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Cosc 2150: Computer Organization Chapter 9, Part 2 Integer multiplication and division

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Multiplication Complex Work out partial product for each digit Take care with place value (column) Add partial products

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Multiplication Example (unsigned) (long hand) 1011 Multiplicand (11 dec) x 1101 Multiplier (13 dec) 1011 Partial products 00000 Note: if multiplier bit is 1 copy 101100 multiplicand (place value) 1011000 otherwise zero 10001111 Product (143 dec) Note: need double length result

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Flowchart for Unsigned Binary Multiplication

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Execution of Example

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Multiplying Negative Numbers This does not work! Solution 1 —Convert to positive if required —Multiply as above —If signs were different, negate answer Solution 2 —Booth’s algorithm

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Booth’s Algorithm

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Example of Booth’s Algorithm 3*7 First setup the columns and initial values. This case 3 is in Q and M is 7. —But could put 7 in Q and M as 3

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Example of Booth’s Algorithm First cycle: Now look at Q 0 and Q -1 With a 10, we Sub (A=A-M), then shift (always to the right) Second cycle: looking at Q 0 and Q -1 —With a 11, we only shift.

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Example of Booth’s Algorithm 2 nd cycle Result Third cycle, Q 0 and Q -1 have 01 —So we will Add (A=A+M), then shift

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3 nd cycle Result 4 th cycle, Q 0 and Q -1 have 00 —So we only shift Example of Booth’s Algorithm

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4 nd cycle Result 5 th cycle, Q 0 and Q -1 have 00 —So we only shift Example of Booth’s Algorithm

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5 th cycle Result Since we are working in 5 bits, we only repeat 5 times

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Example of Booth’s Algorithm Result is A and Q so 0000010101 which is 21. Note: The sign bit is the last bit in A.

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Division More complex than multiplication Negative numbers are really bad! Based on long division

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001111 Division of Unsigned Binary Integers 1011 00001101 10010011 1011 001110 1011 100 Quotient Dividend Remainder Partial Remainders Divisor

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Flowchart for Unsigned Binary Division

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Signed Division 1. Load divisor into M and the dividend into A, Q registers. Dividend must be 2n-bit twos complement number —0111 (7) becomes 00000111 —1001 (-7) becomes 11111001 2. Shift A, Q left 1 bit position 3. If M and A have the same signs, A A –M else A A+M

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4. Step 3 is successful if sign of A is the same as before step 3 and at the end of step 3 A.if successful or (A=0 AND Q=0) then set Q0 1 B.if unsuccessful and (A 0 OR Q 0) then restore the previous value of A 5. Repeat steps 2 through 4 as many times as there are bit positions in Q. 6. The reminder is in A. If the signs are of Divisor and dividend were the same, the quotient is in Q, otherwise the correct quotient is the twos complement of Q.

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Examples of division (signed) AQM=0011AQ 00000111Initial Value11111001Initial Value 00001110shift11110010shift 1101subtract0010Add 00001110restore11110010Restore 00011100shift11100100Shift 1110subtract0001Add 00011100restore11100100Restore 00111000shift11001000Shift 0000subtract1111Add 00001001set Q0 = 111111001Q0 =1 00010010Shift11110010Shift 1110subtract0010Add 00010010restore11110010Restore (a) 7/3(b) -7/3

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Q A &

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