Lec #8: Analysis of Algorithms (part 2) + Assembly Language 1 Analysis of Algorithms (part 2) (45 min) ◮ Quick review from last time ◮ Generating best and worst case expressions for insertion sort. ◮ Take away message (for now) 2 Assembly Language (45 min) ◮ What is a computer? ◮ NIC and NAS ◮ Examples Quick review of Algorithms ◮ Computational problem/algorithms/instances ◮ Running examples: ◮ Finding a maximum element in an array of n integers ◮ Finding the index of string (if it exists) in a sorted array of n strings ◮ Sorting an array of n numbers from least to greatest. ◮ “Finding the visible buildings in a row” ◮ Correctness/efficiency ◮ general cost: T(n) ◮ best case cost: B(n) ◮ worst case cost: W(n) Analyzing Insertion Sort Let us look at the analysis of insertion sort in Cormen, et. al. Best and Worst Time Complexity for Insertion Sort B(n)= W(n)= Big-O Let T (n) and f (n) be positive functions. When we write T (n)= O(f (n)) we say that T (n) is ’ordo’ f (n). This means there exists the positive constants c and n0 such that T (n) ≤ c · f (n) for all n > n0. Let’s do an example: T (n)= .5(n2) + 200(n + 6) ◮ is T (n)= O(n)? (No!) ◮ is T (n)= O(n2)? (Yes!) ◮ is T (n)= O(n3)? (Yes!) Let’s show why exactly! Big-Ω Let T (n) and f (n) be positive functions. When we write T (n)=Ω(f (n)) we say that T (n) is ’omega’ f (n). This means there exists the positive constants c and n0 such that T (n) ≥ c · f (n) for all n > n0. Big-Θ When we write T (n)= θ(f (n)) we say that T (n) is ’theta’ f (n) which means T (n)= O(f (n)) and T (n)=Ω(f (n)) Monotonicity A function f (n) is monotonically increasing if n ≤ m implies f (n) ≤ f (m). We are almost always analyzing cost functions that are monotonically increasing. Take away message ◮ Can apply asymptotic analysis to closed formulas, best or worst case algorithm runs, any algorithm run and even problems. ◮ Doing actual cost calculation for algorithm can be tedious (but in certain cases pays off) ◮ Not all algorithms have a function f (n) where T (n)=Θ(f (n)) ◮ We have not seen algorithms where T (n) has log terms yet, but we will. (Time permitting, a discussion of Binary Search). What is a computer? ◮ the Analytic Engine (Babbage, 1837) ◮ Z3 (Zuse, 1941) ◮ Manchester Baby (1948) CPU + Memory NIC ◮ a simulator (nic) ◮ big endian ◮ 2-byte words, 4-bit bytes ◮ 16 2-byte registers ◮ a fixed 4-byte instruction format ◮ words align on 2-byte boundaries ◮ instructions align on 4-byte boundaries ◮ integers in two’s complement format ◮ an assembler (nas) Assembly Language Instruction Set for NIC Instr Code Operands Description halt 0 000 Halt execution. load 1 rxy Load register r with the bit pattern found in the memory cell whose address is xy. loadc 2 rxy Load register r with bit pattern xy. loadr 3 0st Load register s with the bit pattern found in the memory cell whose address is stored in register t. store 4 rxy Store the bit pattern found in register r in the memory cell whose address is xy. storer 5 0st Store the bit pattern found in register s in the memory cell whose address is stored in register t. move 6 0st Copies the bit pattern found in register s to register t. Instruction Set for NIC (part 2) add 7 rst Add the integers in registers s and t and place the result in register r. addc 8 rxy Add the integer xy to register r. mul 9 rst Multiply the integers in registers s and t and place the result in register r. sub a rst Subtract the integer in register t from the integer in register s and place the result in register r. shift b rst Shift the bit pattern in register s the number of positions given by the integer stored in t and place the result in register r. Positive integers shift to the right, negative to the left. and c rst Do a bit-wise logical and on the patterns in registers s and t and place the result in register r. or d rst Do a bit-wise logical or on the patterns in registers s and t and place the result in register r. xor e rst Do a bit-wise logical xor on the patterns in registers s and t and place the result in register r. Instruction Set for NIC (part 3) jump f 0xy Jump to the instruction in memory cell xy. jumpe f rxy Jump to the instruction in the memory cell at address xy if the bit pattern in register r is equal to the bit pattern in register r0. jumpn f rxy+01b Jump to the instruction in the memory cell at address xy if the bit pattern in register r is not equal to the bit pattern in register r0. jumpl f rxy+10b Jump to the instruction in the memory cell at address xy if the integer in register r is less than the integer in register r0. jumple f rxy+11b Jump to the instruction in the memory cell at address xy if the integer in register r is less than or equal to the integer in register r0. noop f 001 Do nothing. An Example Assembly code (in loop.as) // Writes ff, fe, ..., 2, 1, 0 to memory at address n. word n loadc r0 -1 loadc r1 -1 Loop: store r1 n // write r1 to memory address n addc r1 -1 // r1-- jumpn r1 Loop // if r1 != -1 goto Loop n is set to the address immediately at the end of address space of the program. Binary representation (in loop.bi) 1f1f1f1f20ff21ff411c81fff1090000f00000 Let’s grind this example! Contact (Sagan, 1996): Operating somewhere in the Vega System Assembly Code // beats out the primes on register 9 loadc r3 n loadc r0 0 word n loadr r1 r3 Oloop: noop repeat: loadc r2 2 Loop: addc r1 -1 // r1-- store r2 n loadc r9 0 loadc r2 3 loadc r9 -1 // Thump! store r2 n+2 loadc r9 0 loadc r2 5 jumpn r1 Loop // if r1 != -1 goto Loop store r2 n+4 loadc r2 7 noop store r2 n+6 noop loadc r2 11 noop store r2 n+8 noop loadc r2 13 noop store r2 n+10 noop loadc r2 17 store r2 n+12 addc r3 2 loadc r2 19 loadr r1 r3 store r2 n+14 jumpn r1 Oloop loadc r2 23 jump repeat store r2 n+16 ... Darn only the first 18 primes... we need the first 26. could we do better?.