LeongHW, SoC, NUS (UIT2201: Algorithms) Page 1 Animation of Algorithm Goal: To understand an algorithm by animating its execution, step-by-step. Algorithm:

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Presentation transcript:

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 1 Animation of Algorithm Goal: To understand an algorithm by animating its execution, step-by-step. Algorithm: Sum 1-to-5 (find sum from 1 to 5) (Note: Similar to Sum 1-to-100, but shorter!!) Observe carefully: osequential operations/statements oconditional statements, oiterative statements,

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 2 Simulating an Algorithm ALGORITHM Sum-1-to-5; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 5)? 6. no  goto repeat 7. yes  goto finish Finish: print out the value of sum 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15; Let’s animate the execution of this simple algorithm.

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 3 Simulating an Algorithm ALGORITHM Sum-1-to-5; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 5)? 6. no  goto repeat 7. yes  goto finish Finish: print out the value of sum 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15; k ??? Step 0. CPU sum ??? Our abstract model of the computer Initial state of the algorithm

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 4 Simulating an Algorithm ALGORITHM Sum-1-to-5; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 5)? 6. no  goto repeat 7. yes  goto finish Finish: print out the value of sum 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15; k ??? Step 1. Sum  0; CPU sum 0 Start of execution, at Step 1. Assignment statement; The value of 0 is stored in the storage box called sum. Assignment statement; The value of 0 is stored in the storage box called sum.

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 5 Simulating an Algorithm ALGORITHM Sum-1-to-5; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 5)? 6. no  goto repeat 7. yes  goto finish Finish: print out the value of sum 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15; k 1 Step 2. k  1; CPU sum 0 Assignment statement; The value of 1 is stored in the storage box called k. Assignment statement; The value of 1 is stored in the storage box called k. Executing Step 2.

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 6 Simulating an Algorithm ALGORITHM Sum-1-to-5; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 5)? 6. no  goto repeat 7. yes  goto finish Finish: print out the value of sum 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15; k 1 Step 3. sum + k = CPU sum 1 Assignment statement; The new value of sum is stored; the old value is gone. Assignment statement; The new value of sum is stored; the old value is gone. Executing Step 3. start of “body-of-loop”

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 7 Simulating an Algorithm ALGORITHM Sum-1-to-5; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 5)? 6. no  goto repeat 7. yes  goto finish Finish: print out the value of sum 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15; k 2 Step 4. k + 1 = CPU sum 1 Assignment statement; The new value of k is stored; the old value is gone. Assignment statement; The new value of k is stored; the old value is gone. Executing Step 4. inside “body-of-loop”

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 8 Simulating an Algorithm ALGORITHM Sum-1-to-5; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 5)? 6. no  goto repeat 7. yes  goto finish Finish: print out the value of sum 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15; k 2 Step 5. (k > 5)? (2 > 5)? = NO CPU sum 1 Condition check: evaluate (k > 5)? NO  execute Step 6 next. Condition check: evaluate (k > 5)? NO  execute Step 6 next. Executing Step 5. loop-test

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 9 Simulating an Algorithm ALGORITHM Sum-1-to-5; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 5)? 6. no  goto repeat 7. yes  goto finish Finish: print out the value of sum 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15; k 2 Step 6. goto repeat (Step 3) CPU sum 1 goto “repeat” means to get algorithm to continue at the step labelled “repeat”. goto “repeat” means to get algorithm to continue at the step labelled “repeat”. Executing Step 6.

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 10 Simulating an Algorithm ALGORITHM Sum-1-to-5; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 5)? 6. no  goto repeat 7. yes  goto finish Finish: print out the value of sum 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15; k 2 Step 3. sum + k = CPU sum 3 Add 2 to sum; The new value of sum is stored; the old value is gone. Add 2 to sum; The new value of sum is stored; the old value is gone. Executing Step 3. 2 nd round of loop-body

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 11 Simulating an Algorithm ALGORITHM Sum-1-to-5; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 5)? 6. no  goto repeat 7. yes  goto finish Finish: print out the value of sum 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15; k 3 Step 4. k + 1 = CPU sum 3 Increment k; The new value of k is stored; the old value is gone. Increment k; The new value of k is stored; the old value is gone. Executing Step 4. 2 nd round of loop-body

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 12 Simulating an Algorithm ALGORITHM Sum-1-to-5; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 5)? 6. no  goto repeat 7. yes  goto finish Finish: print out the value of sum k 3 Step 5. (k > 5)? (3 > 5)? = NO CPU sum 3 Condition check: evaluate (k > 5)? NO  execute Step 6 next. Condition check: evaluate (k > 5)? NO  execute Step 6 next. Executing Step 5. 2 nd loop-test 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15;

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 13 Simulating an Algorithm ALGORITHM Sum-1-to-5; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 5)? 6. no  goto repeat 7. yes  goto finish Finish: print out the value of sum k 3 Step 6. goto repeat (Step 3) CPU sum 3 Goto Step 3 and Execute the loop-body again. Goto Step 3 and Execute the loop-body again. Executing Step 6. 2 nd round 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15;

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 14 Simulating an Algorithm ALGORITHM Sum-1-to-5; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 5)? 6. no  goto repeat 7. yes  goto finish Finish: print out the value of sum k 3 Step 3. sum + k = CPU sum 6 Add 3 to sum; The new value of sum is stored; the old value is gone. Add 3 to sum; The new value of sum is stored; the old value is gone. Executing Step 3. 3 rd round of loop-body 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15;

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 15 Simulating an Algorithm ALGORITHM Sum-1-to-5; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 5)? 6. no  goto repeat 7. yes  goto finish Finish: print out the value of sum k 4 Step 4. k + 1 = CPU sum 6 Increment k; The new value of k is stored; the old value is gone. Increment k; The new value of k is stored; the old value is gone. Executing Step 4. 3 rd round of loop-body 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15;

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 16 Simulating an Algorithm ALGORITHM Sum-1-to-5; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 5)? 6. no  goto repeat 7. yes  goto finish Finish: print out the value of sum k 4 Step 5. (k > 5)? (4 > 5)? = NO CPU sum 6 Condition check: evaluate (k > 5)? NO  execute Step 6 next. Condition check: evaluate (k > 5)? NO  execute Step 6 next. Executing Step 5. 3 rd loop-test 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15;

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 17 Simulating an Algorithm ALGORITHM Sum-1-to-5; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 5)? 6. no  goto repeat 7. yes  goto finish Finish: print out the value of sum k 4 Step 6. goto repeat (Step 3) CPU sum 6 Goto Step 3 and Execute the loop-body again. Goto Step 3 and Execute the loop-body again. Executing Step 6. 3 rd round 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15;

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 18 Simulating an Algorithm ALGORITHM Sum-1-to-5; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 5)? 6. no  goto repeat 7. yes  goto finish Finish: print out the value of sum k 4 Step 3. sum + k = CPU sum 10 Add 4 to sum; The new value of sum is stored; the old value is gone. Add 4 to sum; The new value of sum is stored; the old value is gone. Executing Step 3. 4 th round of loop-body 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15;

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 19 Simulating an Algorithm ALGORITHM Sum-1-to-5; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 5)? 6. no  goto repeat 7. yes  goto finish Finish: print out the value of sum k 5 Step 4. k + 1 = CPU sum 10 Increment k; The new value of k is stored; the old value is gone. Increment k; The new value of k is stored; the old value is gone. Executing Step 4. 4 th round of loop-body 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15;

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 20 Simulating an Algorithm ALGORITHM Sum-1-to-5; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 5)? 6. no  goto repeat 7. yes  goto finish Finish: print out the value of sum k 5 Step 5. (k > 5)? (5 > 5)? = NO CPU sum 10 Condition check: evaluate (k > 5)? NO  execute Step 6 next. Condition check: evaluate (k > 5)? NO  execute Step 6 next. Executing Step 5. 4 th loop-test 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15;

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 21 Simulating an Algorithm ALGORITHM Sum-1-to-5; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 5)? 6. no  goto repeat 7. yes  goto finish Finish: print out the value of sum k 5 Step 6. goto repeat (Step 3) CPU sum 10 Goto Step 3 and Execute the loop-body again. Goto Step 3 and Execute the loop-body again. Executing Step 6. 4 th round 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15;

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 22 Simulating an Algorithm ALGORITHM Sum-1-to-5; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 5)? 6. no  goto repeat 7. yes  goto finish Finish: print out the value of sum k 5 Step 3. sum + k = CPU sum 15 Add 5 to sum; The new value of sum is stored; the old value is gone. Add 5 to sum; The new value of sum is stored; the old value is gone. Executing Step 3. 5 th round of loop-body 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15;

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 23 Simulating an Algorithm ALGORITHM Sum-1-to-5; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 5)? 6. no  goto repeat 7. yes  goto finish Finish: print out the value of sum k 6 Step 4. k + 1 = CPU sum 15 Increment k; The new value of k is stored; the old value is gone. Increment k; The new value of k is stored; the old value is gone. Executing Step 4. 5 th round of loop-body 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15;

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 24 Simulating an Algorithm ALGORITHM Sum-1-to-5; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 5)? 6. no  goto repeat 7. yes  goto finish Finish: print out the value of sum k 6 Step 5. (k > 5)? (6 > 5)? = YES CPU sum 15 Condition check: evaluate (k > 5)? YES  execute Step 7 next. Condition check: evaluate (k > 5)? YES  execute Step 7 next. Executing Step 5. 5 th loop-test 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15;

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 25 Simulating an Algorithm ALGORITHM Sum-1-to-5; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 5)? 6. no  goto repeat 7. yes  goto finish finish: print out the value of sum k 6 Step 7. goto finish (Step 8) CPU sum 15 Goto finish (Step 8) (exit the iterative loop!) Goto finish (Step 8) (exit the iterative loop!) Executing Step 7. & exit the iterative loop 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15;

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 26 Simulating an Algorithm ALGORITHM Sum-1-to-5; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 5)? 6. no  goto repeat 7. yes  goto finish finish: print out the value of sum k 6 Step 8. Print to output CPU sum 15 Print statement; print to output the value of sum Print statement; print to output the value of sum Executing Step 8. print output and END 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15; Output of Algorithm: 15

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 27 Summary  Summary of Steps: o1, 2, (3,4,5,6), (3,4,5,6), (3,4,5,6), (3,4,5,6), (3,4,5,7), 8  Note the sequential execution, except for oConditional statements oGoto statements oiterative statements  Questions: oWhere is the “loop-body”? oHow many iteration of the loop-body? oHow many times is the loop-test done?

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 28 Explore further (DIY)  We did Sum 1-to-5 (instead of Sum 1-to-100)  DIY: Simulate the execution for the original algorithm for Sum 1-to-100?  (Use the following “ending”-slides to help you.)

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 29 Simulating an Algorithm ALGORITHM Sum-1-to-100; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 100)? 6. no  goto repeat 7. yes  goto finish Finish: print out the value of sum 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15; k 100 Step 5. (k > 100)? (100 > 100)? = NO CPU sum 5050 Condition check: evaluate (k > 100)? NO  execute Step 6 next. Condition check: evaluate (k > 100)? NO  execute Step 6 next. Executing Step th loop-test

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 30 Simulating an Algorithm ALGORITHM Sum-1-to-5; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 5)? 6. no  goto repeat 7. yes  goto finish Finish: print out the value of sum k 100 Step 6. goto repeat (Step 3) CPU sum 4950 Goto Step 3 and Execute the loop-body again. Goto Step 3 and Execute the loop-body again. Executing Step th round 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15;

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 31 Simulating an Algorithm ALGORITHM Sum-1-to-100; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 100)? 6. no  goto repeat 7. yes  goto finish Finish: print out the value of sum 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15; k 100 Step 3. sum + k = CPU sum 5050 Add 100 to sum; The new value of sum is stored; the old value is gone. Add 100 to sum; The new value of sum is stored; the old value is gone. Executing Step th round of loop-body

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 32 Simulating an Algorithm ALGORITHM Sum-1-to-100; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 100)? 6. no  goto repeat 7. yes  goto finish Finish: print out the value of sum 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15; k 101 Step 4. k + 1 = CPU sum 5050 Increment k; The new value of k is stored; the old value is gone. Increment k; The new value of k is stored; the old value is gone. Executing Step th round of loop-body

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 33 Simulating an Algorithm ALGORITHM Sum-1-to-100; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 100)? 6. no  goto repeat 7. yes  goto finish Finish: print out the value of sum 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15; k 101 Step 5. (k > 100)? (101 > 100)? = YES CPU sum 5050 Condition check: evaluate (k > 100)? YES  execute Step 7 next. Condition check: evaluate (k > 100)? YES  execute Step 7 next. Executing Step th loop-test

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 34 Simulating an Algorithm ALGORITHM Sum-1-to-100; 1. sum  0 2. k  1 repeat: add k to sum 4. add 1 to k 5. Is (k > 100)? 6. no  goto repeat 7. yes  goto finish finish: print out the value of sum 0+1=1; 1+2=3; 3+3=6; 6+4=10; 10+5=15; k 101 Step 7. goto finish (Step 8) CPU sum 5050 Goto finish (Step 8) (exit the iterative loop!) Goto finish (Step 8) (exit the iterative loop!) Executing Step 7. & exit the iterative loop

LeongHW, SoC, NUS (UIT2201: Algorithms) Page 35