# 0000
CHALLENGE 0
> Let's test your knowledge on Caesar's methods
```
Naljnl, Pnrfne jnf n fxvyyrq pbzzhavpngbe, naq ur hfrq n inevrgl bs zrgubqf gb xrrc uvf zrffntrf frperg sebz uvf rarzvrf. Bar bs gurfr zrgubqf jnf gur Pnrfne pvcure, n fvzcyr grpuavdhr gb boshfpngr pbzzhavpngvbaf. SYNT{ebgngr_gung_nycunorg}
```
* The text we are provided is encrypted using [Caesar cipher](https://en.wikipedia.org/wiki/Caesar_cipher) more specifically [ROT13](https://en.wikipedia.org/wiki/ROT13).
* We can easily decode it using [Cyberchef](https://gchq.github.io/CyberChef) which is also known as the Cyber Swiss Army Knife.
* The flag is `FLAG{rotate_that_alphabet}`.
##
## CHALLENGE 1
> This one is not exactly caesar but it might be the key
##
## CHALLENGE 2
> We use our alphabet too frequently, let's try an alternative one
* On clicking on the link, we are met with some encrypted text.
```
PDV KLRBC IOEXQ AEY TLGMF EJVO PDV NSWH ZEU.
PDRF PVYP RF S MSQUOSG, XDRBD GVSQF PDSP RP BEQPSRQF SNN 26 NVPPVOF EA PDV VQUNRFD SNMDSIVP. PDRF GSCVF RP RZVSN AEO AOVKLVQBH SQSNHFRF, SF PDV BOHMPSQSNHFP BSQ BEGMSOV PDV AOVKLVQBH EA NVPPVOF RQ PDV BRMDVOPVYP PE PDV CQEXQ AOVKLVQBH EA NVPPVOF RQ PDV VQUNRFD NSQULSUV.
AEO VYSGMNV, PDV GEFP BEGGEQ NVPPVO RQ PDV VQUNRFD NSQULSUV RF V. RA PDV GEFP BEGGEQ NVPPVO RQ PDV BRMDVOPVYP RF Y, PDVQ PDV BOHMPSQSNHFP BSQ SFFLGV PDSP Y RF NRCVNH PE IV S FLIFPRPLPREQ AEO V.
EPDVO BEGGEQ NVPPVOF RQ PDV VQUNRFD NSQULSUV RQBNLZV P, S, E, R, Q, F, SQZ D. PDV BOHMPSQSNHFP BSQ LFV PDRF RQAEOGSPREQ PE GSCV VZLBSPVZ ULVFFVF SIELP PDV EPDVO FLIFPRPLPREQF RQ PDV BRMDVOPVYP.
ANSU{QEX_RJV_NVSOQVZ_GH_SIBF}
```
* This text is encrypted using Substitution cipher.
* In order to decrypt it we are supposed to substitute the characters based on their frequency.
* As we can see, the letter that has the highest frequency in the given message will be substituted with `E`, the one with the second highest frequency will be substituted with `T` and so on and so forth.
* We can also perform substitution based on known words. For example, we know that the flag format should be `FLAG{}` so we can substitute the following:
```
A with F
N with L
S with A
U with G
```
* The following is the substitution for all alphabets.
```
ANSUVPOFDEQRMIYJZXBCLGKH
||||||||||||||||||||||||
vvvvvvvvvvvvvvvvvvvvvvvv
FLAGETRSHONIPBXVDWCKUMQY
```
* There are online decoders that make out task easier.
```
THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG. THIS TEXT IS A PANGRAM, WHICH MEANS THAT IT CONTAINS ALL 26 LETTERS OF THE ENGLISH ALPHABET. THIS MAKES IT IDEAL FOR FREQUENCY ANALYSIS, AS THE CRYPTANALYST CAN COMPARE THE FREQUENCY OF LETTERS IN THE CIPHERTEXT TO THE KNOWN FREQUENCY OF LETTERS IN THE ENGLISH LANGUAGE. FOR EXAMPLE, THE MOST COMMON LETTER IN THE ENGLISH LANGUAGE IS E. IF THE MOST COMMON LETTER IN THE CIPHERTEXT IS X, THEN THE CRYPTANALYST CAN ASSUME THAT X IS LIKELY TO BE A SUBSTITUTION FOR E. OTHER COMMON LETTERS IN THE ENGLISH LANGUAGE INCLUDE T, A, O, I, N, S, AND H. THE CRYPTANALYST CAN USE THIS INFORMATION TO MAKE EDUCATED GUESSES ABOUT THE OTHER SUBSTITUTIONS IN THE CIPHERTEXT. FLAG{NOW_IVE_LEARNED_MY_ABCS}
```
### Flag
```
FLAG{NOW_IVE_LEARNED_MY_ABCS}
```
##
## CHALLENGE 3
> This tricky one introduces transposition
---
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