Not Sure What has happened here in my use of IMP logical operator

[Dimster]

I'm trying to use IMP in my program and coming up with some odd results. Not sure why the inconsistent results. I do know I can do a work around with NOT, AND and OR . So here is some code to give you an idea of the math I'm using in my program

Code: QB64: [Select]

1. Print
2. Print
3. a = 3: Print "value of a = "; a
4. b = 1: Print "value of b = "; b
5. c$ = ""
6. Print "NOTE: A is Greater than B.....so if number A = 3 and next number B is less than A (at 1 in this case), then this is a TRENDING DOWN situation"
7. Print " Equation is 'If a < b And b > 1 Then c = a Imp b'    "
8. Print
9. If a < b And b > 1 Then c = a Imp b
10. If c = -1 Then c$ = "Trending UP"
11. If c = 0 Then c$ = "Trending Down"
12. Print
13. Print "Result is "; c$;
14. If c$ = "Trending Down" Then Print " ... Correct Result" Else Print " ... Incorrect Result"
15.  
16.  
17.  
18. Sleep
19.  
20.  
21. Print
22. Print
23. a = 1: Print "value of a = "; a
24. b = 8: Print "value of b = "; b
25. c$ = ""
26. Print
27. Print "NOTE: A is Less than B.....so if number A = 1 and next number B is greater than 1 (at 8 in this case), then this is a TRENDING UP situation"
28. Print " Equation is 'If a < b And b > 1 Then c = a Imp b'    "
29.  
30. If a < b And b > 1 Then c = a Imp b
31. If c = -1 Then c$ = "Trending UP"
32. If c = 0 Then c$ = "Trending Down"
33. Print
34. Print "Result is "; c$;
35. If c$ = "Trending UP" Then Print " ... Correct Result" Else Print " ... Incorrect Result"
36.  
37.  
38. 'Sleep
39.  
40.  
41. Sleep
42. Print
43. Print
44. a = 1: Print a; " = value of a"
45. b = 3: Print b; " = value of b"
46. Print "NOTE: A is Less than B.....so if number of A = 1 and next number of success is greater than 1 (at 3 in this case), then this is a TRENDING UP situation"
47. Print " Equation is 'If a < b And b > 1 Then c = a Imp b'    "
48.  
49. If a < b And b > 1 Then c = a Imp b
50. If c = -1 Then c$ = "Trending UP"
51. If c = 0 Then c$ = "Trending Down"
52. Print
53. Print "Result is "; c$;
54.  
55. If c$ = "Trending UP" Then Print " ... Correct Result" Else Print " ... Incorrect Result"
56.  
57.  

The equation is the same in each of the 3 cases, only the value of "8" is different. I'm not sure why equation 1 and 3 are producing the correct result and equation 2 is not.

[TempodiBasic]

Hi  Dimster
I think that you have loosen the point because you types

Code: QB64: [Select]

1. THEN c = a IMP b

  buuutttt you mean

Code: QB64: [Select]

1. THEN  c = (a<b) IMP (b>1)

.
Give a try because for now you have calculated   True IMP True as any positive number is true.

[Dimster]

@TempodiBasic  Thank You - that is indeed what I was missing.

[Dimster]

@TempodiBasic  - so I thought the matter was solved but here is that same routine with the changes in the IMP format. The result were ok until I added the last one where a=8 and b =1. This is virtually the identical calculation as the 1st one where a = 3 and b =1 but for some reason the math formula is failing.

What appears to be happening is... the previous value for c$ is simply being repeated and the evaluation of the formula is either being skipped or producing a sum with is not a -1 or 0. The logic of the equation seems to be correct. I must be missing something in the formula again.

Code: QB64: [Select]

1. Print
2. Print
3. a = 3: Print "value of a = "; a
4. b = 1: Print "value of b = "; b
5. Print "NOTE: A is Greater than B.....so if number A = 3 and next number B is less than A (at 1 in this case), then this is a TRENDING DOWN situation"
6. Print " Equation is 'If a < b And b > 1 Then c = (a<b) Imp (b>1)'    "
7. Print
8. If a < b And b > 1 Then c = (a < b) Imp (b > 1)
9. If c = -1 Then c$ = "Trending UP"
10. If c = 0 Then c$ = "Trending Down"
11. Print
12. Print "Result is "; c$;
13. Sleep
14. If c$ = "Trending Down" Then Print " ... Correct Result" Else Print " ... Incorrect Result"
15.  
16.  
17.  
18. Sleep
19.  
20.  
21. Print
22. Print
23. a = 1: Print "value of a = "; a
24. b = 8: Print "value of b = "; b
25. Print
26. Print "NOTE: A is Less than B.....so if number A = 1 and next number B is greater than 1 (at 8 in this case), then this is a TRENDING UP situation"
27. Print " Equation is 'If a < b And b > 1 Then c = (a<b) Imp (b>1)'    "
28.  
29. If a < b And b > 1 Then c = (a < b) Imp (b > 1)
30. If c = -1 Then c$ = "Trending UP"
31. If c = 0 Then c$ = "Trending Down"
32. Print
33. Print "Result is "; c$;
34. Sleep
35. If c$ = "Trending UP" Then Print " ... Correct Result" Else Print " ... Incorrect Result"
36.  
37.  
38. 'Sleep
39.  
40.  
41. Sleep
42. Print
43. Print
44. a = 1: Print a; " = value of a"
45. b = 1: Print b; " = value of b"
46. Print "NOTE: A is Equal B.....so if number of A = 1 and next number of B is equal to  1 (as in this case), then this is TRENDING Steady but as Steady is not a choice and is NOT Down then TRENDING UP is the answer"
47. Print " Equation is 'If a < b And b > 1 Then c = (a<b) Imp (b>1)'    "
48. If a < b And b > 1 Then c = (a < b) Imp (b > 1)
49. If c = -1 Then c$ = "Trending UP"
50. If c = 0 Then c$ = "Trending Down"
51. Print
52. Print "Result is "; c$;
53. Sleep
54. If c$ = "Trending UP" Then Print " ... Correct Result" Else Print " ... Incorrect Result"
55.  
56. Sleep
57.  
58. Print
59. Print
60. a = 8: Print "value of a = "; a
61. b = 1: Print "value of b = "; b
62. Print
63. Print "NOTE: A is Greater than B.....so if number A = 8 and next number B is less than A (at 1 in this case), then this is a TRENDING Down situation"
64. Print " Equation is 'If a < b And b > 1 Then c = (a<b) Imp (b>1)'    "
65.  
66. If a < b And b > 1 Then c = (a < b) Imp (b > 1)
67. If c = -1 Then c$ = "Trending UP"
68. If c = 0 Then c$ = "Trending Down"
69. Print
70. Print "Result is "; c$;
71. Sleep
72. If c$ = "Trending Down" Then Print " ... Correct Result" Else Print " ... Incorrect Result"
73.  
74.  

[TempodiBasic]

Hi Dimster
Let:s debug
Using the table True False

a.  b   (a<b)   (b>1)      IMP.    Trending 
3   1    0         0         0 imp 0 =-1    UP
1   8    - 1     - 1      - 1 imp - 1 =- 1  UP
1   1     0        0        0 imp 0 =-1      UP
8   1     0        0        0  imp 0=-1    UP

Well it seems that we always got UP  and so you must correct the lines 14 and 72 that are wrong.

[TempodiBasic]

Hi Dimster
this evening I can use the QB64 Ide so I can test and MOD your code to verify the goal

Well in the first algorythm there is a bug...
when in the code we test

Code: QB64: [Select]

1.  If a < b And b > 1 Then c = (a < b) Imp (b > 1)

we have not thought that the part

Code: QB64: [Select]

1.  c = (a < b) Imp (b > 1)

will be executed only when the 2 conditions are true, so we can get only the Trending UP or a null action (using the flag security tip, otherwise we don't reset at each pass the c variable).

I found it when I wanted to write in modular way your code using a main and a SUB.
Here mod that shows the differences

Code: QB64: [Select]

1. 'Print
2. 'Print
3. 'a = 3: Print "value of a = "; a
4. 'b = 1: Print "value of b = "; b
5. 'Print "NOTE: A is Greater than B.....so if number A = 3 and next number B is less than A (at 1 in this case), then this is a TRENDING DOWN situation"
6. 'Print " Equation is 'If a < b And b > 1 Then c = (a<b) Imp (b>1)'    "
7. 'Print
8. 'If a < b And b > 1 Then c = (a < b) Imp (b > 1)
9. 'If c = -1 Then c$ = "Trending UP"
10. 'If c = 0 Then c$ = "Trending Down"
11. 'Print
12. 'Print "Result is "; c$;
13. 'Sleep
14. 'If c$ = "Trending Down" Then Print " ... Correct Result" Else Print " ... Incorrect Result"
15.  
16.  
17.  
18. 'Sleep
19.  
20.  
21. 'Print
22. 'Print
23. 'a = 1: Print "value of a = "; a
24. 'b = 8: Print "value of b = "; b
25. 'Print
26. 'Print "NOTE: A is Less than B.....so if number A = 1 and next number B is greater than 1 (at 8 in this case), then this is a TRENDING UP situation"
27. 'Print " Equation is 'If a < b And b > 1 Then c = (a<b) Imp (b>1)'    "
28.  
29. 'If a < b And b > 1 Then c = (a < b) Imp (b > 1)
30. 'If c = -1 Then c$ = "Trending UP"
31. 'If c = 0 Then c$ = "Trending Down"
32. 'Print
33. 'Print "Result is "; c$;
34. 'Sleep
35. 'If c$ = "Trending UP" Then Print " ... Correct Result" Else Print " ... Incorrect Result"
36.  
37.  
38. ''Sleep
39.  
40.  
41. 'Sleep
42. 'Print
43. 'Print
44. 'a = 1: Print a; " = value of a"
45. 'b = 1: Print b; " = value of b"
46. 'Print "NOTE: A is Equal B.....so if number of A = 1 and next number of B is equal to  1 (as in this case), then this is TRENDING Steady but as Steady is not a choice and is NOT Down then TRENDING UP is the answer"
47. 'Print " Equation is 'If a < b And b > 1 Then c = (a<b) Imp (b>1)'    "
48. 'If a < b And b > 1 Then c = (a < b) Imp (b > 1)
49. 'If c = -1 Then c$ = "Trending UP"
50. 'If c = 0 Then c$ = "Trending Down"
51. 'Print
52. 'Print "Result is "; c$;
53. 'Sleep
54. 'If c$ = "Trending UP" Then Print " ... Correct Result" Else Print " ... Incorrect Result"
55.  
56. 'Sleep
57.  
58. 'Print
59. 'Print
60. 'a = 8: Print "value of a = "; a
61. 'b = 1: Print "value of b = "; b
62. 'Print
63. 'Print "NOTE: A is Greater than B.....so if number A = 8 and next number B is less than A (at 1 in this case), then this is a TRENDING Down situation"
64. 'Print " Equation is 'If a < b And b > 1 Then c = (a<b) Imp (b>1)'    "
65.  
66. 'If a < b And b > 1 Then c = (a < b) Imp (b > 1)
67. 'If c = -1 Then c$ = "Trending UP"
68. 'If c = 0 Then c$ = "Trending Down"
69. 'Print
70. 'Print "Result is "; c$;
71. 'Sleep
72. 'If c$ = "Trending Down" Then Print " ... Correct Result" Else Print " ... Incorrect Result"
73. For d = 1 To 2 Step 1
74.     If d = 1 Then m$ = "Dimster" Else If d = 2 Then m$ = "Tempo"
75.     Cls
76.     Print " New IMP test  it has result False and Trending down"; " mode "; m$
77.     ImpAB 0, 1, m$ ' True Imp False  --> False
78.     Print " New IMP test  it has result True and Trending up"; " mode "; m$
79.     ImpAB 3, 1, m$ ' False Imp False --> True
80.     Print " New IMP test  it has result True and Trending up"; " mode "; m$
81.     ImpAB 3, 4, m$ ' True Imp True --> True
82.     Print " New IMP test  it has result True and Trending up"; " mode "; m$
83.     ImpAB 3, 2, m$ ' False Imp True --> True
84. Next
85.  
86. End
87.  
88. Sub ImpAB (a As Integer, b As Integer, Mode As String)
89.     Print
90.     Print
91.     Print "value of a = "; a
92.     Print "value of b = "; b
93.     If a < b Then
94.         Print "NOTE: A is Less than B....."
95.         If Mode = "Dimster" And (b <= 1) Then
96.             Print "  and if b<=1 "
97.             Print " the Equation 'If a < b And b > 1 Then c = (a<b) Imp (b>1)' is never executed   "
98.         End If
99.     ElseIf a > b Then
100.         Print "NOTE: A is Greater than B.....and with both if B>1 both B<1 IMP gives True "
101.         If b > 1 Then
102.             Print " the Equation 'If a < b And b > 1 Then c = (a<b) Imp (b>1)' is True "
103.         Else
104.             Print " the Equation 'If a < b And b > 1 Then c = (a<b) Imp (b>1)' is never executed   "
105.         End If
106.     End If
107.     If Mode = "Dimster" Then Print " so the condition Trending Down cannot be reached!"
108.     c = 99 ' flag of initialization
109.     If Mode = "Dimster" Then
110.         If a < b And b > 1 Then c = (a < b) Imp (b > 1)
111.     ElseIf Mode = "Tempo" Then
112.         c = (a < b) Imp (b > 1)
113.     End If
114.     If c = -1 Then c$ = "Trending UP"
115.     If c = 0 Then c$ = "Trending Down"
116.     Print
117.     Print "Result is "; c$;
118.     Print
119.     Sleep
120. End Sub

I hope that you find interesting the bug of no reset of variable c and the bug of no calculate c when b>1 is false.
Good Coding

[Dimster]

@TempodiBasic - Thank You very much for your examples - I am studying them closely. In my main program I am using the very simple, tried & true, formula of " If A => B then "Trending UP" else "Trending Down".

When there are only the two variables, perhaps IMP is not the best choice (ie in terms of accuracy of results or flexibility of use Logically). And I do see your point where that very first Logical decision ( ie A > B AND B > 1) leads to a failure to engage the IMP condition in the  C = (A < B) IMP (B > 1) in a number of scenarios where the values of A & B vary. Steve gave me a really good example using only two variables ( Your alive or Not alive ) a little while ago. Since then I have come across a number of occasions where I can see one result produced in my program would imply a Trend. Which has me exploring IMP again.

So its back to the drawing board. Thanks again Tempodi

[luke]

For reference, a IMP b is equivalent to (NOT a) OR b.

[Dimster]

@luke - Thanks Luke. I was aware of that, and that IMP returns a value of -1 meaning all bits are set, or 0 meaning no bits are set (according to the wiki). Steve gave me an example and he was using -1 and 0 in the results. But when I research IMP beyond our wiki, I'm getting info that the bit wise comparison for IMP results in 1 and 0, not -1 and 0. Which is subtly confusing. Plus in terms of Logical Operator (AND, OR, XOR), the results are True = 1 and False = 0. I am working on the Eureka Moment - making IMP work if the goal.

[bplus]

With Boolean True or False (only), anything not = 0 is True. So -1 or 1 or 1,000,000 no difference, all True!

[TempodiBasic]

Hi guys and gals

Hi Dimster

Here I post a demonstration to show how to use IMP in the logical managing of the move of a character on the screen and the alternative more known OR.  In the top left corner you can see the result of detector FUNCTION, in the Top Right corner you can see the type of detector FUNCTION at work, in the bottom of the window you can read the help for user, the title of window remembers you that it is a Demo.

Code: QB64: [Select]

1. ' Demo in two languages  / Dimostrazione in due linguaggi
2. Screen 0 'screen text mode / modalit… schermo solo testo
3. _Title "Demo OR and IMP to move a character on screen" ' setting title of window / impostare il titolo della finestra
4. ' defining useful constants  / definire costanti utili
5. Const X = 1, Y = 2, Life = 3, True = -1, False = 0, ModeOr = 10, ModeImp = 20
6. ' defining variables and arrays  / definire variabili ed array
7. Dim Hero(1 To 3) As Integer, Limit(1 To 2, 1 To 2) As Integer, ModeEvaluation As Integer
8.  
9. 'initialization of variables  / inizializzazione delle variabili
10. Hero(X) = (Rnd * 79) + 1
11. Hero(Y) = (Rnd * 22) + 1
12. Hero(Life) = True
13. Limit(1, X) = 10
14. Limit(1, Y) = 2
15. Limit(2, X) = 70
16. Limit(2, Y) = 19
17. ModeEvaluation = ModeImp
18.  
19. ' main loop  /ciclo principale
20. Do
21.     Cls ' refresh screen  / ripulisce lo schermo
22.  
23.     ' manage evaluation mode and its results / gestisce la modalit… di valutazione e i suoi risultati
24.     Locate 1, 1
25.     If ModeEvaluation = ModeImp Then
26.         If outheRange(Limit(1, X), Limit(2, X), Hero(X)) Or outheRange(Limit(1, Y), Limit(2, Y), Hero(Y)) Then Print "Out of game" Else Print " In game"
27.  
28.     ElseIf ModeEvaluation = ModeOr Then
29.         If IntheRange(Limit(1, X), Limit(2, X), Hero(X)) And IntheRange(Limit(1, Y), Limit(2, Y), Hero(Y)) Then Print " In Game" Else Print "Out Of Game"
30.     End If
31.  
32.     ' border of game field  / bordo del campo di gioco
33.     Locate Limit(1, Y), Limit(1, X)
34.     Print String$(Abs(Limit(1, X) - Limit(2, X)), Chr$(178));
35.     For a = Limit(1, Y) To Limit(2, Y) Step 1
36.         Locate a, Limit(1, X): Print Chr$(178)
37.         Locate a, Limit(2, X): Print Chr$(178)
38.     Next
39.     Locate Limit(2, Y), Limit(1, X)
40.     Print String$(Abs(Limit(1, X) - Limit(2, X)), Chr$(178))
41.  
42.     'hero / l'eroe
43.     Locate Hero(Y), Hero(X)
44.     Print "
    ";
45.  
46.     'Helpscreen   / la guida a schermo
47.     Locate 1, 60: Print "ModeEvaluation: ";
48.     If ModeEvaluation = ModeImp Then Print "IMP"; Else Print " OR";
49.     Locate 24, 1: Print " cursors to  move Hero, Enter key to switch ModeEvaluation, Space to quit";
50.  
51.     ' input and command manager     / gestore dell'input e dei comandi
52.     a$ = InKey$
53.     Select Case a$
54.         Case Chr$(0) + "M"
55.             If Hero(X) < 80 Then Hero(X) = Hero(X) + 1
56.         Case Chr$(0) + "H"
57.             If Hero(Y) > 1 Then Hero(Y) = Hero(Y) - 1
58.         Case Chr$(0) + "P"
59.             If Hero(Y) < 25 Then Hero(Y) = Hero(Y) + 1
60.         Case Chr$(0) + "K"
61.             If Hero(X) > 1 Then Hero(X) = Hero(X) - 1
62.         Case "K"
63.             Hero(Life) = False
64.         Case Chr$(32)
65.             End
66.         Case Chr$(13)
67.             If ModeEvaluation = ModeImp Then ModeEvaluation = ModeOr Else ModeEvaluation = ModeImp
68.     End Select
69.     _Limit 20 ' save CPU  / risparmia la CPU
70. Loop
71. End ' Endo of program  /fine del programma
72.  
73. ' SUB and FUNCTION area  / area delle SUB e delle FUNCTION
74. Function outheRange (min As Integer, max As Integer, value As Integer)
75.     outheRange = (min < value Imp value > max)
76. End Function
77.  
78. Function IntheRange (min As Integer, max As Integer, Value As Integer)
79.     If min > Value Or max < Value Then IntheRange = 0 Else IntheRange = -1
80. End Function
81.  

  [ You are not allowed to view this attachment ]  

[Dimster]

I am impressed and confused @TempodiBasic

Here is my equation :  "If a < b And b > 1 Then c = (a<b) Imp (b>1)'    " If you remove the IF statement the equation becomes "c = (a<b) Imp (b>1)"

And here is your equation : "outheRange = (min < value Imp value > max)".

I do see a lot of difference between them. In your coding min, max and value are all calculate, whereas in mine I have fixed the values of a,b and fixed the value of 1. If I substitute a = min and b= max, then my equation becomes (min < max IMP max > 1) whereas yours converts to (a <1 IMP 1 > b)

In my mind, my equation is much more logical than yours in as much as min being less than max ( ie my a's & b's convert to your equation becomes ..(min < max IMP max > 1)... is very logical and max great than 1 pretty well a "must be obvious statement" but my equation doesn't work whereas your equation does works.

In your equation which I converted to reflect my a's and b's ...(a <1 IMP 1 > b)... i'm having difficult seeing how a < 1 , which indicates a value of a as zero or negative, can IMPLY 1 will be greater than b.

I will get it - one day - I  hope.

[Pete]

@TempodiBasic

Well, it has SCREEN 0 and Hero in it, so I had to run it! Hey, this takes me back. We had a challenge the QB64 forum 20 years ago to write this type of movement app in the fewest lines possible. I won! Michael Calkins hates me to this day, and he didn't even enter the contest, go figure. Anyway, I can't recall if I used the IMP function in that or not. I think it took just a little over 20 lines to code, no colons.

Well back to work, but thanks for a walk down QB memory lane.

Pete

[bplus]

@TempodiBasic

Well, it has SCREEN 0 and Hero in it, so I had to run it! Hey, this takes me back. We had a challenge the QB64 forum 20 years ago to write this type of movement app in the fewest lines possible. I won! Michael Calkins hates me to this day, and he didn't even enter the contest, go figure. Anyway, I can't recall if I used the IMP function in that or not. I think it took just a little over 20 lines to code, no colons.

Well back to work, but thanks for a walk down QB memory lane.

Pete

Nice challenge Pete, since I wasn't there then here is my hat in the ring:

Code: QB64: [Select]

1. hero = Int(Rnd * 25 * 80) + 1
2. Do
3.     Cls
4.     For y = 2 To 19
5.         For x = 10 To 70
6.             If y = 2 Or y = 19 Or x = 10 Or x = 70 Then
7.                 Locate y, x
8.                 Print Chr$(178)
9.             End If
10.             If (y - 1) * 80 + x - 1 = hero Then Locate 25, 32: Print "Hero is inbounds.";
11.         Next
12.     Next
13.     Locate hero \ 80 + 1, (hero Mod 80) + 1
14.     Print Chr$(1);
15.     a$ = InKey$
16.     If a$ = Chr$(0) + "M" Then hero = hero - (((hero Mod 80) + 1) <> 80)
17.     If a$ = Chr$(0) + "K" Then hero = hero + (((hero Mod 80)) <> 0)
18.     If a$ = Chr$(0) + "H" Then hero = hero + 80 * ((hero \ 80) > 0)
19.     If a$ = Chr$(0) + "P" Then hero = hero - 80 * ((hero \ 80) < 24)
20.     _Limit 10
21. Loop
22.  

Sorry one colon. :(

Edit: 4 less lines swapping variables for literals.

[bplus]

@TempodiBasic

Seems this is slightly better way to write InRange:

Code: QB64: [Select]

1. Function InRange% (min As Integer, max As Integer, Value As Integer)
2.     If Value >= min And Value <= max Then InRange% = -1
3. End Function
4.  

exactly to the point, though doing everything in Long Type covers more ground.

What do you think?