Today, in class we were discussing about how s-box is used in feistal ciphers. Our sir went on to say that for s-box substitution, we take the 1st and last bits as one part and the remaining bits as another, following which we find the matching 4 bits.
So if we were to have the 6bits as: "100100", it would map to 1110 (left column: 10 , right column: 0010)
My doubt now is if it possible for us to take different positions of bits together, like the 1st and 3rd bit? If not, is there a reason why we can only take the 1st and last bits?
Related
Could you explain how to convert from lz77 to huffman on the example in the below picture?
Easy:
In the first step your output is essentially 3 numbers:
prev index
number of characters to repeat
next character (be it ascii or unicode)
The algorithm demands that you specify a sliding window up front. That means you know how big (1) and (2) can be at most.
In other words, you know how many bits (1) and (2) will take up.
Since (3) is essentially also a character from a fixed length alphabet, you also know the bit-length of (3)
That means it's safe to simply concatenate them.
So, the output of the first algorithm can be thought of as outputting a bit-sequence, where every item in the sequence has a fixed length.
That's ideal for applying huffman.
Of course the specifics are not mentioned, and you can choose from a lot of options.
normalized huffman table
1 on left-branch vs 0 on left-branch
priorities when merging items of similar count
etc
So I can not readily explain the exact output values you are showing.
But I hope I can at least explain how to get from A to B.
You can't. The coding shown is, well, figurative. Not literal. The symbols A, B, and C are all coded to the single bit 0. Obviously that's not going to be very helpful on the decoding end.
I came across the follow question while reading a CS book, can someone please explain it to me? >"The Little Man computer can have ten operation codes (0-9) and address 100 words of storage (0-99). If binary numbers are to replace decimal numbers, what must the minimum number of bits in each word of the LMC be?"
Since you need to be able to distinguish 10 codes for operation, the minimum word size would have to be 4 bits. Using 4 bits, you can represent up to 2^4 = 16 possible codes (since each bit can be 0 or 1). Anything less (2^3 = 8) will not allow a separate binary number for each code.
The Little Man Computer is an architecture where one instruction is held in one word, therefore a word has to contain both the op code and the address. That means you have to hold 000 to 999 so my answer would be 10 bits. You could assume that the question implies the op code and address in separate fields - in that case you need 4 bits for the op code and 7 bits for the address making 11 in total.
Note that the LMC has a "jump if greater than or equal to zero" instruction and for this to mean anything you must be able to represent negative numbers - so that implies that memory has a sign bit. My own simulation allows -999 to +999 as numbers in memory.
I am working on a school project (if you couldn't figure that out just by the fact that I'm using MIPS and QTSpim), and my group chose to make a calculator for large (128-bit) numbers. We know how to do operations on 128-bit numbers, but what we're having trouble with is having the user input.
The professor doesn't quite know how to do it, so does anyone know if there is a way to load a 128-bit integer using MIPS and QTSpim?
MIPS registers hold 32-bit integers, so the result would have to be stored in 4 registers, but is there a way to make that happen?
Thanks!
I would:
Read the user input as a string
Convert the ASCII codes of the each digit to a number 0-9 (i.e. subtract '0')
Apply a radix conversion from base 10 to base 2, and hold the results in four 32 bit words
Why is there a difference between 8, 16, 32, 64, 128 bits? As gusbro described you validate the character string, for every new number character multipy by 10 and add the new number. You already mentioned you know how to do operations on 128 bit numbers so...just do the operations, multiply and add. If you dont know how to do the operations then you are mulitplying by 10 which 0xA which is 0b1010. Using elementary school math, starting with the ones column 0 times anything is zero. the base to the power 1 column (10s in elementary school the twos column here) 1 times anything is itself but you move over one location. the fours column is a zero, the eights column is a 1 so add in abcd shifted left three columns
abcd
x1010
=====
0000
abcdx
0000xx
abcdxxx
So multiplying by 10 is the same as taking the number shifted left one plus the number shifted left three, shifting and adding an infinite number of bits using 32 bit registers is fairly easy. If need be do it 16 or 24 bits at a time leaving bit 17 or bit 24 as the carry bit.
if you dont have a way to multiply and add 128 bits you wont get very far with a 128 bit calculator, so perhaps the above was un necessary.
The problem is as follows:
By replacing each of the letters in the word CARE with 1, 2, 9, and 6 respectively, we form a square number: 1296 = 36^(2). What is remarkable is that, by using the same digital substitutions, the anagram, RACE, also forms a square number: 9216 = 96^(2). We shall call CARE (and RACE) a square anagram word pair and specify further that leading zeroes are not permitted, neither may a different letter have the same digital value as another letter.
Using words.txt (right click and 'Save Link/Target As...'), a 16K text file containing nearly two-thousand common English words, find all the square anagram word pairs (a palindromic word is NOT considered to be an anagram of itself).
What is the largest square number formed by any member of such a pair?
NOTE: All anagrams formed must be contained in the given text file.
I don't understand the mapping of CARE to 1296? How does that work? or are all permutation mappings meant to be tried i.e. all letters to 1-9?
All assignments of digits to letters are allowed. So C=1, A=2, R=3, E=4 would be a possible assignment ... except that 1234 is not a square, so that would be no good.
Maybe another example would help make it clear? If we assign A=6, E=5, T=2, then TEA = 256 = 16² and EAT = 625 = 25². So (TEA=256, EAT=625) is a square anagram word pair.
(Just because all assignments of digits to letters are allowed, does not mean that actually trying out all such assignments is the best way to solve the problem. There may be some other, cleverer, way to do it.)
In short: yes, all permutations need to be tried.
If you test all substitutions letter for digit, than you are looking for pairs of squares with properties:
have same length
have same digits with number of occurrences as in input string.
It is faster to find all these pairs of squares. There are 68 squares with length 4, 216 squares with length 5, ... Filtering all squares of same length by upper properties will generate 'small' number of pairs, which are solutions you are looking for.
These data is 'static', and doesn't depend on input strings. It can be calculated once and used for all input strings.
Hmm. How to put this. The people who put together Project Euler promise that there is a solution that is under one minute for every problem, and there is only one problem that I think might fail this promise, but this is not it.
Yes, you could permute the digits, and try all permutations against all squares, but that would be a very large search space, not at all likely to be the (TM) right thing. In general, when you see that your "look" at the problem is going to generate a search that will take too long, you need to search something else.
Like, suppose you were asked to determine what numbers would be the result of multiplying two primes between 1 and a zillion. You could factor every number between 1 and a zillion, but it might be faster to take all combinations of two primes and multiply them. Since you are looking at combinations, you can start with two and go until your results are too large, then do the same with three, etc. By comparison, this should be much faster - and you don't have to multiply all the numbers out, you could take logs of all the primes and then just add them and find the limit for every prime, giving you a list of numbers you could add up.
There are a bunch of innovative solutions, but the first one you think of - especially the one you think of when Project Euler describes the problem, is likely to be wrong.
So, how can you approach this problem? There are probably too many permutations to look at, but maybe you can figure out something with mappings and comparing mappings?
(Trying to avoid giving it all away.)
I don't understand where the extra bits are coming from in this article about s-boxes. Why doesn't the s-box take in the same number of bits for input as output?
It is the way s-boxes work. They can be m * n ==> m bit input , n bit output.
For example, in the AES S-box the number of bits in input is equal to the number of bits in output.
In DES, m=6 and n=4.
The input is expanded from 32 to 48 bits in the first stages of DES. So it is be reduced to 32 bits again by applying one round of S-box substitution. Thus no information is lost here.
The Wikipedia article on itself can be a bit confusing. It will make people think that information is lost. You should read the article in conjuncture with implementation details of some encryption algorithm using s-boxes.
What extra bits? They are going from 6 to 4.
EDIT: Whoops! I'm an idiot. This is kinda like a 2nd grade multiplication table. They strip the outer bits off of the 6-bit block to be encypted, and leave the middle 4. Just like a table for an arithmatic operation, they go down one side, and find the outer bit sequence, then across the top and find the middle ones. To answer your question, it could input and output the same number of bits, but this s-box is just set up to do it the way it does. Its arbitrary.