I need help breaking this code.

[scguy]

My maths teacher gave me this code that is based on an alphabetic numbering system, like A=1, B=2 etc. It is then encoded using a quadratic i dont know where to begin

 

(70406535) (6825358770) (23136720258287) (175202) (70406535)

 

(15026273520253570) (40673542535) (26236742535582) (735202582)

 

(447175202535). (7040636787) (36787) (150106175175490).

 

Each bracket represents a word.

[1veedo]

there are 26 letters in the alphebet.

 

I think a good bet would be that 71 = 7 - 1, two letters. But 13 = only one letter.

 

Maybe they're even in order, but that would make it too hard. And then again, maybe 42 is actually aquin to 21?

[Dak]

THIS WAS FUN TO BREAK, SO IV REPOSTED THE CODE IN THE BRAINTEASERS FORUM. ANYONE WHO WISHES TO SOLVE IT FOR THEMSELVES SHOULD NOT READ THIS POST OR THE NEXT ONE

 

lol, i always wanted to be a spy when i was a kid, so i recon i could give this code breaking malakay a go. i hope, by your mentioning of the word 'quadratic' and 'maths teacher', your not expecting me to use maths tho. im gonna rite everystep so i can cum bak and change it if its wrong and so u can see how it was done, just skip to the end for the answre if thats all your after.

 

(70406535) (6825358770) (23136720258287) (175202) (70406535)

 

(15026273520253570) (40673542535) (26236742535582) (735202582)

 

(447175202535). (7040636787) (36787) (150106175175490).

 

--------------------------------------------------------------------

 

look for common repititions and seperate them out. eg, 202 occour alot and 02 and/or 20 dont occour atall outside of 202, so assume that 202 is one letter. this gives

 

(70406535) (6825358770) (231367 202 58287) (175 202) (70406535)

 

(150262735 202 53570) (40673542535) (26236742535582) (735 202 582)

 

(447175 202 535). (7040636787) (36787) (150106175175490).

 

letters:202

 

----------------------------------------------------------------------

 

note word four has been split into 175 202. a chek reveals that there are no 17s or 75s outside 175, so assume that 175 is a letter to get

 

(70406535) (6825358770) (231367 202 58287) (175 202) (70406535)

 

(150262735 202 53570) (40673542535) (26236742535582) (735 202 582)

 

(447 175 202 535). (7040636787) (36787) (150106 175 175 490).

 

letters:

202

175

 

-----------------------------------------------------------------------

 

this now isolates two interesting triplets: 535 and 735. note that every time 35 occours, it is preceded by either 5 or 7 (note that it is not correct to assume that all numbers are represented as triplets, as not all words contain a number of numbers that is divisible by three). so treating 535 and 735 as letters gives

 

(70406 535) (682 535 8770) (231367 202 58287) (175 202) (70406 535)

 

(150262 735 202 535 70) (406 735 42 535) (26236742 535 582) (735 202 582)

 

(447 175 202 535). (7040636787) (36787) (150106 175 175 490).

 

letters:

202

175

735

535

 

---------------------------------------------------------------------

 

hmm. words one and five start with 70406 and end with 535. word 11 starts with 70406, and ends with 36787 -- which is what word 12 is. this makes 70406 a likely candidate for TH, 535 a likely candidate for E, and 36787 a likely candidate for IS (the only 2-letter word i can think of that can be put on the end of TH to make a word), so:

 

 

(70406 535) (682 535 8770) (231367 202 58287) (175 202) (70406 535)

 

(150262 735 202 535 70) (406 735 42 535) (26236742 535 582) (735 202 582)

 

(447 175 202 535). (70406 36787) (36787) (150106 175 175 490).

 

letters:

202

175

735

535 (E?)

70406 (TH?)

36787 (IS?)

 

-------------------------------------------------------------------

 

if 70406 is two letters (which seems likely, as its probably TH, plus its alot longer than the other codes), a reasanoble split would be 70 406, as 70 has already been isolated on the end of word six, and 406 at the beginning of word seven

 

(70 406 535) (682 535 87 70) (231367 202 58287) (175 202) (70 406 535)

 

(150262 735 202 535 70) (406 735 42 535) (26236742 535 582) (735 202 582)

 

(447 175 202 535). (70 406 36787) (36787) (150106 175 175 490).

 

letters:

202

175

735

535 (E?)

70 (T?)

406 (H?)

36787 (IS?)

 

---------------------------------------------------------------------

 

150, 262, 367 and 87 can also be isolated

 

(70 406 535) (682 535 87 70) (231 367 202 582 87) (175 202) (70 406 535)

 

(150 262 735 202 535 70) (406 735 42 535) (262 36742 535 582) (735 202 582)

 

(447 175 202 535). (70 406 367 87) (367 87) (150 106 175 175 490).

 

letters:

202

175

735

535 (E?)

70 (T?)

406 (H?)

367 (I?)

87 (S?)

150

262

 

--------------------------------------------------------------------

 

this leaves us with the following terms, in bold, which we are not sure about

 

 

(70 406 535) (682 535 87 70) (231 367 202 582 87) (175 202) (70 406 535)

 

(150 262 735 202 535 70) (406 735 42 535) (262 367 42 535 582) (735 202 582)

 

(447 175 202 535). (70 406 367 87) (367 87) (150 106 175 175 490).

 

letters:

202

175

735

535 (E?)

70 (T?)

406 (H?)

367 (I?)

87 (S?)

150

262

 

------------------------------------------------------------------------

 

well, 582 and 682 both end the same, so we dont know wether its 682 and 582, or 6 5 and 82, so we will use a cryptographical technique known as 'ignoring them'. likewize, 447 and 490 get ignored for now as, with only one occourance of each, it is inposible to know wether they are one, two or three words. we canot tell where the split in 231367 comes, so sod it for now.

 

(70 406 535) (682 535 87 70) (231 367 202 582 87) (175 202) (70 406 535)

 

(150 262 735 202 535 70) (406 735 42 535) (262 367 42 535 582) (735 202 582)

 

(447 175 202 535). (70 406 367 87) (367 87) (150 106 175 175 490).

 

letters:

202

175

735

535 (E?)

70 (T?)

406 (H?)

367 (I?)

87 (S?)

150

262

 

---------------------------------------------------------------------

 

ok, onto stage two - working out what the actual numbers mean. count up the occourances of the letters in the code to get their frequencies, useful cos most frequent usually = E. also note double-letters, and if any letter usually follows another, as these can be useful.. also note that a two letter word is formed by 175 and 202, thus one of them must be a vowel; same with 367-87

 

 

letters: occourances: notes:

202:.............5.........................(vowel?)

175:..............4..................(double letter)(vowel?)

735:.............3

535:..............7(E?)(most common)(follows 70 406 alot)(never starts a word)

70:................5...................(T?)(followed by 406 alot)

406:..............4.................(H?)(preceded by 70 lot)

367:..............4.................(I?)(forms 2 letter word with 87, thus

87:................4................(S?)either 367 or 87 = vowel)(3/4 end a word)

150:.............2...................(both start words)

262:.............2

either 202 or 175 = vowel

either 367 or 87 = vowel

 

lets go ahead and assume that 535 = E, as there seems to be a lot of indicators that this is the case (bold still = undecided numbers)

 

(70 406 e) (682 e 87 70) (231 367 202 582 87) (175 202) (70 406 e)

 

(150 262 735 202 e 70) (406 735 42 e) (262 367 42 e 582) (735 202 582)

 

(447 175 202 e). (70 406 367 87) (367 87) (150 106 175 175 490).

 

----------------------------------------------------------------

 

cleaning up our table of info gives

 

letters: occourances: notes:

202:.............5........................(vowel?)

175:.............4.............(double letter)(vowel?)

735:.............3

535:.............7..........................E

70:.............5.............(T?)(followed by 406 alot)

406:.............4.............(H?)(preceded by 70 lot)

367:.............4.............(I?)(forms 2 letter word with 87, thus

87:.............4.............(S?)either 367 or 87 = vowel)(3/4 end a word)

150:.............2.............(both start words)

262:.............2

either 202 or 175 = vowel

either 367 or 87 = vowel

 

now lets assume, du to the fact that they start off a three letter word that ends with e twice and are found together 3/4 times that 406 occours, that 70 = t and 406 = h

 

(t h e) (682 e 87 t) (231 367 202 582 87) (175 202) (t h e)

 

(150 262 735 202 e t) (h 735 42 e) (262 367 42 e 582) (735 202 582)

 

(447 175 202 e). (t h 367 87) (367 87) (150 106 175 175 490).

 

---------------------------------------------------------------------

 

letters: occourances: notes:

202:.............5..........................(vowel?)

175:.............4.............(double letter)(vowel?)

735:.............3

535:.............7...............................E

70:..............5.............................T

406:.............4.............................H

367:.............4....................(I?)(forms 2 letter word with 87, thus

87:.............4......................(S?)either 367 or 87 = vowel)(3/4 end a word)

150:.............2.................(both start words)

262:.............2

either 202 or 175 = vowel

either 367 or 87 = vowel

 

checking this list of two letter words confirms that 'is' is the only two letter word that can complete the word T H x y, so accept that 367 = i and 87 = s

 

(t h e) (682 e s t) (231 i 202 582 s) (175 202) (t h e)

 

(150 262 735 202 e t) (h 735 42 e) (262 i 42 e 582) (735 202 582)

 

(447 175 202 e). (t h i s) (i s) (150 106 175 175 490).

 

----------------------------------------------------------------------

 

checking the last reference again, and ignoring two letter words that contain t,h,e,i or s (as we already have assighned them), and all the two letter words that mean egyptian god etc and the ones that dont make sence followed by 'the', then 175 202 can = go, of, on, or, up

 

letters: occourances: notes:

202:.............5.............(vowel?)(o,f,n,r or p)

175:.............4.......(double letter)(vowel?)(g,o or u)

735:.............3

535:.............7................................E

70:...............5..............................T

406:.............4..............................H

367:.............4...............................I

87:...............4..............................S

150:.............2.................(both start words)

262:.............2

either 202 or 175 = vowel

175 202 can = go, of, on, or, up

 

note that 175 = g, o or u, and also that it exists as a double letter. as gg and uu are very uncommon, we can tell that 175 = o (told you that noting repititions would come in useful )

 

 

letters: occourances: notes:

202:.............5..........................(f,n or r)

175:.............4..............................O

735:.............3

535:.............7...............................E

70:...............5..............................T

406:.............4.............................H

367:.............4.............................I

87:.............4..............................S

150:.............2.............(both start words)

262:.............2

175 202 can = of, on, or

 

(t h e) (682 e s t) (231 i 202 582 s) (o 202) (t h e)

 

(150 262 735 202 e t) (h 735 42 e) (262 i 42 e 582) (735 202 582)

 

(447 o 202 e). (t h i s) (i s) (150 106 o o 490).

 

--------------------------------------------------------------------

 

letters: occourances: notes:

202:.............5..........................(f,n or r)

175:.............4..............................O

735:.............3

535:.............7...............................E

70:...............5..............................T

406:.............4.............................H

367:.............4.............................I

87:.............4..............................S

150:.............2.............(both start words)

262:.............2

175 202 can = of, on, or

 

 

now examine word ten, "447 o 202 e". we know that 202 = f, n or r, so word ten = -ofe, -one, or -ore. by asessing their suitibility simply by going "aofe: no, bofe:no cone:yes etc), we get the following possibilities for word ten:cofe, bone, cone, done, gone, hone, lone, zone, bore, core, gore, lore, more, pore; which is absolutely no use at the moment but could be useful later.

 

also, word 7 has an h. hs are always followed by a vowel other than y, and as we have used up e,i and o, 735 = a or u.

 

now i admit that this is branching off a little into non-standard-cryptography, but using this site we can assume for the moment that a logical completion to the phrase "this is **oo*", is "this is proof", and "h**e" is have. this turns "150 262 735 202 e t" into "p 262 a 202 e t" which, using the same site, gives either planet or placet. ill chose planet. this gives

 

(t h e) (682 e s t) (231 i 202 582 s) (o 202) (t h e)

 

(p l a n e t) (h a v e) (262 i 42 e 582) (735 202 582)

 

(447 o 202 e). (t h i s) (i s) (p r o o f).

 

i know that step was a bit dodgy, but it seems to make sence so far

 

---------------------------------------------------------------------

 

now our table reads

 

202: N

175: O

735: A

535: E

70: T

406: H

367: I

87: S

150: P

262: L

42: V

106: R

490: F

 

now to sub newly defined numbers in:

 

 

(t h e) (682 e s t) (231 i n 582 s) (o n) (t h e)

 

(p l a n e t) (h a v e) (l i v e 582) (a n 582)

 

(447 o n e). (t h i s) (i s) (p r o o f).

 

horray! tho i non-cryptographically 'solved' words 6, 7 and 13, the results of substituting letters --> numbers in the other words seems to make sence. huzza!

 

----------------------------------------------------------------------

 

note that words 8 and 9 are almost complete, bar the last letter, and also that this last letter is the same. using the crossword completer, the only viable single word that will complete both live- and an- is, rather obviously (but its worth checking), 'd' so 582 = d.

 

202: N

175: O

735: A

535: E

70: T

406: H

367: I

87: S

150: P

262: L

42: V

106: R

490: F

582: D

now to sub newly defined number in:

 

 

(t h e) (682 e s t) (231 i n d s) (o n) (t h e)

 

(p l a n e t) (h a v e) (l i v e d) (a n d)

 

(447 o n e). (t h i s) (i s) (p r o o f).

 

the remaining numbers are all different, and so the letters must be too. so the phrase reads

 

The best minds on our planet have lived and gone. This is proof.

 

dont savvy why thats proof, but there u go.

 

PHEW!

 

yay! i did it eat your heart out MI6, it took me f***ing ages, but i finally did it alri-ight, alri-ight, wahoo! IN YOUR FACE CODE, I BROKE YOU MWAHAHA, do do, do do, do - HEY, adodo do do, do - HAY

 

yay

[Dak]

just to completely trounce the code, i'll even do the math bit

 

code value:letter:postion in alphabet(x)

735:A:1

682:B:2

582: D :4

535:E:5

490:F:6

406:H:7

367:I:9

262:L:12

202:N:14

175:O:15

150:P:16

106:R:18

87:S:19

70:T:20

42:V:22

 

ok, so its quadratic:

 

ax^2 + bx + c = y

 

taking A (which should = 1) and B (which should = 2) gives

 

a + b + c = 735

4a + 2b + c = 682

 

divide second by 2 to get

 

a + b + c = 735

2a + b + 1/2c = 341

 

take second from first to get

 

1/2c -a = 394

 

now take D(which should be 4) and F(which should be 6) to get

 

16a + 4b + c = 582

36a + 6b + c = 490

 

times top by 3 and bottom by 2 to get

 

48a + 12b + 3c = 1746

72a + 12b + 2c = 980

 

take second from first to get

 

-24a + c = 766

 

put the two equasions together to get

 

1/2c -a = 394

c - 24a = 766

 

times top by 2 to get

 

c -2a = 788

c - 24a = 766

 

take first from second to get

 

-22a = -22

 

times by -1 and divide by 22 to get

 

a = 1

 

so, using the first dedused equasion

 

c -2a = 788

 

becomes

 

c - 2 = 788

 

add 2

 

c = 790

 

now take A again

 

a + b + c = 735

 

becomes

 

1 + b + 790 = 735

 

subtract 791

 

b = -56

 

so, the quadratic is

 

x^2 - 56x + 790 = y

 

where x = letters position in alphabet, and y = number that it is assighned in the code.

 

test:

 

Vs position in alphabet is 22, and its code is 42.

 

x^2 - 56x + 790 = y = 42

22^2 - (56*22) + 790 = y = 42

484 -1232 + 790 = y = 42

42 = y = 42

 

yeah, suck that code! i beat u

[RedAlert]

Wow he's smart isn't he?

[1veedo]

Never done that beofre. My math teachers always suck. I sleep all class, get writen up for "disturbing others," and still manage A's

 

That actually looks like a lot of fun. Maybe I'll teach myself some of that.

[RedAlert]

I know me too.

[BenSon]

That was seriously impressive ....

 

~Scott

[scguy]

Very impressive Dak:eek: , the teacher gave the answer and u were correct 100% i really am very surprised that it was broken by somebody. Maybe somebody could post another and i could write it down before it gets answered so i can do it myself

[scguy]

If anybody is still interested i got the system he used. It was a reverse numbering system of the alphabet ie Z=1 , Y=2 etc, it was then coded through the use of this quadratic: x^2+2x+7

[1veedo]

I dont get it then...

 

Y = 2 so

 

x^2+2x+7

4 + 4 + 7 = 15? I thought these were three diget numbers.

so, the quadratic is

 

x^2 - 56x + 790 = y

...

4 - 112 + 790

682?

 

Ok, anyway. If you have a message like that and the formular (ie, key), how would somebody (the entended reciever of the message) quickly find out what it is?

[Dak]

<-- read far too many spy books as a kid

 

Maybe somebody could post another and i could write it down before it gets answered so i can do it myself

sorry, didnt know you just wanted a clue.

[Newtonian]

Ha Dak you didnt get the meaning of the phrase though....

The teacher fully expected scguy to fail. And in his revealing the answer sought to both astonish and humble scguy.Hence the 'this is proof'.

Unfortunately reading his reply scguy doesnt seem to have milked it.I would have entered the class with crooked eyebrows,a sickeness grimace of evil sniggering,one side of my mouth creased with a sadistic grin.Planted the answer infront of the teacher without a glance and walked to my seat "mwuahahahahahahahahahhaha".

[RedAlert]

I agree with Newtonian.

[scguy]

I dont get it then...

 

Y = 2 so

 

x^2+2x+7

4 + 4 + 7 = 15? I thought these were three diget numbers....

4 - 112 + 790

682?

 

Ok' date=' anyway. If you have a message like that and the formular (ie, key), how would somebody (the entended reciever of the message) quickly find out what it is?[/quote']

 

 

Take B for example, thats 25^2+50+7= 625+57= 682. They are not all 3 digits, that is what makes it even harder.

[Dak]

its interesting how theres no potential for confusion...

 

take N for example, which is 202 in the code. now look at L, whos value is 262. there are no letters whos code value ends with 20... if there were, there could be confusion if plased before to L, eg imagine F was 120, then FL would be 120262, which has 202 in the middle, which would be confusing in a long line of numbers.

 

ie, 783251202625498654 could be interpreted as having an N in the middle,

 

eg 783251202625498654 or as having an FL in the middle,

 

eg 783251202625498654

 

but as F doesnt = 120, and as no letters code value ends with 20, this confusion does not arise and as far as i can see there are no two letters which can be plased next to each other to create a confusion...

 

is that something to do with the fact its a quadratic, or something to do with the numbers chosen for a b and c? or did the teacher just keep choosing different quadratics untill he found one which didnt result in confusion?

[scguy]

Hmm that is very interseting dak, now i have a question for u. How the heck would one go about solving this code or just finding the quadratic purely by mathematical means, is it even possible? Could there be some kind of procedure or logical way to do this or are the numbers in to many permutations thus requiring the use of a computer?

[Dak]

actually, if id been paying attention i could have worked it out as soon as id worked out two letters. eg, nowing what letters numbers Y and Z code for, then wed get:

 

ax² + bx + c = X

ax² + bx + c = Y

 

and we could plonk in the alphabetic value for the letters coded for by z and y as x, eg if y = a, x = 1; and if z = b, x = 2:

 

a + b + c = y

a4 + b2 + c = z

 

then just simultaniously solve them from there, work out the quadratic and then work out the code value of each word and so take it from there. but my aversion to maths made me do it the long winded way. (doh)

 

as for any purely mathematical way, ie with no cryptology involved whatsoever... i dunno. maybe if you PM dave and ask him nicely, hell move this to the mathmatics thread and the resident mathmatical geneuses can have a bash. (if they do it in, like, 2 seconds i will not be a happy bunny)