■ 🕑 4. Chapter 1
│  SICP chapter 1 
│  > Building Abstractions with Procedures
│  
│  https://mitp-content-server.mit.edu/books/content/sectbyfn/books_pres_0/6515/sicp.zip/full-text/book/book-Z-H-9.html#%_chap_1
│   
├─■ 🕑 13. exercise 1.3
│ │   (define (square x) 
│ │     (* x x)) 
│ │   (define (sum-squares x y) 
│ │     (+ (square x) (square y))) 
│ │    
│ │   (define (sum-greater-squares x y z) 
│ │     (if (> x y) 
│ │         (if (> y z) 
│ │             (sum-squares x y) 
│ │             (sum-squares x z)) 
│ │         (if (> x z) 
│ │             (sum-squares y x) 
│ │             (sum-squares y z)))) 
│ │    
│ │   (sum-greater-squares 1 2 3) 
│ │   (sum-greater-squares 4 2 3) 
│ │   (sum-greater-squares 5 1 0) 
│ │  
│ │   
│ └─■ 🕑 14. Excercise 1.3
│   │  My attempt at exercise 1.3. it took me a long time as most of my attempts made code that excluded anything that wasn't the maximum number, so i guess finding the "middle" number is what makes this challenging. my first attempts were checking each variable against the others individually (mistake). if x>y then output x otherwise 0, if x>z output x otherwise 0 then multiply both outputs, repeat for each variable (rather silly in retrospect). After this i decided to only compare to the third variable if the first check failed and directly compute the square if the variable wasn't the minimum. Code below:
│   │  
│   │  #lang sicp
│   │  (define (sicp_ex_1.3 x y z)
│   │    (+
│   │      (if (< x y) (if (< x z) 0 (* x x)) (* x x))
│   │      (if (< y x) (if (< y z) 0 (* y y)) (* y y))
│   │      (if (< z y) (if (< z x) 0 (* z z)) (* z z))))
│   │      
│   │  (sicp_ex_1.3 1 2 3)
│   │  result was 13 so it worked
│   │   
│   ├─■ 🕑 15. corrected exercise 1.3
│   │   Updated to catch PhatCat's fix -- if x, y, and z are the same
│   │   value, the code does not work
│   │   
│   │   (define (square x) 
│   │     (* x x)) 
│   │   (define (sum-squares x y) 
│   │     (+ (square x) (square y))) 
│   │     
│   │   (define (sum-greater-squares x y z) 
│   │     (if (> x y) 
│   │         (if (> y z) 
│   │             (sum-squares x y) 
│   │             (sum-squares x z)) 
│   │         (if (> x z) 
│   │             (sum-squares y x)
│   │             (if (= x y z)
│   │                 (sum-squares x x)
│   │                 (sum-squares y z)))))
│   │    
│   └─■ 🕑 16. fix ex 1.3
│     │  summed all 3 squares if x y z were the same. added a check to just force one variable to zero if all the numbers were equal so theres probably a more elegant solution :P. Code:
│     │  
│     │  #lang sicp
│     │  
│     │  (define (square x) (* x x))
│     │  
│     │  (define (sicp_ex_1.3 x y z)
│     │    (+
│     │      (if (= x y) 0 (if (= x z) 0
│     │                        (if (< x y) (if (< x z) 0 (square x)) (square x))))
│     │      (if (< y x) (if (< y z) 0 (square y)) (square y))
│     │      (if (< z y) (if (< z x) 0 (square z)) (square z))))
│     │     
│     │  (sicp_ex_1.3 1 2 3)
│     │  (sicp_ex_1.3 2 2 2)
│     │  (sicp_ex_1.3 2 1 0)
│     │   
│     └─■ 🕑 17. epic phail
│         Fixed the fix
│         
│         #lang sicp
│         
│         (define (square x) (* x x))
│         
│         (define (sicp_ex_1.3 x y z)
│           (+
│             (if (= x y z) 0
│                 (if (< x y) (if (< x z) 0 (square x)) (square x))))
│             (if (< y x) (if (< y z) 0 (square y)) (square y))
│             (if (< z y) (if (< z x) 0 (square z)) (square z)))
│            
│         (sicp_ex_1.3 1 2 3)
│         (sicp_ex_1.3 2 2 2)
│         (sicp_ex_1.3 2 1 0)
│          
└─■ 🕑 18. exercise 1.3
    [(define (square a)
        (* a a))
    
    (define (smallest x y z)
        (if (< x y)
            (if (< x z) x z)
            (if (< y z) y z)))
    
    (define (procedure x y z)
        (- (+ (square x)(square y)(square z))(square (smallest x y z))))
    
    (procedure 7 10 10)]