使用 SMTLIB2 查找 z3 中的最大数字

Question

I have 7 cups which contains some water. I need to program these cups to have different amounts of water. Once this is done I need to measure the cup which has the highest amount of water and then remove some quantity (say 2 units of water).

c implementations:

float c1=2.0, c2= 2.6, c3 = 2.8, c4=4.4 , c5 = 2.4, c6 = 2.1, c7 = 5.8;

if((c1 > c2) && (c1 > c3) && (c1 > c4) && (c1 > c5) && (c1 > c6) && (c1 > c7)); c1=c1-2;

if((c2 > c1) && (c2 > c3) && (c2 > c4) && (c2 > c5) && (c2 > c6) && (c2 > c7)); c2=c2-2;

if((c3 > c2) && (c3 > c1) && (c3 > c4) && (c3 > c5) && (c3 > c6) && (c3 > c7)); c3=c3-2;

if((c4 > c2) && (c4 > c3) && (c4 > c1) && (c4 > c5) && (c4 > c6) && (c4 > c7)); c4=c4-2;

if((c5 > c2) && (c5 > c3) && (c5 > c4) && (c5 > c1) && (c5 > c6) && (c5 > c7)); c5=c5-2;

if((c6 > c2) && (c6 > c3) && (c6 > c4) && (c6 > c5) && (c6 > c1) && (c6 > c7)); c6=c6-2;

if((c7 > c2) && (c7 > c3) && (c7 > c4) && (c7 > c5) && (c7 > c6) && (c7 > c1)); c7=c7-2; 

This will give an answer as c7 = 3.8

I was trying to implement this in z3 and I assigned the values to c1....c7

ite( (and((> c1  c2) (> c1  c3) (> c1  c4) (> c1  c5) (> c1  c6) (> c1  c7)))  (= c1_1 (- c1 2)  (= c1_1 c1))
.
.
.repeated till c7_1

and when I get the model values it should give c7_1 as 3.8

Is it possible to define this in z3? When I am using the and's of different conditions in if condition (in ite) its giving me an error. Can it not be defined like this? Is there someway around this?

Thanks in advance

[problem description][1]

I am experimenting with the Z3 tool, It was easy to get he first part However for the second part is being a bit difficult.

Answer 1

Sure. Here it is in SMTLib:

; declare the cups
(declare-const c1 Real)
(declare-const c2 Real)
(declare-const c3 Real)
(declare-const c4 Real)
(declare-const c5 Real)
(declare-const c6 Real)
(declare-const c7 Real)

; each cup has a non-negative units of water
(assert (>= c1 0))
(assert (>= c2 0))
(assert (>= c3 0))
(assert (>= c4 0))
(assert (>= c5 0))
(assert (>= c6 0))
(assert (>= c7 0))

; each amount is different
(assert (distinct c1 c2 c3 c4 c5 c6 c7))

; find maximum, helper function
(define-fun max ((a Real) (b Real)) Real (ite (> a b) a b))

; find the cup with maximum water in it
(define-fun maxC () Real (max c1 (max c2 (max c3 (max c4 (max c5 (max c6 c7)))))))

; make sure there's at least 2 units in the max, per the problem
(assert (>= maxC 2))

; final value
(define-fun finalRes () Real (- maxC 2))

; solve
(check-sat)
(get-value (c1 c2 c3 c4 c5 c6 c7 maxC finalRes))

z3 says:

sat
((c1 2.0)
 (c2 (/ 11.0 6.0))
 (c3 (/ 19.0 12.0))
 (c4 (/ 7.0 4.0))
 (c5 (/ 3.0 2.0))
 (c6 (/ 23.0 12.0))
 (c7 (/ 5.0 3.0))
 (maxC 2.0)
 (finalRes 0.0))

So, looks like it put 2 units in c1 , less than 2 on all the others, so you ended up with a final value of 0 left.

Your question is rather vague in what other constraints might be at play here, but hopefully this will get you started.

Answer 2

Here's an alternative approach for finding max that doesn't use ITE .

; declare the cups
(declare-const c1 Real)
(declare-const c2 Real)
(declare-const c3 Real)
(declare-const c4 Real)
(declare-const c5 Real)
(declare-const c6 Real)
(declare-const c7 Real)

; each cup has a non-negative units of water
(assert (>= c1 0))
(assert (>= c2 0))
(assert (>= c3 0))
(assert (>= c4 0))
(assert (>= c5 0))
(assert (>= c6 0))
(assert (>= c7 0))

; each amount is different
(assert (distinct c1 c2 c3 c4 c5 c6 c7))

(declare-fun max () Real)

(assert (and (<= c1 max) (<= c2 max) (<= c3 max) (<= c4 max) (<= c5 max) (<= c6 max) (<= c7 max)))

(assert (or (<= max c1) (<= max c2) (<= max c3) (<= max c4) (<= max c5) (<= max c6) (<= max c7) ))

; make sure there's at least 2 units in the max, per the problem
(assert (<= 2 max))

; final value
(define-fun finalRes () Real (- max 2))

; solve
(check-sat)
(get-value (c1 c2 c3 c4 c5 c6 c7 maxC finalRes))

And the result:

sat
((c1 2.0)
 (c2 (/ 4.0 3.0))
 (c3 0.0)
 (c4 (/ 1.0 3.0))
 (c5 1.0)
 (c6 (/ 2.0 3.0))
 (c7 (/ 5.0 3.0))
 (max 2.0)
 (finalRes 0.0))