Z3

Empty model in z3

由 扶醉桌前 提交于 2021-02-08 15:44:27
问题 z3py snippet: x = Int('x') s = Solver() s.add(x <= x) print s.check() print s.model() print s.model().sexpr() http://rise4fun.com/Z3Py/mfPU Output: sat [] Any value of x would do but z3 returns empty model. Does a missing free variable x in the model indicates that any integer value is a valid model? 回答1: Yes, in Z3, if a constant (such as x ) does not appear in the model, then it is a "don't care". That is, any value of x will satisfy the formula. When evaluating the value of a constant, we

Empty model in z3

由 冷暖自知 提交于 2021-02-08 15:44:10
问题 z3py snippet: x = Int('x') s = Solver() s.add(x <= x) print s.check() print s.model() print s.model().sexpr() http://rise4fun.com/Z3Py/mfPU Output: sat [] Any value of x would do but z3 returns empty model. Does a missing free variable x in the model indicates that any integer value is a valid model? 回答1: Yes, in Z3, if a constant (such as x ) does not appear in the model, then it is a "don't care". That is, any value of x will satisfy the formula. When evaluating the value of a constant, we

Z3 Conditional Statement

由 放肆的年华 提交于 2021-02-08 10:45:25
问题 How to write a conditional statement in Z3. eg: if (a%2==0){ value=1 } I am trying to achieve this in Z3 Solver by Microsoft Research but so far no luck 回答1: Look up SSA form: https://en.wikipedia.org/wiki/Static_single_assignment_form Essentially, you'll have to change your program to look something like: value_0 = 0 value_1 = (a%2 == 0) ? 1 : value_0 Once it is in this so called static single assignment form, you can now translate each line more or less directly; with the latest assignment

Getting proof from z3py

由 醉酒当歌 提交于 2021-02-08 02:26:13
问题 I've been skimming through the documentation of Z3Py and for the likes of me have not been able to figure out how to get the proof from a Solver (E.g. if I start from an instance of De Morgan's Laws how can I extract the proof from Z3Py of the instance, step by step). The only reference I saw was for proof(self) in the Solver class which supposedly gets the proof of the last check if proof construction is enabled, but I keep getting back the very vague error: Traceback (most recent call last)

Getting proof from z3py

由 拟墨画扇 提交于 2021-02-08 02:22:39
问题 I've been skimming through the documentation of Z3Py and for the likes of me have not been able to figure out how to get the proof from a Solver (E.g. if I start from an instance of De Morgan's Laws how can I extract the proof from Z3Py of the instance, step by step). The only reference I saw was for proof(self) in the Solver class which supposedly gets the proof of the last check if proof construction is enabled, but I keep getting back the very vague error: Traceback (most recent call last)

【每日算法/刷穿 LeetCode】1423. 可获得的最大点数（中等）

由 半世苍凉 提交于 2021-02-06 14:36:55
点击 这里 可以查看更多算法面试相关内容~ 题目描述 几张卡牌 排成一行，每张卡牌都有一个对应的点数。点数由整数数组 nums 给出。 每次行动，你可以从行的开头或者末尾拿一张卡牌，最终你必须正好拿 k 张卡牌。 你的点数就是你拿到手中的所有卡牌的点数之和。 给你一个整数数组 nums 和整数 k，请你返回可以获得的最大点数。 示例 1： 输入：nums = [1,2,3,4,5,6,1], k = 3 输出：12 解释：第一次行动，不管拿哪张牌，你的点数总是 1 。 但是，先拿最右边的卡牌将会最大化你的可获得点数。 最优策略是拿右边的三张牌，最终点数为 1 + 6 + 5 = 12 。 示例 2： 输入：nums = [2,2,2], k = 2 输出：4 解释：无论你拿起哪两张卡牌，可获得的点数总是 4 。 示例 3： 输入：nums = [9,7,7,9,7,7,9], k = 7 输出：55 解释：你必须拿起所有卡牌，可以获得的点数为所有卡牌的点数之和。 示例 4： 输入：nums = [1,1000,1], k = 1 输出：1 解释：你无法拿到中间那张卡牌，所以可以获得的最大点数为 1 。 示例 5： 输入：nums = [1,79,80,1,1,1,200,1], k = 3 输出：202 提示： 1 <= nums.length <= 10^5 1 <= nums

Modeling a small programming language and analysis in SMT-LIB using datatypes and forall

由 无人久伴 提交于 2021-02-05 11:58:20
问题 I am trying to model a small programming language in SMT-LIB 2. My intent is to express some program analysis problems and solve them with Z3. I think I am misunderstanding the forall statement though. Here is a snippet of my code. ; barriers.smt2 (declare-datatype Barrier ((barrier (proc Int) (rank Int) (group Int) (complete-time Int)))) ; barriers in the same group complete at the same time (assert (forall ((b1 Barrier) (b2 Barrier)) (=> (= (group b1) (group b2)) (= (complete-time b1)

Modeling a small programming language and analysis in SMT-LIB using datatypes and forall

由 五迷三道 提交于 2021-02-05 11:58:02
问题 I am trying to model a small programming language in SMT-LIB 2. My intent is to express some program analysis problems and solve them with Z3. I think I am misunderstanding the forall statement though. Here is a snippet of my code. ; barriers.smt2 (declare-datatype Barrier ((barrier (proc Int) (rank Int) (group Int) (complete-time Int)))) ; barriers in the same group complete at the same time (assert (forall ((b1 Barrier) (b2 Barrier)) (=> (= (group b1) (group b2)) (= (complete-time b1)

Getting all solutions of a boolean expression in Z3Py never ends

由 风格不统一 提交于 2021-02-05 08:43:05
问题 Probably a basic question related to Z3: i am trying to get all solutions of a boolean expression, e.g. for a OR b , i want to get {(true, true),(false,true),(true,false)} Based on other responses found, e.g. Z3: finding all satisfying models, i have the following code: a = Bool('a') b = Bool('b') f1=Or(a,b) s=Solver() s.add(f1) while s.check() == sat: print s s.add(Not(And(a == s.model()[a], b == s.model()[b]))) The issue is that it enters an infinite loop as at the second iteration: the

Counterexample output of Z3

由 有些话、适合烂在心里 提交于 2021-02-04 18:11:26
问题 When a formula in Z3 is unsat and (get-proof) is specified there is an output which I don't find any information about what it is. Where can I find any documentation about that? Seems to me quite unreadable, is there possibly any tool that takes this as an input? Cheers, Matt 回答1: The "proofs" produced by Z3 are not for human consumption. An outdated version of the format is described in the paper: Proofs and Refutations, and Z3. The z3_api.h file has a long description of each one of the
订阅 Z3