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Aug 25 In-Class Exercise.

Please post your solutions to the Aug 25 In-Class Exercise tot his thread.

Best,

Chris

Please post your solutions to the Aug 25 In-Class Exercise tot his thread. Best, Chris

-- Aug 25 In-Class Exercise

 
Scheme:
(gcd 14 30): 30 != 0, gcd 30 14
(gcd 30 14): 14 != 0, gcd 14 2
(gcd 14 2): 2 != 0, gcd 2 0
(gcd 2 0): 0 == 0, return 2 
 
Prolog:
gcd(14, 30, U) -> 30 != 0, not binding
gcd(14, 30, X) -> 30 != 0 (good), Y is 14, gcd(30, 14, X)
gcd(30, 14, U) -> 14 != 0, not binding
gcd(30, 14, X) -> 14 != 0 (good), Y is 2, gcd(14, 2, X)
gcd(14, 2, U) -> 2 != 0, not binding
gcd(14, 2, X) -> 2 != 0 (good), Y is 0, gcd(2, 0, X)
gcd(2, 0, U) -> 0 == 0, U = 2 so result is 2

(Edited: 2021-08-25)

<pre> Scheme: (gcd 14 30): 30 != 0, gcd 30 14 (gcd 30 14): 14 != 0, gcd 14 2 (gcd 14 2): 2 != 0, gcd 2 0 (gcd 2 0): 0 == 0, return 2 Prolog: gcd(14, 30, U) -> 30 != 0, not binding gcd(14, 30, X) -> 30 != 0 (good), Y is 14, gcd(30, 14, X) gcd(30, 14, U) -> 14 != 0, not binding gcd(30, 14, X) -> 14 != 0 (good), Y is 2, gcd(14, 2, X) gcd(14, 2, U) -> 2 != 0, not binding gcd(14, 2, X) -> 2 != 0 (good), Y is 0, gcd(2, 0, X) gcd(2, 0, U) -> 0 == 0, U = 2 so result is 2 </pre>

-- Aug 25 In-Class Exercise

Scheme:

	Define:	(gcd 14 30)
	Since 	30 != 0
			(14 mod 30) = 14
	Then	(gcd 30 14)
	Since 	14 != 0
	Then 	(gcd 30 2)
	Since 	2!= 0
	Then	(gcd 2 0)
	Since 	0 == 0
	Return 2

Prolog:

	l? gcd(14 30 0)
	Since -30 != 0
		(14 mod 30) = 14
	Then gcd (30, 14, x)
	Since -14 != 0
	Then  gcd (14, 2, x)
	Since -2 != 0
	Then gcd (2, 0, x)
	Since -0 == 0
	Return 2

(Edited: 2021-08-30)

Scheme: Define: (gcd 14 30) Since 30 != 0 (14 mod 30) = 14 Then (gcd 30 14) Since 14 != 0 Then (gcd 30 2) Since 2!= 0 Then (gcd 2 0) Since 0 == 0 Return 2 Prolog: l? gcd(14 30 0) Since -30 != 0 (14 mod 30) = 14 Then gcd (30, 14, x) Since -14 != 0 Then gcd (14, 2, x) Since -2 != 0 Then gcd (2, 0, x) Since -0 == 0 Return 2

-- Aug 25 In-Class Exercise

Scheme:

gcd(14, 30)

 30 != 0 so recursive gcd(30, 14 mod 30)

gcd(30, 14)

gcd(14, 2)

gcd(2, 0)

v=0 so return 2

Prolog |?-gcd(14, 30, U)

  V = 30, doesn't = 0, first rule not satisfied
  Y=14 mod 30=14, |?-gcd(30, 14, X)

|?-gcd(14, 2)

|?-gcd(2, 0)

  V = 0 = 0 so the first rule is satisfied and return U=2

(Edited: 2021-08-25)

Scheme: gcd(14, 30) 30 != 0 so recursive gcd(30, 14 mod 30) gcd(30, 14) gcd(14, 2) gcd(2, 0) v=0 so return 2 Prolog |?-gcd(14, 30, U) V = 30, doesn't = 0, first rule not satisfied Y=14 mod 30=14, |?-gcd(30, 14, X) |?-gcd(14, 2) |?-gcd(2, 0) V = 0 = 0 so the first rule is satisfied and return U=2

-- Aug 25 In-Class Exercise

Scheme:

gcd(14 30) v is not 0 , so modulo 14 30 is 14 so

--> gcd(30, 14)

gcd(30,14) v is not 0 , so modulo 30 14 is 2 so

--> gcd(14,2)

gcd(14,2) v is not 0, so modulo 14 2 is 0 so

--> gcd(2,0)

gcd(2,0) v is 0, so return 2

(Edited: 2021-08-25)

Scheme: gcd(14 30) v is not 0 , so modulo 14 30 is 14 so --> gcd(30, 14) gcd(30,14) v is not 0 , so modulo 30 14 is 2 so --> gcd(14,2) gcd(14,2) v is not 0, so modulo 14 2 is 0 so --> gcd(2,0) gcd(2,0) v is 0, so return 2

-- Aug 25 In-Class Exercise

Scheme:

    (gcd 14 30) 30 = 0 is false. 14 modulo 30 = 14. Return (gcd 30 14)
    (gcd 30 14) 14 = 0 is false. 30 modulo 14 = 2. Return (gcd 14 2)
    (gcd 14 2) 2 = 0 is false. 14 modulo 2 = 0. Return (gcd 2 0)
    (gcd 2 0) 0 = 0 is true. Return 2.

Prolog:

    gcd(14, 30, U) 30 = 0 is false. gcd(30, 14, x) not(30 = 0) is true. Y = 14 modulo 30 = 14. gcd(30, 14, x)
    gcd(30, 14, U) 14 = 0 is false. gcd(30, 14, x) not(14 = 0) is true. Y = 30 modulo 14 = 2. gcd(14, 2, x)
    gcd(14, 2, U) 2 = 0 is false. gcd(14, 2, x) not (2 = 0) is true. Y = 14 modulo 2 = 0. gcd(2, 0, x)
    gcd(2, 0, U) 0 = 0 is true. Return U = 2.

Scheme: (gcd 14 30) 30 = 0 is false. 14 modulo 30 = 14. Return (gcd 30 14) (gcd 30 14) 14 = 0 is false. 30 modulo 14 = 2. Return (gcd 14 2) (gcd 14 2) 2 = 0 is false. 14 modulo 2 = 0. Return (gcd 2 0) (gcd 2 0) 0 = 0 is true. Return 2. Prolog: gcd(14, 30, U) 30 = 0 is false. gcd(30, 14, x) not(30 = 0) is true. Y = 14 modulo 30 = 14. gcd(30, 14, x) gcd(30, 14, U) 14 = 0 is false. gcd(30, 14, x) not(14 = 0) is true. Y = 30 modulo 14 = 2. gcd(14, 2, x) gcd(14, 2, U) 2 = 0 is false. gcd(14, 2, x) not (2 = 0) is true. Y = 14 modulo 2 = 0. gcd(2, 0, x) gcd(2, 0, U) 0 = 0 is true. Return U = 2.

-- Aug 25 In-Class Exercise

scheme:

   (define (gcd 14 30 ) 
     (if (= 30 0)
         14
         (gcd 30 14)
             (gcd 14 16)
                 (gcd 16 14)
                    (gcd 14 2)
                      (gcd 2 0)

Prolog 1? - gcd(14, 30, U)

       gcd(U, V, X) :-not(V=0), Y is U mod V
       gcd(14, 30, X) :-not(V=0), 14 is 14 mod 30
       gcd(V, Y, X) = gcd (30, 14, X)

scheme: (define (gcd 14 30 ) (if (= 30 0) 14 (gcd 30 14) (gcd 14 16) (gcd 16 14) (gcd 14 2) (gcd 2 0) Prolog 1? - gcd(14, 30, U) gcd(U, V, X) :-not(V=0), Y is U mod V gcd(14, 30, X) :-not(V=0), 14 is 14 mod 30 gcd(V, Y, X) = gcd (30, 14, X)

-- Aug 25 In-Class Exercise

Scheme

 gcd(14,30). 14 = u and 30 = v
 v is not 0 so we pass gcd(30,modulo(14,30) 14 mod 30 = 14, so gcd(30,14)
 gcd(14, modulo(30,14) so gcd(14, 2)
 gcd(2, modulo(14,2)) so gcd(2,0)
 since v = 0, we return 2 which is the gcd of 14 and 30.

prolog

 gcd(14,30,U) U=14, V=30, V is not 0 so go to next 
 gcd(14,30,X) U=14, V=30, V is not 0 so we solve Y = 14 mod 30 = 14. gcd(30,14,X)
 gcd(30,14,X) 14!=0 so next, U=30, V=14, Y = 30 mod 14 = 2. gcd(14,2,X)
 gcd(14,2,X) 2!=0 so next, U=14, V=2, Y = 14 mod 2 = 0. gcd(2,0,X)
 gcd(2,0,X) since v = 0, result is 2 which is the gcd of 14 and 30.

Scheme gcd(14,30). 14 = u and 30 = v v is not 0 so we pass gcd(30,modulo(14,30) 14 mod 30 = 14, so gcd(30,14) gcd(14, modulo(30,14) so gcd(14, 2) gcd(2, modulo(14,2)) so gcd(2,0) since v = 0, we return 2 which is the gcd of 14 and 30. prolog gcd(14,30,U) U=14, V=30, V is not 0 so go to next gcd(14,30,X) U=14, V=30, V is not 0 so we solve Y = 14 mod 30 = 14. gcd(30,14,X) gcd(30,14,X) 14!=0 so next, U=30, V=14, Y = 30 mod 14 = 2. gcd(14,2,X) gcd(14,2,X) 2!=0 so next, U=14, V=2, Y = 14 mod 2 = 0. gcd(2,0,X) gcd(2,0,X) since v = 0, result is 2 which is the gcd of 14 and 30.

-- Aug 25 In-Class Exercise

|?-gcd(14, 30, u) :-V =0 gcd(u, v, X): - not(v=0), Y is u mod v, gcd(v, Y, X)

    14,30                                   30 14

v(30) is not 0 u(14) mod v(30) = 14, so Y=14 14 is not 0 gcd(v, Y, X) v(30) mod Y(14) = 2, so Y=2 2 is not 0 gcd(v, Y, X) v(30) mod Y(2) = 0 0 = 0 return 2

|?-gcd(14, 30, u) :-V =0 gcd(u, v, X): - not(v=0), Y is u mod v, gcd(v, Y, X) 14,30 30 14 v(30) is not 0 u(14) mod v(30) = 14, so Y=14 14 is not 0 gcd(v, Y, X) v(30) mod Y(14) = 2, so Y=2 2 is not 0 gcd(v, Y, X) v(30) mod Y(2) = 0 0 = 0 return 2

-- Aug 25 In-Class Exercise

Scheme

(gcd 14 30) -> since v != 0, gcd (30 14) (gcd 30 14) -> 30%14 = 2 (gcd 14 2) -> 14%2 = 0 (gcd 2 0) -> since v = 0, returns 2

Prolog

|? - gcd(14, 30, X) -> since V != 0, Y = 14 mod 30, gcd(30, 14, X) gcd(30,14,X) -> since V != 0, Y = 30 mod 14, gcd(14, 2, X) gcd(14,2,X) -> since V != 0, Y = 14 mod 2, gcd(2,0,X) gcd(2,0,X) -> since V = 0, returns 2

(Edited: 2021-08-25)

;Scheme : ;(gcd 14 30) -> since v != 0, gcd (30 14) (gcd 30 14) -> 30%14 = 2 (gcd 14 2) -> 14%2 = 0 (gcd 2 0) -> since v = 0, returns 2 Prolog : |? - gcd(14, 30, X) -> since V != 0, Y = 14 mod 30, gcd(30, 14, X) gcd(30,14,X) -> since V != 0, Y = 30 mod 14, gcd(14, 2, X) gcd(14,2,X) -> since V != 0, Y = 14 mod 2, gcd(2,0,X) gcd(2,0,X) -> since V = 0, returns 2