[Guest]

Part TWO

Here is a code breaking problem set by my son`s Engineering Maths Lecturer as “something to keep your mind active” during the Summer Vacation…..and he can only solve half of it so far.

Can anyone help?

Note: This is the other half of the previous Codebreaker problem, (...which has now been solved.see starting Aug 7 2008 .)

This time we are dealing with Boxes A - M .

However , some of the code calculations appear to be different to the ones used for Boxes N - Z in the previous problem.

===========================================================

A master criminal has broken into a jeweller`s strong room which has the walls lined with several thousand storage boxes protected by individual four digit combination lock codes.. Each storage box contains small bags of precious stones (diamonds etc.) so the total value of all the contents of the locked boxes in this room is several million of ££s .

Each box is labelled with a letter and three digits and the thief knows that the combination lock code for each individual box can be calculated from this sequence.

Eg ...........Box No A164 has a Combination Lock Code of 0055

However , each box has this Combination Lock Code engraved on the INSIDE and fortunately several of these boxes are empty and have been left OPEN . This means that the criminal can now list each open Box Number with its own Security Code , and , using this information will attempt to work out the method of calculating any Security Code from its Box Number.

The aim is to open two particular boxes which contain the most valuable bag of diamonds

So the question is .

"What is the code for boxes B522 and K664 ?"

Here is the list of open BOX numbers with their security CODES from inside.

Can you workout ( and explain ! ) the method to determine the combination lock code of Boxes B522 and K664

BOX CODE BOX CODE BOX CODE BOX CODE BOX CODE

A164 0055 A272 0181 A429 0313 A532 0495 A891 0339

A902 0777 B286 0768 B052 0943 B362 1252 B504 0433

B622 0366 B813 0771 B913 1242 C020 1908 C111 0742

C353 0623 C594 0933 C679 2553 C803 0771 C901 2701

D295 0398 D355 2861 D340 2811 D380 2819 D540 1452

D621 0799 D893 3171 D940 2535 E097 1995 E162 1312

E225 0796 E472 3921 E542 3490 E611 2999 E734 2476

E979 1469 F061 0977 F091 0978 F235 5190 F354 4794

F440 4385 F536 3973 F774 3171 F954 2319 G027 1990

G155 1521 G156 1591 G239 1170 G331 0785 G567 6442

G602 6167 G718 5859 G853 5574 G954 5216 H087 4999

H109 4606 H218 4303 H581 3429 H687 3114 H705 2847

H958 2295 I109 1603 I206 1361 I303 1020 I489 0711

I635 0174 I851 8337 I854 8349 I965 8187 J125 7721

J219 7577 J367 7302 J572 6996 J643 6737 J653 6734

J726 6516 J845 6310 J966 6155 K071 5979 K170 5779

K412 5121 K581 4920 K639 4751 K741 4563 K749 4529

L028 3966 L195 3724 L105 3721 L211 3516 L373 3320

L465 3135 L574 2932 L579 2985 L693 2736 L764 2576

L898 2321 L983 2141 M195 1721 M237 1531 M456 1106

M546 0970 M665 0773 M701 0569 M846 0381 M931 0102

[Guest]

Background information: These "Codebreaker Wanted " (parts 1 and 2 ) problems were originally a single brainteaser question using a large database. It didn`t take long to realise that the calculations used to deal with Boxes A-M and Boxes N-Z were different so we split them into two questions for this forum to increase the chance of them being solved.

[Guest]

This one is a bit harder than the previous code breaker problem.

Edited August 19, 2008 by brillo

[Guest]

the third code digit uses the same calculation as the previous problem

first box digit + third box digit multiplied by 7 = the last digit in the answer is the third code digit

Edited August 19, 2008 by brillo

[Guest]

the third code digit uses the same calculation as the previous problem

first box digit + third box digit multiplied by 7 = the last digit in the answer is the third code digit

The first box letter decode is a bit confusing as it could respond to more than one number depending on the letter.

Any theories for this ?

[Guest]

Ok. So far for the life of me I can't figure out how the first to code numbers work. But this is what I have found.

(I combined part one with part two... to get a bigger sample space.... you said they where part of the same riddle)

If box2 is equal to 0 then code2 is equal to 9 (*0** = *9**)

Eg:

B052 0943

C020 1908

E097 1995

F061 0977

F091 0978

G027 1990

H087 4999

K071 5979

L028 3966

Q068 9969

R002 8940

R012 8947

R017 8990

R025 8956

R030 8900

R031 8970

R032 8941

R033 8912

R037 8994

R054 8986

R082 8946

The strange thing about this is that their is really only one way to make 9 from 0..... that is to add. How ever it is possible that some type of sequence is used. Eg: the 9 times table

09 18 27 36

0+9=9 ; 1+8 =9 ; 2+7=9 ; 3+6=9 .....

If box2 is equal to 5 then code2 is a 4 or a nine (*5** = *4or9**) however it seems to be at relatively random intervals.

Eg:

A532 0495

B504 0433

C594 0933

D540 1452

E542 3490

F536 3973

G567 6442

H581 3429

J572 6996

K581 4920

L574 2932

L579 2985

M546 0970

P541 0427

R544 8430

R560 8454

R568 8410

S526 7477

V555 4409

Another observation I have made is that the first two box values give you the first two code numbers at a constant value. (AAxx = BByy)

Eg:

D355 2861

D340 2811

D380 2819

D3 = 28

F061 0977

F091 0978

F0 = 09

G155 1521

G156 1591

G1 = 15

I851 8337

I854 8349

I8 = 83

......

and last but not least I have noticed that the algorithm get less chaotic and more predictable as A gets closer to Z . I have made truth tables of the first two code numbers to display this.

Box1,Box2 = Code1 (Eg I854 (8349) I,8 = 8)

  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

0 * 0 1 * 1 0 1 4 * * 5 3 * * * * 9 8 * * * * * * * *

1 0 * 0 * 1 * 1 4 1 7 5 3 1 * * 0 * 8 * * * 4 * * * *

2 0 0 * 0 0 5 1 4 1 7 * 3 1 * * * * 8 * * * * * * * *

3 * 1 0 2 * 4 0 * 1 7 * 3 * * * 0 * 8 * 6 * * * * * *

4 0 * * * 3 4 * * 0 * 5 3 1 * * * 9 8 7 * * * * 2 * *

5 0 0 0 1 3 3 6 3 * 6 4 2 0 * * 0 * 8 * * * 4 * * * *

6 * 0 2 0 2 * 6 3 0 6 4 2 0 * * * 9 8 * 6 * * * 2 * *

7 * * * * 2 3 5 2 * 6 4 2 0 * * * * 8 * * * * * * * *

8 0 0 0 3 * * 5 * 8 6 * 2 0 * * * * 8 * 6 * * * * * * 

9 0 1 2 2 1 2 5 2 8 6 * 2 0 * * * * 8 7 * * 4 * * * *


Box1,Box2 = Code2 (Eg I854 (8349) I,8 = 3)

  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 

0 * 9 9 * 9 9 9 9 * * 9 9 * * * * 9 9 * * * * * * * *

1 0 * 7 * 3 * 5 6 * 7 7 7 7 * * * * 8 * * * 8 * * * * 

2 1 7 * 3 7 1 1 3 3 5 * 5 5 * * * * 7 * * * * * * * *

3 * 2 6 8 * 7 7 * 0 3 * 3 * * * 6 * 6 * 6 * * * * * *

4 3 * * * 9 3 * * 7 * 1 1 1 * * * 5 5 * * * * * 5 * *

5 4 4 9 4 4 9 4 4 * 9 9 9 9 * * 4 * 4 4 * * 4 * * * *

6 * 3 5 7 9 * 1 1 1 7 7 7 7 * * * 3 3 * 3 * * * 3 * *

7 * * * * 4 1 8 8 * 5 5 5 5 * * * * 2 2 * * * * * * *

8 3 7 7 1 * * 5 * 3 3 * 3 3 * * * * 1 * 1 * * * * * *

9 0 2 7 5 4 3 2 2 1 1 * 1 1 * * * * 0 0 * * 0 * * * *[/codebox]

But..... I have found a "loop hole" that I am 99% sure will have the correct answer for all of the numbers given to figure out. If you wish I can post it... but I think their is a way to figure it out with 100% certainty.

Edited August 20, 2008 by toxicoxyde

[Guest]

From above, we know that the 3rd digit in the code depends on the 2nd and fourth digit of the box

also that the first two digits of the code depend on the first two digits of the box

From D340:2811 and D380:2819 it would seem that the fourth digit of the code depends on the 3rd of the box; and from L574:2932 and L579:2985 the 4th code digit also depends on the 4th box digit.

So, I leave it as an exercise for the reader to determine the precise relationships.

Another observation is that for 1st code digit is never 9 - does that imply a mod9 has been used somewhere?

[Guest]

and last but not least I have noticed that the algorithm get less chaotic and more predictable as A gets closer to Z . I have made truth tables of the first two code numbers to display this.

I am of the opinion that this second part of the puzzle (Boxes A - M) is a separate set of calculations ( albeit with possible duplicate methods ?) so I reckon that adding the data from Boxes N - Z might only confuse the issue.

But there again it was you that cracked Code Digit Four from the first part so perhaps you are right !

{)

[Guest]

Just a few thoughts on box letter = code digit one

code digit one for A is always 0.

code digit one for M going backwards to I ( ie M/L/K/J/I) runs from 0 to 8 , possibly depending on box digit one in groups of 5.

eg

M931 = 0102

M195 = 1721

L983 = 2141

L028 = 3966

K749 = 4529

K071 = 5979

J966 = 6155

J125 = 7721

I965 = 8187

...but then the early box I numbers are not in this sequence but run

I635 = 0174

I109 = 1603

H958 = 2295

H687 = 3114

H087 = 4999

G954 = 5216

G602 =6167

...and then run a little wild until

A+++ = 0+++

There must be a sequence here somehow

[Guest]

But..... I have found a "loop hole" that I am 99% sure will have the correct answer for all of the numbers given to figure out. If you wish I can post it... but I think their is a way to figure it out with 100% certainty.

I`d be interested in your thoughts if you`d care to post it.

[Guest]

I`d be interested in your thoughts if you`d care to post it.

I have found that I was a wrong on my solution for the other puzzle half.

To get the last digit of the code we have to make a table from A-Z. We get this by assigning a value to the letter then multiplying it by three(and we use only the last digit).

A=0*3 B=1*3 C=2*3 D=3*3 .......

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

0 3 6 9 2 5 8 1 4 7 0 3 6 9 2 5 8 1 4 7 0 3 6 9 2 5

Now to solve the last number we have to solve the second to last number so we do the multiply by seven thing, keep the first digit and place the last digit in our code.

A164 00; (1+4)*7=35

B622 03; (6+2)*7=56

E542 34; (5+2)*7=49

G155 15; (1+5)*7=42

R017 89; (0+7)*7=49

so we have

A164 005;3

B622 036;5

E542 349;4

G155 152;4

R017 899;4

Next we multiply box digit 3 by 7, then add 10(to keep if from going negative) and subtract the number we associated with the letter.

A164 005;3 ; (6*7)+10-(A0)=52

B622 036;5 ; (2*7)+10-(B3)=21

E542 349;4 ; (4*7)+10-(E2)=36

G155 152;4 ; (5*7)+10-(G8)=37

R017 899;4 ; (1*7)+10-(R1)=16

A164 005;3 ; 52

B622 036;5 ; 21

E542 349;4 ; 36

G155 152;4 ; 37

R017 899;4 ; 16

Now we add them together and the last digit is the last digit of our code.

A164 005;3+52=55

B622 036;5+21=26

E542 349;4+36=40

G155 152;4+37=11

R017 899;4+16=10

A164 0055

B622 0366

E542 3490

G155 1521

R017 8990

This method works for all boxes!

Now with that in mind the "loop hole" works like this.....

We can make use of the information that we have. We know that the first two Box numbers make the first two code numbers

Eg:

D355 2861

D340 2811

D380 2819

D3 = 28

F061 0977

F091 0978

F0 = 09

G155 1521

G156 1591

G1 = 15

I851 8337

I854 8349

I8 = 83

with that in mind we can use substitution to get the answers for B522 and K664

B504 0433

B5=04

K639 4751

K6=47

so...

B522 04

K664 47

then we just solve the rest.

B522 04;(5+2)*7=49

B522 049;4; (2*7)+10-(B3)=21

B522 049;4+21=25

B522 0495

K639 47;(6+3)*7=63

K639 473;6; (3*7)+10-(K0)=31

K639 473;6+31=37

K639 4737

You might want to double check it to make sure I did it all right. But that is the idea. The only problem I have with this method is it can't be used for the other puzzle. We can't be 100% sure of W456 because we don't have a sample of W4xx. (and I am 99% positive that when the method is found to solve the first part of the code it will work across the board).

[Guest]

I've been looking through it and I agree with all the formula so far but I think Toxic has it a little bit wrong.

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

0 7 4 1 8 5 2 9 6 3 0 7 4 1 8 5 2 9 6 3 0 7 4 1 8 5

I think to decode this puzzle box it relies heavily on the 7 times multiplier, we have seen it about 3 other times or so.

that being said.

last digit of ... (3rd digit * 7) + (number equiv to letter) + carryover from 3rd box

it's basicly the same as toxic's except there is no adding 10 to it and it does make a different in some of the answers if we need to use the carry over from this.

to me it makes more sense and is a little bit cleaner.

[Guest]

Now with that in mind the "loop hole" works like this.....

We can make use of the information that we have. We know that the first two Box numbers make the first two code numbers

Eg:

D355 2861

D340 2811

D380 2819

D3 = 28

F061 0977

F091 0978

F0 = 09

G155 1521

G156 1591

G1 = 15

I851 8337

I854 8349

I8 = 83

with that in mind we can use substitution to get the answers for B522 and K664

B504 0433

B5=04

K639 4751

K6=47

so...

B522 04

K664 47

then we just solve the rest.

B522 04;(5+2)*7=49

B522 049;4; (2*7)+10-(B3)=21

B522 049;4+21=25

B522 0495

K639 47;(6+3)*7=63

K639 473;6; (3*7)+10-(K0)=31

K639 473;6+31=37

K639 4737

You might want to double check it to make sure I did it all right. But that is the idea. The only problem I have with this method is it can't be used for the other puzzle. We can't be 100% sure of W456 because we don't have a sample of W4xx. (and I am 99% positive that when the method is found to solve the first part of the code it will work across the board).

The two Code Digit Four solutions both look convincing , especially using the x7 multiplyer for the letters being that little bit easier.( as suggested by militious )

For Code Digits One and Two , your "Loop Hole" theory works providing you have the relevant combination in your database to work from ..but as you have already pointed out , if a combination is not already in the database then it is unanswerable at the moment.

A few thoughts re Code Digit One and Two

As stated at the start of this second part of this puzzle, we initially thought that the overall problem (Boxes A - Z ) was based on two different sets of theories . Boxes A - M and Boxes N - Z.

On looking further into it , the Code Digits One and Two methods in the previous puzzle was true for ALL the boxes M-Z whereas in this second puzzle (as Brillo mentioned) there could be at least three different methods used to calculate Code Digits One and Two ..... Box A , Boxes B-I(part) and Boxes I(part) - M

[Guest]

Here is the database for Box Letter + Box Digit one = Code Digits One and Two showing the pattern/regular sequences of Code Digits one and Two

The ( brackets) denote missing code digits from the original database , but I have substituted the logical number into them for that particular sequence

From C5 onwards there are regular patterns that can be seen (as noted )

A1 - 5 start out as a normal sequence 00 - 04....then changes

A0 (09) All A = Code Digit One=1. All 0 =Code Digit Two = 9

A1 00

A2 01

A3 (02)

A4 03

A5 04 Boxes A B C (up to C5 ) showing no recognisable sequence

A6 ( )

A7 ( )

A8 03

A9 07

B0 09

B1 06

B2 07

B3 12

B4 ( )

B5 04

B6 03

B7 ( )

B8 07

B9 12

C0 19

C1 07

C2 17

C3 06

C4 ( )

C5 09

C6 25…Difference of nine descending START

C7 (16)

C8 07….Difference of nine STOP

C9 27…Difference of eight descending START

D0 (19)

D1 11

D2 03…Difference of eight STOP

D3 28…Difference of seven descending START

D4 21

D5 14

D6 07

D7 00…Difference of seven STOP

D8 31…Difference of six descending START

D9 25

E0 19

E1 13

E2 07…Difference of six STOP

E3 (01 )…Given as 29 …but I reckon it should be 01 or 44 (six stop or five start)

E4 39….Difference of five descending START (or maybe start at E3 )

E5 34

E6 29

E7 24

E8 (19)

E9 14

F0 09

F1 04..Difference of five STOP

F2 51..Difference of four descending START

F3 47

F4 43

F5 39

F6 ( )..Given as 51 …but I reckon should be 35

F7 31

F8 (27)

F9 23

G0 19

G1 15

G2 11

G3 07

G4 (03 )….Difference of four STOP

G5 64….....Difference of three descending START ?

G6 61

GT 58

G8 55

G9 (52)

H0 49

H1 46

H2 43

H3 (40)

H4 (37)

H5 34

H6 31

H7 28

H8 (25)

H9 22

I0 (19)

I1 16

I2 13

I3 10

I4 07

I5 (04)

I6 01….Difference of three STOP

I7 85….Difference of two descending START

I8 83

I9 81

JO (79)

J1 77

J2 75

J3 73

J4 (71)

J5 69

J6 67

J7 65

J8 63

J9 61

K0 59

K1 57

K2 55

K3 (53)

K4 51

K5 49

K6 47

K7 45

K8 (43)

K9 (41)

LO 39

L1 37

L2 35

L3 33

L4 31

L5 29

L6 27

L7 25

L8 23

L9 21

M0 (19)

M1 17

M2 15

M3 (13)

M4 11

M5 09

M6 07

M7 05

M8 03

M9 01… Difference of two STOP

Here is the separate database for Box Letter + Box Digit One = Code Digits One and Two

The ( brackets) denote missing code digits, but I have substituted the logical number into some of them for that particular sequence.

Can anyone help with Boxes A B and C and provide answers for the remaining gaps? ( or deduce the method used for calculating Code Digits One and Two...in case it is not just a basic look-up table for calculating these two digits.)

All ideas most welcome !

[Guest]

Here is the separate database for Box Letter + Box Digit One = Code Digits One and Two

Here is the database for Box Letter + Box Digit one = Code Digits One and Two showing the pattern/regular sequences of Code Digits one and Two

The ( brackets) denote missing code digits from the original database , but I have substituted the logical number into them for that particular sequence

From C5 onwards there are regular patterns that can be seen (as noted )

A1 - 5 start out as a normal sequence 00 - 04....then changes

A0 (09) All A = Code Digit One=1. All 0 =Code Digit Two = 9

A1 00

A2 01

A3 (02)

A4 03

A5 04 Boxes A B C (up to C5 ) showing no recognisable sequence

A6 ( )

A7 ( )

A8 03

A9 07

B0 09

B1 06

B2 07

B3 12

B4 ( )

B5 04

B6 03

B7 ( )

B8 07

B9 12

C0 19

C1 07

C2 17

C3 06

C4 ( )

C5 09

C6 25…Difference of nine descending START

C7 (16)

C8 07….Difference of nine STOP

C9 27…Difference of eight descending START

D0 (19)

D1 11

D2 03…Difference of eight STOP

D3 28…Difference of seven descending START

D4 21

D5 14

D6 07

D7 00…Difference of seven STOP

D8 31…Difference of six descending START

D9 25

E0 19

E1 13

E2 07…Difference of six STOP

E3 (01 )…Given as 29 …but I reckon it should be 01 or 44 (six stop or five start)

E4 39….Difference of five descending START (or maybe start at E3 )

E5 34

E6 29

E7 24

E8 (19)

E9 14

F0 09

F1 04..Difference of five STOP

F2 51..Difference of four descending START

F3 47

F4 43

F5 39

F6 ( )..Given as 51 …but I reckon should be 35

F7 31

F8 (27)

F9 23

G0 19

G1 15

G2 11

G3 07

G4 (03 )….Difference of four STOP

G5 64….....Difference of three descending START ?

G6 61

GT 58

G8 55

G9 (52)

H0 49

H1 46

H2 43

H3 (40)

H4 (37)

H5 34

H6 31

H7 28

H8 (25)

H9 22

I0 (19)

I1 16

I2 13

I3 10

I4 07

I5 (04)

I6 01….Difference of three STOP

I7 85….Difference of two descending START

I8 83

I9 81

JO (79)

J1 77

J2 75

J3 73

J4 (71)

J5 69

J6 67

J7 65

J8 63

J9 61

K0 59

K1 57

K2 55

K3 (53)

K4 51

K5 49

K6 47

K7 45

K8 (43)

K9 (41)

LO 39

L1 37

L2 35

L3 33

L4 31

L5 29

L6 27

L7 25

L8 23

L9 21

M0 (19)

M1 17

M2 15

M3 (13)

M4 11

M5 09

M6 07

M7 05

M8 03

M9 01… Difference of two STOP

There were a couple of possibly incorrect entries mentioned in the above database.

This could be due to typo errors in the original listing ....but it is also rumoured that the originator inserted a few "wrong" codes as red herrings in order to spice up the decoding exercise. I had already excluded at least three that had been noticed prior to posting here ...and it looks as though there are a couple more in the Code Digit One/Two database above that don`t follow the rules .

note: If there were any typos then they were definitely in the orginal listing as all the entries on here were checked prior to posting here.

As if it wasn`t hard enough anyway !

[Guest]

There were a couple of possibly incorrect entries mentioned in the above database.

This could be due to typo errors in the original listing ....but it is also rumoured that the originator inserted a few "wrong" codes as red herrings in order to spice up the decoding exercise. I had already excluded at least three that had been noticed prior to posting here ...and it looks as though there are a couple more in the Code Digit One/Two database above that don`t follow the rules .

note: If there were any typos then they were definitely in the orginal listing as all the entries on here were checked prior to posting here.

As if it wasn`t hard enough anyway !

I thought there might be some errors as I also made a list of the code digits 1+2 digit and there were a couple of digits that didnt really fit in.It`s just the early box letters (A/B/C)that dont seem to follow any rigid pattern.Am wondering if there might be more errors there as well

[Guest]

I've been looking through it and I agree with all the formula so far but I think Toxic has it a little bit wrong.

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

0 7 4 1 8 5 2 9 6 3 0 7 4 1 8 5 2 9 6 3 0 7 4 1 8 5

I think to decode this puzzle box it relies heavily on the 7 times multiplier, we have seen it about 3 other times or so.

that being said.

last digit of ... (3rd digit * 7) + (number equiv to letter) + carryover from 3rd box

it's basicly the same as toxic's except there is no adding 10 to it and it does make a different in some of the answers if we need to use the carry over from this.

to me it makes more sense and is a little bit cleaner.

I do like this version for working out code digit 4 as it is easier to use being an addition instead of subtraction method.

[Guest]

I do like this version for working out code digit 4 as it is easier to use being an addition instead of subtraction method.

Dido. When I figured out the original with multiples of 3, I failed to realise that (10-3)=7 , (10-6)=4 , (10-9)=1 ..... all multiples of 7.

[Guest]

Dido. When I figured out the original with multiples of 3, I failed to realise that (10-3)=7 , (10-6)=4 , (10-9)=1 ..... all multiples of 7.

I missed that as well despite making a list of the 3 , 7 and 9 multipliers and failing to notice that the *7 list is a reversal of the *3 list.

3 : 3 6 9 2 5 8 1 4 7 0

7 : 7 4 1 8 5 2 9 6 3 0

I`m pleased that the theory is virtually complete and using the same *7 mulltiplier is the common thread for both Code Digits Three and Four

This has been an intriguing exercise , especially now we`re near the very end.

However , I still can not see too much of a pattern trying to deduce the Code Digit One and Two from Box Letter and Box Digit One .

It looks like a straightforward lookup table in this A - M section (but not as straightforward/obvious as it was in the N - Z part ) ..but there are still a few missing digits , especially for a few in Box A , B & C section .(A6 A7 B4 B7 C4 ?)

Any thoughts on these last few ....just to finish it off?