please help Knob and Switch Computer Exercises see httpusers

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Knob and Switch Computer Exercises (see http://users.dickinson.edu/~braught/kands/kands.html):

Q3: Using the full “Machine Simulation” to work out your response, list the assembly language program that will cause the smaller of two values stored in memory locations 10 and 11 to be stored into memory location 12.

Solution

Knob and computer simulator is its innovative use of knowbs and switches as \"cognitive hooks\" that encourage student interaction.This simulator is a complete stored program computer.Instructions entered in main memory locations in assembly language are automatically assembled into machine language.The assembly/machine language instructions are documented on the

Assembly Language Instruction: example Meaning: Machine Language Instruction:

LOAD [REG][MEM] LOAD R2 12 R2=M[12] 1 000 0001 0 RR MMMMM
STORE [MEM][REG] STORE 8 R3 M[8]=R3 1 000 0010 0 RR MMMMM
MOVE [REG1][REG2] MOVE R2 R0 R2=R0 1 001 0001 0000 RR RR

The final increment of the K&S computer ,as described in the previous section, also supports assembly language programming.

Machine and Assembly Language Instructions for the K&S computer.
This can be accomplished by
entering a sequence of 8 machine language instructions in main memory. First, the contents of
the desired memory locations must be loaded into registers (1000000100001010 loads the
contents of memory location 10 into R0, and 1000000100101011 loads the contents of
memory location 11 into R1). The next instruction (1010001000100001) subtracts R1 from
R0 and stores the result in R2. Next, a branch-if-negative instruction (0000001100000110)
causes the PC to jump to location 6 in memory if the negative flag in the ALU is set, i.e., if
R1 > R0. If this occurs, the instruction at memory location 6 (1000001000101100) will store
the value from R1 into memory location 12, followed by a halt instruction
(1111111111111111). If a jump does not occur, the instruction in memory location 4
(1000001000001100) will store the value from R1 into memory location 12, followed by a
halt instruction (1111111111111111). Figure 4 shows the K&S Computer