Binary subtraction with numbers in 2’s complement

How can i perform binary subtraction of two numbers that are already in 2’s complement? I have to subtract 01010011 from 10100110,both numbers are in 2’s complement. I know that 10100110 is -90,and 01010011 is 83,so the result should be -173. So if i use 8 bits that means that there’s overflow. But i don’t get how can i perform the subtraction when both the numbers are in 2’s complement. If i just perform the subtraction i get the same number 01010011. How do i show the result and that there’s overflow?

Is the complement of this decision problem in $P$?

Are there any two primes that are NOT a factor of $ M$ that multiply up to $ M$ ?

Fact: Any two primes that multiply up to $ M$ . Must be factors of $ M$ !

Thus because of the fact above an $ O(1)$ algorithm exists. It always outputs $ NO$

Complement

Are there any two primes that are a factor of $ M$ that multiply up to $ M$ ?

Fact: A complement of a decision problem does not always require to always return $ YES$ or $ NO$ . It can be either one!

(eg. $ M$ = 6 and two primes that multiply up to $ M$ are $ 3$ ,$ 2$ .)

Well, this I find interesting this is deciding $ Semi-Primes$ .

Question

Shouldn’t $ Semi-Primes$ be in $ P$ , because what was shown above?

Must a decision problem in $NP$ have a complement in $Co-NP$, if I can verify the solutions to in polynomial-time?

Goldbach’s Conjecture says every even integer $ >$ $ 2$ can be expressed as the sum of two primes.

Let’s say $ N$ is our input and its $ 10$ . Which is an integer > 2 and is not odd.

Algorithm

1.Create list of numbers from $ 1,to~N$

2.Use prime-testing algorithm for creating a second list of prime numbers

3.Use my 2_sum solver that allows you to use primes twice that sum up to $ N$

for j in range(list-of-primes)):   if N-(list-of-primes[j]) in list-of-primes:    print('yes')    break 

4.Verify solution efficently

if AKS-primality(N-(list-of-primes[j])):     if AKS-primality(list-of-primes[j]):         print('Solution is correct') 

5.Output

yes 7 + 3 Solution is correct 

Question

If the conjecture is true, then the answer will always be Yes. Does that mean it can’t be in $ Co-NP$ because the answer is always Yes?

Write a MIPS Program to perform the conversion of 32-bit sign/magnitude binary to 2s complement numbers?

So i have been working on this and this is what i got so far Please help me if i did it wrong or i can fix something here. Thank You.

$ LC0: .ascii “Enter %d bit binary value: 0” $ LC1: .ascii “Original binary = %s20” $ LC2: .ascii “Ones complement = %s20” $ LC3: .ascii “Twos complement = %s20” main: addiu $ sp,$ sp,-144 sw $ 31,140($ sp) li $ 5,32 # 0x20 lui $ 4,%hi($ LC0) addiu $ 4,$ 4,%lo($ LC0) jal printf nop

    addiu   $  4,$  sp,24     jal     gets     nop      addiu   $  3,$  sp,24     addiu   $  2,$  sp,60     addiu   $  5,$  sp,92     li      $  6,49                 # 0x31     li      $  7,48                 # 0x30     li      $  9,49                 # 0x31     li      $  8,48                 # 0x30 

$ L4: lb $ 4,0($ 3) nop bne $ 4,$ 6,$ L2 nop

    b       $  L3     sb      $  8,0($  2) 

$ L2: bne $ 4,$ 7,$ L3 nop

    sb      $  9,0($  2) 

$ L3: addiu $ 2,$ 2,1 bne $ 2,$ 5,$ L4 addiu $ 3,$ 3,1

    sb      $  0,92($  sp)     addiu   $  4,$  sp,91     addiu   $  2,$  sp,127     addiu   $  5,$  sp,95     li      $  8,1                        # 0x1     li      $  6,49                 # 0x31     li      $  7,48                 # 0x30     li      $  9,1                        # 0x1     li      $  11,49                  # 0x31     li      $  10,48                  # 0x30 

$ L8: lb $ 3,0($ 4) nop bne $ 3,$ 6,$ L5 nop

    bne     $  8,$  9,$  L6     nop      b       $  L7     sb      $  10,0($  2) 

$ L5: bne $ 3,$ 7,$ L6 nop

    bne     $  8,$  9,$  L6     nop      sb      $  11,0($  2)     b       $  L7     move    $  8,$  0 

$ L6: sb $ 3,0($ 2) $ L7: addiu $ 2,$ 2,-1 bne $ 2,$ 5,$ L8 addiu $ 4,$ 4,-1

    sb      $  0,128($  sp)     addiu   $  5,$  sp,24     lui     $  4,%hi($  LC1)     addiu   $  4,$  4,%lo($  LC1)     jal     printf     nop      addiu   $  5,$  sp,60     lui     $  4,%hi($  LC2)     addiu   $  4,$  4,%lo($  LC2)     jal     printf     nop      addiu   $  5,$  sp,96     lui     $  4,%hi($  LC3)     addiu   $  4,$  4,%lo($  LC3)     jal     printf     nop      move    $  2,$  0     lw      $  31,140($  sp)     nop     j       $  31     addiu   $  sp,$  sp,144 

Complement of a DFA wihtout final states


Let $ L_1=\{Q,\Sigma,q_0,\delta,Q\}$ be a DFA that accepts a language $ L$ and where all the states are also final states. If we want a DFA that accepts the complement of $ L$ , we swap its accepting states with its non-accepting states, that is $ \overline{L_1}=\{Q,\Sigma,q_0,\delta,Q-Q\}$ . In this case we have a DFA without final states. Is this still a DFA? Is it regular?