Codebreaker Wanted~Find the diamonds

[redrooster]

I have been attempting this very difficult codebreaker problem for the past few weeks and can get no further than finding the answer for just the first digit of the code.

Can anyone on here help either with suggestions or answers?

This was set as a holiday project for University Engineering Students and as far as I know , nobody managed to solve it !

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


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.. Many ( but not all ) of the storage boxes contains small bags of precious stones (diamonds etc.) so the total value of all the contents of the locked boxes in this room will be several million of ££s .

Each box is labelled with a letter and four digits and the thief has inside information and knows that the combination lock code for each individual box can be mathematically calculated using the Box Number .

He also knows that the only digits used for the code will be 1 , 2 , 3 , 4 and 5.

Eg ...Box No J 0023 has a Combination Lock Code of 1215

However , each box has its 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.

His inside information has told him that there is one particular box containing the master list of what is contained in each box in the strong room...
Box J 6383 ... so this is the box to open first.

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

Can you work out how the codes can be calculated and explain the method used to determine the combination lock code of
Box J 6383 ?

.BOX = CODE... BOX = CODE... BOX = CODE

J 0023 = 1215... E 2182 = 5225... Z 6163 = 1445
H 0124 = 4441... Z 2271 = 1552 ... J 6242 = 3545
C 0138 = 4513... K 2380 = 5322 ... A 6322 = 4525
Y 0209 = 5554... A 2408 = 4124 ... K 6481 = 3513
E 0215 = 3235... Z 2494 = 4221 ... J 6519 = 1525
H 0262 = 1234... Z 2812 = 5332 ... K 6620 = 3341
R 0448 = 5132... A 2994 = 4425 ... J 6709 = 5355
R 0464 = 2135... C 3010 = 5414 ... E 6999 = 4515
W 0496 = 2125 ... Z 3229 = 3321 ... A 7238 = 3222
R 0545 = 2545 ... D 3631 = 1152 ... E 7402 = 5454
A 0704 = 5122 ... L 3865 = 2545 ... A 7535 = 3555
A 0859 = 4453... A 4089 = 4455 ... A 7713 = 5542
R 0871 = 5545 ... Z 4108 = 2414 ... A 7962 = 4431
Y 0926 = 4254 ... E 4111 = 3222 ... U 8056 = 5543
Z 1043 = 4455 ... A 4166 = 3325 ... E 8108 = 2332
V 1044 = 3442 ... K 4431 = 3552 ... E 8121 = 2522
U 1095 = 2245 ... A 4476 = 4343 ... E 8250 = 4531
U 1096 = 1142 ... S 4705 = 4133 ... Z 8596 = 1335
V 1192 = 4521 ... A 4707 = 2455 ... E 8391 = 3232
R 1244 = 1355 ... Z 4725 = 2325 ... A 8844 = 5432
H 1381 = 4255 ... J 5264 = 4524 ... A 8930 = 4545
E 1390 = 4251 ... Z 5271 = 1225 ... A 9024 = 5153
R 1479 = 1523 ... J 5362 = 5535 ... K 9139 = 3443
K 1533 = 5244 ... E 5567 = 3123 ... K 9356 = 2424
J 1853 = 5322 ... Z 5970 = 5452 ... A 9751 = 3242
A 6053 = 3245 ... Z 9800 = 3134


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


The only hints I can offer are that the letters may not be involved in the calculations (but I`m not 100% sure ) and also each digit of the code may be solved separately. ..so there might be four separate methods to calculate each digit of the code (but they may be interlinked).

Good luck !

redrooster

 

[regor60]

A further hint is that there can be only 625 different combinations, so there will be repetitions. For example, Z 1043 and A 4089 have the same combination, 4455, so I would focus on that path.

 

[davee123]

It's pretty open-ended, so I have to admit I'm a little reluctant to bother with it.

If the plausible numbers were 0-5, then I might think it has something to do with a conversion to base 5. It could admittedly be a conversion to base 4, with +1 to each digit, but that seems unlikely to me (not that I've tried that idea).

In light of that, my guess would be that since each digit only has 5 possibilities, there's a good chance that each digit is derived independently. That seems a bit odd, though-- it would mean that whatever function is being performed, there are 5 different possible outcomes for each digit. I could see 3 with "greater than, less than, equal to", but 5 seems strange. Not that it's impossible by any means, but that it's open to more arbitrarity.

Anyway, if you assume that each digit is derived independently, then we know that at least ONE digit is derived using ONLY the 4-digit code and (possibly) the letter. Hence, order the list in 4 different ways, according to each of the 4 digits, and see if any patterns emerge with the corresponding 4-digit codes/letters.

Oh, and given the sheer volume of examples that are provided, it seems that the problem is designed to be EXTREMELY HARD-- so I'm thinking that a complex and convoluted expression is being used. Honestly, it makes me think that solving this would involve a lot of time spent just guessing and having a lot of trial-and-error, rather than making headway, so I'm not jumping at solving it, short of a small foray. Perhaps I'll test the "base 4" theory, but beyond that, I have to say it's a little daunting.

DaveE

 

[redrooster]

It's pretty open-ended, so I have to admit I'm a little reluctant to bother with it.

If the plausible numbers were 0-5, then I might think it has something to do with a conversion to base 5. It could admittedly be a conversion to base 4, with +1 to each digit, but that seems unlikely to me (not that I've tried that idea).

In light of that, my guess would be that since each digit only has 5 possibilities, there's a good chance that each digit is derived independently. That seems a bit odd, though-- it would mean that whatever function is being performed, there are 5 different possible outcomes for each digit. I could see 3 with "greater than, less than, equal to", but 5 seems strange. Not that it's impossible by any means, but that it's open to more arbitrarity.

Anyway, if you assume that each digit is derived independently, then we know that at least ONE digit is derived using ONLY the 4-digit code and (possibly) the letter. Hence, order the list in 4 different ways, according to each of the 4 digits, and see if any patterns emerge with the corresponding 4-digit codes/letters.

Oh, and given the sheer volume of examples that are provided, it seems that the problem is designed to be EXTREMELY HARD-- so I'm thinking that a complex and convoluted expression is being used. Honestly, it makes me think that solving this would involve a lot of time spent just guessing and having a lot of trial-and-error, rather than making headway, so I'm not jumping at solving it, short of a small foray. Perhaps I'll test the "base 4" theory, but beyond that, I have to say it's a little daunting.

DaveE

I don`t think it is going to be as complicated as that.

I think that a large number of the examples are there as a bit of a smokescreen as you are expected to pick out the necessary examples to logically work out the answer.

I`ve managed to work out the first Code digit using a combination of the Box number digits and then converted that straightforward calculation to the correct Code digit using a simple series.

I have worked out similar codebreaking problems in the past and they have always used basic maths without having to resort to anything too complicated.(no binary , hex , logs etc) The deduction of any answers has been down to logic + maths and did not involve any guessing or trial and error.

For example the series 7 4 1 8 5 2 9 6 3 0 has been used previously to translate digits 1 2 3 4 5 6 7 8 9 0 ...this series being derived from a x7 multiplication and using the units only.

Similar examples involve the x3 and x9 multiples.

 

[redrooster]

I don`t know whether anyone is still looking at this codebreaker problem...so here are the method and answer that I worked out for the FIRST Code Digit

Spoiler

Method used

After making a list of all the Codes that began with the SAME FIRST Code digit it was noticed that adding the 2nd + 3rd + 4th Box Number digits together gave totals that were related.

ie...If a multiple of 5 was subtracted from this total , the answer always gave the same FINAL digit.(but this was NOT the first Code Digit)

Example :Using a FIRST Code digit of 4

H0124 = 4441 Adding 1+2+4 = 7 minus (1x5) =2
C0138 = 4513 Adding 1+3+8 = 12 minus (2x5) = 2
Z2494 = 4221 Adding 4+9+4 = 17 minus (3x5) = 2
A2994 = 4425 Adding 9+9+4 = 22 minus (4x5) = 2
E6999 = 4515 Adding 9+9+9 = 27 minus (5x5) = 2

So it can be seen that if the initial total is 7 or 12 or 17 or 22 or 27 the final total (after subtracting a multiple of 5) is always 2 ...and this always corresponds to a FIRST Code Digit of 4

(Remember that the Code Number consists of numbers 1, 2, 3, 4, and 5 so this is why the "subtraction of multiples of 5 " is relevant .)

Similarly applying the same method to other Box Numbers/Code Numbers with different FIRST code digits of 1, 2, 3 and 5, from the list produces the following table

If the FINAL total digit = 1 the FIRST Code digit is 5
if 2 = 4
if 3 = 3
if 4 = 2
if 0 = 1

 

[mugaliens]

I grew quite tired of the poorly presented data and decided to codify it in standard, comma-delimited format:

Box_Ltr , Char_num, Box_Num , Box_Code
A , 1 , 704 , 5122
A , 1 , 859 , 4453
A , 1 , 2408 , 4124
A , 1 , 2994 , 4425
A , 1 , 4089 , 4455
A , 1 , 4166 , 3325
A , 1 , 4476 , 4343
A , 1 , 4707 , 2455
A , 1 , 6053 , 3245
A , 1 , 6322 , 4525
A , 1 , 7238 , 3222
A , 1 , 7535 , 3555
A , 1 , 7713 , 5542
A , 1 , 7962 , 4431
A , 1 , 8844 , 5432
A , 1 , 8930 , 4545
A , 1 , 9024 , 5153
A , 1 , 9751 , 3242
C , 3 , 138 , 4513
C , 3 , 3010 , 5414
D , 4 , 3631 , 1152
E , 5 , 215 , 3235
E , 5 , 1390 , 4251
E , 5 , 2182 , 5225
E , 5 , 4111 , 3222
E , 5 , 5567 , 3123
E , 5 , 6999 , 4515
E , 5 , 7402 , 5454
E , 5 , 8108 , 2332
E , 5 , 8121 , 2522
E , 5 , 8250 , 4531
E , 5 , 8391 , 3232
H , 8 , 124 , 4441
H , 8 , 262 , 1234
H , 8 , 1381 , 4255
J , 10 , 23 , 1215
J , 10 , 1853 , 5322
J , 10 , 5264 , 4524
J , 10 , 5362 , 5535
J , 10 , 6242 , 3545
J , 10 , 6519 , 1525
J , 10 , 6709 , 5355
K , 11 , 1533 , 5244
K , 11 , 2380 , 5322
K , 11 , 4431 , 3552
K , 11 , 6481 , 3513
K , 11 , 6620 , 3341
K , 11 , 9139 , 3443
K , 11 , 9356 , 2424
L , 12 , 3865 , 2545
R , 18 , 448 , 5132
R , 18 , 464 , 2135
R , 18 , 545 , 2545
R , 18 , 871 , 5545
R , 18 , 1244 , 1355
R , 18 , 1479 , 1523
S , 19 , 4705 , 4133
U , 21 , 1095 , 2245
U , 21 , 1096 , 1142
U , 21 , 8056 , 5543
V , 22 , 1044 , 3442
V , 22 , 1192 , 4521
W , 23 , 496 , 2125
Y , 25 , 209 , 5554
Y , 25 , 926 , 4254
Z , 26 , 1043 , 4455
Z , 26 , 2271 , 1552
Z , 26 , 2494 , 4221
Z , 26 , 2812 , 5332
Z , 26 , 3229 , 3321
Z , 26 , 4108 , 2414
Z , 26 , 4725 , 2325
Z , 26 , 5271 , 1225
Z , 26 , 5970 , 5452
Z , 26 , 6163 , 1445
Z , 26 , 8596 , 1335
Z , 26 , 9800 , 3134

 

[mugaliens]

I could provide leading zeros for the numbers less than four digits, but I'm tired, and Excel isn't that accomodating.

 

[hamster143]

I could provide leading zeros for the numbers less than four digits, but I'm tired, and Excel isn't that accomodating.

Format cells -> number -> category: custom -> type: 0000.

 

[redrooster]

OBSERVATIONS..This is assuming that the prefix LETTER is not part of any calculation

(i) Changing the FIRST Box Digit affects the SECOND , THIRD & FOURTH Code Digits only (and not the FIRST Code Digit)

So this means that the FIRST Code Digit is dependent on the SECOND,THIRD & FOURTH Box Digits only

Z 5271 = 1225
Z 2271 = 1552

(ii) Changing the SECOND Box digit affects ALL four Code Digits

V 1044 = 3442
R 1244 = 1335

(iii ) Changing the THIRD Box digit affects ALL four Code Digits

S 4705 = 4133
S 4725 = 2325

(iv) Changing the FOURTH Box digit affects ALL four Code Digits

S 4705 = 4133
A 4707 = 2455

note...the examples below show that it only affects three of the four Code Digits

Z 1043 = 4455
V 1044 = 3442

U 1095 = 2245
U 1096 = 1142


From the above (ii), (iii) and (iv) it looks as though Code Digits TWO , THREE and FOUR will involve using all the four Box Number digits in any calculations

Finally

(v) Different Boxes having the SAME CODE

E 4111 = 3222
A 7238 = 3222

A 4089 = 4455
Z 1043 = 4455

 

[mugaliens]

Format cells -> number -> category: custom -> type: 0000.

Thanks, hamster! Here you go, with the comma-delimiter replaced with space-delimitation (easier to import back into Excel and most other programs):

A 01 0704 5122
A 01 0859 4453
A 01 2408 4124
A 01 2994 4425
A 01 4089 4455
A 01 4166 3325
A 01 4476 4343
A 01 4707 2455
A 01 6053 3245
A 01 6322 4525
A 01 7238 3222
A 01 7535 3555
A 01 7713 5542
A 01 7962 4431
A 01 8844 5432
A 01 8930 4545
A 01 9024 5153
A 01 9751 3242
C 03 0138 4513
C 03 3010 5414
D 04 3631 1152
E 05 0215 3235
E 05 1390 4251
E 05 2182 5225
E 05 4111 3222
E 05 5567 3123
E 05 6999 4515
E 05 7402 5454
E 05 8108 2332
E 05 8121 2522
E 05 8250 4531
E 05 8391 3232
H 08 0124 4441
H 08 0262 1234
H 08 1381 4255
J 10 0023 1215
J 10 1853 5322
J 10 5264 4524
J 10 5362 5535
J 10 6242 3545
J 10 6519 1525
J 10 6709 5355
K 11 1533 5244
K 11 2380 5322
K 11 4431 3552
K 11 6481 3513
K 11 6620 3341
K 11 9139 3443
K 11 9356 2424
L 12 3865 2545
R 18 0448 5132
R 18 0464 2135
R 18 0545 2545
R 18 0871 5545
R 18 1244 1355
R 18 1479 1523
S 19 4705 4133
U 21 1095 2245
U 21 1096 1142
U 21 8056 5543
V 22 1044 3442
V 22 1192 4521
W 23 0496 2125
Y 25 0209 5554
Y 25 0926 4254
Z 26 1043 4455
Z 26 2271 1552
Z 26 2494 4221
Z 26 2812 5332
Z 26 3229 3321
Z 26 4108 2414
Z 26 4725 2325
Z 26 5271 1225
Z 26 5970 5452
Z 26 6163 1445
Z 26 8596 1335
Z 26 9800 3134

 

[redrooster]

Is anyone still working on this problem ?

If so , are you getting anywhere with it?

Any suggestions or ideas most welcome.

 

[RazorRose]

I'm still working on it... but I'm like way behind you atm lol.
(actually i just started working on it.)

 

[cronxeh]

Z 1043 = 4455

The only coincidence I found is that 1043 in base 10 is same as 4455 in base 6. I don't know if its the nature of number 1043, but certainly other numbers under Z don't follow the same conversion.

 

[redrooster]

I don`t know whether anyone is still working on this Codebreaking problem.

I am having absolutely no success in working it out.
This was set as a project for a group of Engineering Students in 2008 ..and only one student managed to crack it. (The method used was not divulged)
The same problem was set again in 2009 for another group of students , and to make it easier , the creator added a further list of clues to the original list.

I`ve managed to obtain a copy of this second list , so if anyone is interested then here it is.

Let me know if anyone has any suggestions.
BOX...CODE

R 0040 = 2551
A 0158 = 2255
E 0843 = 1455
Z 0869 = 3324
C 1071 = 3252
Q 1167 = 2232
R 1335 = 5153
Z 1795 = 5452
A 2356 = 2324
A 2593 = 4553
A 3643 = 3411
E 3695 = 1545
A 4163 = 1452
A 4953 = 4531
A 5209 = 5121
A 5454 = 3255
E 5862 = 5223
J 6998 = 5235
Z 8101 = 4214
J 8102 = 3155
E 9153 = 2515
A 9396 = 3354
J 9774 = 3253

 

[redrooster]

EXTRA CLUES


Is anyone still working on this difficult Codebreaker problem ?

If so , then you may be interested in the following.

The original compiler of this problem has admitted that this is a difficult one to solve , so he has given a few extra clues for the series of Boxes C0200 > C 0209

Code Digit FOUR (C4) looks interesting.


C 0200 = 4134
C 0201 = 3555
C 0202 = 2551
C 0203 = 1442
C 0204 = 5423
C 0205 = 4314
C 0206 = 3245
C 0207 = 2231
C 0208 = 1133
C 0209 = 5554

Also , it is 99% certain that the first letter is not relevant ...so is no part of the solution

 

[cronxeh]

The suspense is killing me :grumpy: