Q9.2 1 Point This is the church encoding for if false then false else true: (Ax.Ay.y) (Ax.Ay.y) (Ax.Ay.x). What is the beta-normal form of this expression? (Ax.Ay.y) (Ax.Ay.y) (Ax.Ay.x) λκ.λγ.κ Ax.Ay.y Ay-y It produces an infinite loop. Save Answer Q9.3 1 Point Which of the following lambda expressions correctly represents if false then (if false then false else true) else false? ○ (Ax.Ay.y)(x.Ay.y)(λx.λy.x)(λx.λy.x) (\x.\y.x)(\x.\y.y)((\x.\y.y)(\x.\y.y)(^x.Лy.x)) © (Ax.Ay.y)((Ax.Ay.y)(Ax.Ay.y)(Ax.Ay.x))(Ax.λy.y) ○ (Ax.Ay.y)(x.Ay.x)(λx.λy.y)(^x.λy.x) Q9 3 Points Given the following Church encodings: True: Ax.Ay.x False: Ax.Ay.y If a Then Else c: abc Q9.1 1 Point Which of the following represents if true then false else true ? (Ax.Ay.y) (Ax.Ay.x) (Ax.Ay.y) (Ax.Ay.x) (Ax.Ay.y) (Ax.Ay.x) (Ax.Ay.x) (Ax.Ay.y) (Ax.Ay.y) (Ax.Ay.y) (Ax.Ay.y) (Ax.Ay.x) Save Answer

Question Compute a Huffman trie for the following string of length 61. UUJCUJUVAJHHVUJJUUHEUVJJAVHUJEJEJVUVEUAVJVAAVHJEJJJEHVJJEJHHU Here are the frequencies of the characters in the string: char freq А C 1 7 H 8. J 18 U 12 V 10 Using the encodings from the Huffman trie you computed, how many bits are needed to encode the above string? Do not count the bits to represent the encoding table (or any bits used for padding and byte alignment). Answer Your answer should be a positive integer.

(2) {w = {0, 1]* : w corresponds to the binary encoding, without leading O's, of natural numbers that are powers of 4} Testing strings: €, 0, 1, 01, 10, 11, 100, 101, 1000, 1010, 10000 The accepted strings are: L [ E 01 11 100 101 1000 1010 10000

15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 # # v1: v2: This program gives you some practice working with bitwise, shift, and rotate operators main: .global .data .word .word .text .set .ent nop 12 main 0xFEEDABEE lw $s1, v1 lw $s2, v2 ## $t0 = v2 with highest ## $t1 = v2 with lower 1 byte set to 1, other bits staying the same ## $t2 = v1 multiplied by 4 *without* using a multiply instruction ## $t3 = v2 devided by 7 *without* using a multiply instruction .end noreorder main ## $t4 = remainder of v1 divided by 5 ## $t5 = v2 with the lowest 6 bits flipped (inverted) and the upper 26 bits staying the same ## $t6 = 0x1401FED0 without using any pseudo-instructions (i.e., shouldn't use "li $t6, 0xABCD1357") ## $s1 = v1 ## $s2 = v2 12 bits cleared (set to 0), other bits staying the same # This code simply loops infinitely spin: j spin main