 6, 30
Same conventions as before. How many did you get right from the diagram?
Posted by Harry at 07:09 PM | Comments (40)
December 19, 2005
Rotor 2 wirings coming tomorrow
Sometime in the afternoon/evening of Tuesday 20th December I will publish the rotor 2 wirings! By the way, the Punchcard needs to be turned over and rotated so that the first two holes are in the top left corner, this then gives 0-> 0, 1->1 and so on.
Posted by Harry at 08:40 PM
December 15, 2005
Music maestro please
Harry listened to Flutter by Bonobo quite a lot while writing this year's Challenge so he will give free copies of Flutter (from iTunes) to the first 20 correct submissions for Challenge 8. Meanwhile for the rest of you here are two mp3's which you can think of as theme tunes for the Challenge. One of them, Harry's theme, comes from the 2004 Challenge, while the other was written for the 2003 Challenge, but we have renamed it the Michael Five as it has the right kind of feel for this story. Enjoy:
Posted by Harry at 03:03 PM | Comments (48)
The Pringle Enigma plus first rotor wiring (WARNING - EDITED)
No not the mystery of who ate all of Harry's Pringles (Harry did), but the mystery of how to turn a Pringles can into an Enigma machine. Well almost. You can download the (Blue Peter style) plans
They will be more use to you once all the settings have been revealed. So far we have had four correct submissions for Challenge 8B and we are hoping to get a lot more. For now here is the wiring for Rotor
21, 26, 9, 23, 28, 14, 5, 20, 13, 25, 31, 27, 33, 8, 1, 0, 18, 30, 11, 4, 6, 16, 24, 3, 7, 15, 2, 17, 10, 29, 32, 35, 19, 12, 22, 34
Remember there are 36 pins on the rotor and I have numbered them 0 to 35. The first one (one) is connected to pin 21 on the other side, while the second (two) is connected to 26 and so on. The numbering is given round the edge of the wheel so it goes clockwise on one face and anticlockwise on the other (i.e., pin 1 on one side is adjacent to (but not connected to) pin 1 on the other side, and pin 2 is adjacent to pin 2 on the other side but connected to pin 26 and so on.
I will publish some info about the other rotor and the punchcard later on.
Seasons greetings , enjoy the Pringles,
Posted by Harry at 02:12 PM | Comments (23)
December 12, 2005
The Punchcard
There was a small mistake in my drawing of the punchcard. The correct version is now up at
Hope no one was too inconvenienced.
Posted by Harry at 08:33 PM | Comments (69)
CHALLENGE 8B SECOND HINT
The photo is back from the lab and they've got rid of all of the glare. The rotor is still a pretty tangled maze, but I'll get out my jeweller's loup and see if I can disentangle any of it for Thursday. Meanwhile I was reading through my notes on the machine:
1. The rotor nearest the plugboard is the fast one. It turns away from the keyboard once every key press, after the key is pressed and the lamp has lit.
2. The next rotor turns once every 36 key presses (again after the lamp has lit) but the first time it turns in each message does depend on how the machine is set up. It turns towards the keyboard.
3. When the key is pressed the current flows through the punchcard assembly, through the first rotor from right to left, through the second rotor from right to left, through the reflector wiring, back through the second rotor from left to right through the first rotor from left to right, back through the punchcard assembly and to the lamps.
4. The punchcard is hard to read so I had a cipher clerk put a grid on it to help you. Here it is again:
Posted by Harry at 02:53 PM | Comments (28)
December 09, 2005
Winners for Challenge 7A
And the lucky winners are:
Thomas Cappleman and Robert Harman, the The Joker Of No Trumps from Reading School, Reading,
Timothy Palmer from King Edward VI College Stourbridge,
Francesca Golding from North London Collegiate School,
Mark Dessain, the Shadow from St John Fisher Catholic High School, Harrogate,
Becci Rushton from Endon High School, Stoke-on-Trent,
Charmaine Law, Jenny, Zoe, Roberta and Rosi, the Dark Knights from Clarendon House Grammar School, Ramsgate,
Jun Kawakami, Max Ankers, Charles Barry, George Kiff and Tsubasa Sato from The King's School, Cheshire,
Emma Forth from The Grange School, Cheshire, Northwich, Cheshire,
Graham Smith, Brett Hornby, Sasha Brookesbank, Mark Freeman, Huw Foden, Matthew Powell, Neil Hayes and Alex Dawson, the Mathemagicians from Harry Carlton Comprehensive School, Loughborough.
You should have got an email from us. Get in touch to confirm your details. All the best,
Posted by Harry at 04:33 PM | Comments (20)
FIRST HINT FOR CHALLENGE 8B
[For clarification: this is NOT a change to the text of Challenge 8B, it
is extra information about it. You still need to submit your decryption
of the ORIGINAL 8B text.]
I just got this from signals intelligence. Apparently immediately after
we received the encrypted transmission yesterday from the Weather
Station we picked up a ghost transmission from the same source on a
different frequency. It was very faint but we were close enough to get
all of it I think. Here it is:
MDY5H YI2Y1 D8EV4 H08WU O1E7K 2502D CG8DV Z5X4T DT5IO GW2KI G7HJA GUYZW
SI68B I5OSY HAS03 DVZMK MNARO 7O8I9 R1YR4 839DH ORMUR SO4VJ THO2P ZR7PZ
SJZHO T3WPH CK3NL V9MM9 IY5VZ B8B4Z 02VNM W4M3L D3XRO OX2WZ 5ULZM KXKY5
0H14O VJH4N OZ6U7 RF1IX 8VQLN K3W5U GO4Y6 EMWFN NY47G A9IYK FL06C MYYME
R7AR7 90RVT GB0T3 CSA3C CYS9F NWH0V R4YBL Y5HE1 MBXI9 UGG1U ECI5H SKUGA
PR7CG BLED4 KN7FU QLZMC 839EV ACSVL 9U5II QN8V1 029ZD BIVTO DLMC9 XXOP0
T60FA U2LUH ZMSW4 PNY3B 5NG8I 7H773 KEQG1 IYWNR U4RPK Z6BJO ZZJSV P27WY
ZKV4P NFMTH TQ5OW EKZQD VDFM1 9XQDO PGRW8 3L4DD HFPOF 7121M TLCW3 HXZ75
XOK7M SM8C0 SC2KS D1OYR TLLW6 O715Z WVWEJ XN4VU 3MP2T U5Q1Q F4MYF XFMQK
OPKLU 3GK94 2MZHM YOE46 AYGJ2 IS2L2 QAL30 TF9O6 2QDMJ 4X5JC SWJ1W IIEQ1
NAL1T S0J06 C2XMX NKWDL XFMHG E6EQI Z3UX9 PAJAK JSP5S WPPK4 WBM8P 89BZW
UY6WL FN4PW 3BXBZ 92X5N GNF4Q KY8HR XSUVC WTXWW HDZFR M5ZKG FVAGP DGIDR
POJXA 03D8D 3MTG4 1IKVQ T60OT AKE2L 0CWEF YEY8Z 9Q4DJ IEN8J JF0KC PLM36
RD72V RHSJ9 9VPWR OWNI0 D1PED 6ZMNK FNIYH EWJF4 ID4XI EFU76 3MYDA 4C8OT
AFLQ9 PHKXD FNWSS NZQCX SOB3P BKBMO 9WOD8 WQIBJ NIS5O T5UKS C84JQ L1WOB
XK0UX MJ6IF END1X H07XM WM2RZ QVVC5 2MFS2 2CYBW N6MKT 8JMQD TCJEH 4HQFF
UFKLM YWF5C YMK18 0X68X Z79R1 O3081 XHPJT 4URB0 IR0QX 8W6E1 GLQP2 D2TR8
C7DCS WFSS7 BLXCU OWY0R 027IL YNYRM 7GBAP IHEO5 N2ZUN P5V3S X94RO PR5SP
6PJMQ PDOKQ 4PFWF VPXBJ XXDYO V8EEO YJG6I CU66R YKNZY 6SFH8 KPRJ6 XYCH7
EFXWP GQII1 LFQR2 T3Y35 86W7V NO7YN 993NO GH4ML XPPNE OO3P8 SV7ZJ SC0FA
3Q08Z PNMWH 2L7BL NEGKJ MPF77 DVYF2 9BF4L 8JO4L 6JKFK QXIDH VP9O8 HZVUX
6FWEC NDAAQ 62OLJ X6R7Q KHOWF Z0YM7 WZCDC I6DYI GP8Z0 SZMGA UH087 NS0NF
YS60L 13JOG V8KO6 Z3XDF XJW84 TZH8G H7UQ3 259H1 1ZTUQ ISKIC 9PW3H 6LHZQ
N1VPI A215S G8JB5 NKIBN OIODE 0K1FP GKXA4 P91OR 72TDW XHAJC IH2WG EN55R
JM0XH LZWZ1 MAJ09 9OIYA U42KN ENUP5 TFGCG GY1SW QJ3YO 0I1QL 5WA03 78P2Y
CK51C GLO46 6HJG6 XWRFJ AVKH7 1L3BG 7QHD6 BJSKB EXLRN TZR7Y 9BSQI TU4SI
Z8S8Y 4BHF5 ONVHR 1Z0CZ GONKP K6VH0 N50J9 HVBTC VULD2 TPA9Z XXQXX 6NZN7
1E0YF UKM2U NQTC6 84IO1 0SAYM H0KWZ A4XG2 9W3UI EOBJU ZBSTN 1ALA1 9WZKI
BV9YK HYE8I LICF3 YXBUT ACLJY D0KRW GPSBT J9E5K SNFM2 J9AS5 3MMK4 BZYSP
WN3H9 HUK3K UNOC0 XUH97 SA7AD FSE46 SL6M7 THS3K WHM8C 5GUD3 52Z1X WMMJN
76F5G 1NNW0 RH0W3 G1JQJ 3OJ4A PMTRG 7K6UT P71KZ 00DZ1 G5JXP U8U14 DX00T
TIKBQ B2A9D G11IO J6IYE MNB9Z J4J04 IFE8Y F263T L45C7 DEWZZ J0MHO 399QD
GH0DD PN915 07JGE YK905 SW87U Q595M NXCLJ EKY97 PWQXE RTBX3 RQZQD 1X8WZ
5IC00 Z0USF LVTX4 VYS6F 5SH4Q 6PZET IDGHF 9BPVN DQA7P JFLQC X196U WX04J
136YK V186A EUV1H XZZ4R CVWHG G0YVF 9AZ75 UOT14 DSACA NF1XX IDDCV Y0V6E
JA02G NLH2S JN9KW I1QO6 N19YI EZSKW 4R9JZ WPGOY IVSG8 2PA6V JRWTC KIP9G
I6PSL YPVMD HOGWF U2PIX 1ADX6 Z9JU5 9YUPL DM1OR 0YIA6 1M6T7 J2P41 9LBBK
4B9WC 820DQ G1Q7R 81QC1 CX2MU IDPPK 06RQ4 0BQ1B AKKMH CP2BF YSGO6 S6V9Q
ALXCW OLD40 T5GLN 0QJMS 1GSGV 3L80Z W48VG AXPRS HJFS1 NMMU9 U5DVG HUJ6F
2MCKN OSGEO ATQTG JOGE7 LPGY7 9R5V9 P0M3S XCSO6 90JOP F6RIQ SKXH3 BX3VH
S91V7 DCVGO NGMQZ TLQU
As you can see it starts off the same as the transmission we received
but has a whole chunk extra at the end. I asked the radio guys if they
have any idea what was happening and one of them suggested that it might
be a secure long range transmission using a directional mast. That's why
we found it hard to pick up. But he made an interesting remark. If it
was long range then the main part of the message (after any headers,
weather reports etc) would normally be sent twice in case of
transmission errors. If my hunch is right this extra text will be the
report from the captain of the Leninsky Komsomov. Now looking at the text we can see they haven't just transmitted the encrypted text twice, so I figure they have a made a classic mistake and instead have encrypte part of the message twice!. If my hunch is right the extra text is just the captain's report and I think we all might be able
to guess how that starts.
Now it is almost certainly a coincidence that the initial message had
1296 characters in it but it is highly significant for us. It means that
at the end of encrypting the original whole message the rotors were back at their
initial position. If the clerk then re-entered the captain's report he
would have been doing so from the SAME INITIAL SETTINGS as the machine
started with.
If we can guess how the captain's report starts then we might be able to
reduce the number of possible wirings of the rotors.
I had a look at my notes, and have a couple more observations to make on
the machine. The rotors are driven using a similar mechanism to the
Enigma. The first rotor turns fast and every complete revolution of it
drives the second rotor one notch in the opposite direction. The catch
engaging the second rotor is adjacent to the Z on the wheel, but of
course if the first rotor is inserted with the Z at the top then the
second rotor could turn when the first character is entered. By putting the
first rotor in at different positions you can get the first turn of the
second rotor at any of the first 36 characters of the ciphertext.
Presumably the Soviet cipherclerk had been given initial settings for
the rotors. Maybe they were written down and I didn't notice.
Our photolab guys tell me that the other rotor image should be back from
processing at 3.30pm on Monday.
More later.
Posted by Harry at 03:03 PM | Comments (21)
Winnners for Challenge 6
The winners for Challenge 6A were
Jon Dawson from The Gryphon School, Sherborne, Dorset
Rachel from Fairham Community College, Nottingham, Nottinghamshire
John Harrison and Simon Scott, the The Dark Horses from Lakeview Sixth Form Centre, Nottingham, Nottinghamshire
Eliza, Dani, EJ, Bonzo,and Bex from The John Bentley School, Calne, Wiltshire
Martyn Compton from Paston College, North Walsham, Norfolk
Jose Mirza from Brighton College, Brighton, East Sussex
Geaorge Lilley and the Cipher Soldiers from William Edwards School, Grays, Essex
David Royal from St Cuthbert's High School, Newcastle upon Tyne, Tyne and Wear
Charlotte Frost and Alice Godwin, from The Folkestone School for Girls, Folkestone, Kent
Robert Edwards from Noadswood School, Southampton, Hampshire
We did email you so if you haven't replied please get in touch to confirm your details as soon as possible.
Posted by Harry at 09:51 AM | Comments (5)
A new thread for comments
use it as you will.
Posted by Harry at 09:19 AM | Comments (8)
December 08, 2005
Challenge 8 and timing
Well, here it is, the last part of the Challenge. Part A should be straightforward, but part B is horrrible! I have written a report on the cipher machine and those of you who can programme can use the report to start coding up a simulator and trying it out. If I get any other info about the machine (and I will) I will post it here.
If you have no hope of writing a programme you could still have a go at deciphering the message by hand, by building a "Pringle tube" Enigma machine. I'll explain how later, but you will need to know more or less all the settings to do this as it would too long to change the wiring diagrams. Anyway it might be a fun Christmas project.
The scoring for Challenge 8 will be the same as the scoring for Challenge 7.
There is a pdf document with a report on the cipher machine that you will need to download. If you are having trouble email me and I will email you a copy, but please do try first.
Good luck to all of you.
(I'll announce the part 7 winners tomorrow)
Posted by Harry at 04:35 PM | Comments (55)
December 07, 2005
last chance for 7B
Here are the first two row fo the decryption matrix for Challenge 7B
2 21 17
10 4 9
You should be able to do the rest by looking at what this gives you.
[By the way, if you are doing the decryption as vM, where M is the matrix and v is a row vector then these are the first two columns instead.]
Posted by Harry at 03:32 PM | Comments (27)
December 06, 2005
A reader writes .....
"Harry, is the following matrix correct?
12 5 11
26 20 15
4 3 10"
Apart from the question of whether this is the encryption or decryption matrix I will say that the first two entries in the top row and the last one in the bottom row are correct. Hope that helps some of you.
Posted by Harry at 04:53 PM | Comments (14)
December 04, 2005
Extended crib
For those of you working on Challenge 7B here is an extended crib I have worked out
The first eighteen characters of the cipher text are
SSODTUQOEBIEFKYJXM
The first eighteen characters of the plaintext are
COMRADESWEAREUNDER
You might even take a guess at the next six characters if you are feeling lucky, Anyway it gives you something more to check your calculations against.
Good luck,
Posted by Harry at 09:44 PM | Comments (23)
December 02, 2005
Second hint for Challenge 7B
So far (if you have been following the comments, and you should have been) you know it is a Hill cipher. In this case given by a 3 x 3 matrix. That means there are nine numbers a,b,c,d,e,f,g,h,i so that if the first three letters are represented by numbers x,y,z (taken mod 26 so that a is represented by 0, b by 1 and so on) then the first three characters of the cipher text are represented by the numbers
ax+by+cz, dx+ey+fz, gx+hy+iz
where the numbers are again taken mod 26, so for example if ax+by+cz=27 then we regard that as 1, representing b (1 is the remainder when you divide 27 by 26) and if ax+by+cz=59 you regard it as 7 representing h.
Now looking at the ciphertext you see the first three letters are SSO, represented by the numbers 18, 18, 14, so you know that in fact
ax+by+cz=18
dx+ey+fz=18
gx+hy+iz=14
Now looking at previous challenges you can guess that the first three letters of the plain text are COM, so that gives you a guess for x,y,z and gives you
2a+14b+12c=18
2d+14e+12f=18
2g+14h+12i=14
That gets you started. Guessing the next three characters of the plaintext gives you another bit of information, and looking for other cribs (what is the most common triple of letters in english?) helps to give enough data to start cracking the cipher.
Good luck!
Posted by Harry at 10:21 PM | Comments (36)
Your wish is my command ...
A new thread is hereby provided.
Posted by Harry at 04:36 PM | Comments (6)