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Verification Engineer Interview Questions

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Sep 9, 2010

Digital Design/Verification Engineer at SerialTek was asked...

Mar 16, 2011
 Create a 8 input AND gate using 3 4:1 muxes7 AnswersWithout an enable bit on at least one of the mux's the maximum inputs would be 7.Not so. You only need 2 4:1 muxes. Have the output of the first be the select to the second. 8 input and gate.tie 3 0's to the three inputs of initial 2 4x1 mux, the 3rd input be an actual input, 2 sel be 2 inputs. feed the output of the results of the two muxes as sel to the 3rd mux and tie the last inputs to actual inputs and top two inputs to 0's.Show More ResponsesI can only make it 7 bits with that explenation.I don't see it being possible with three standard 4-1 muxes... Using 4, this question is straight forward... The two selects of each mux are your 8 inputs... tie out put of each mux to the (11) case input to the mux.We need 3 4:1 MUX and a And gate. Are we allowed to use 'and' gate?A to H are the 8 inputs. For the first 2 muxes we can have GH as select bits with all their inputs tied to 0. Connect output of these muxes to the first 2 input lines of third mux. Tie the third input to 0. Now we care only about the 4th input line when EH are both 1s. We can derive an expression and connect it to the 4th input line of third mux. job done.

Design Verification & Test Engineer at Marvell Semiconductor was asked...

Sep 20, 2010
 You have seven stones and a weighing scale. Six of the stones are equal in weight and one is lighter. How will you figure out which one is lighter ? Minimum tries required to do so ? 5 AnswersTrail 1: At random weigh two stones vs. two stones (3 sitting on the side) A: Of the 4 on the scales if one side weighs more then the other then weigh one on each side (since one of them must be heavier) B. If the 2 vs 2 are equal then at random weigh 2 (one on each side) of the three left on the side. If they are the same then the 3rd one that never got weighed is the heaviest. Simple case of process of elimination by grouping (Divide and Test)I would first weigh in one stone, say stone 1, and assume the weight is say 2 lbs(try 1). Then separate the 6 remaining stones into 2 piles, 2,3,4 and 5,6,7. Weigh in either 2,3,4 or 5,6,7, it doesn't matter. Say 2,3,4, if the sum of these 3 is 6 then the lighter stone has to be in the 5,6,7(try 2). Weigh in 5,6, if the sum of the two is 4 then the lighter is stone 7(try3). If sum is less than 4 then weigh in either 5 or 6 to find out. So, the maximum number of tries is 4 and least is 3.Needs two weighing at most: 1. Put {1, 2, 3} on LHS and {4, 5, 6} on RHS. 2. If LHS and RHS are equal 7 is the lighter one. else discard heavier of previously weighed group. Now we have a group 3 stones left. Lets call them A, B, C. 3. Put A on LHS and B on RHS. 4. If LHS and RHS are equal C is the lighter one. else lighter or LHS or RHS is the lighter one. Voila!Show More ResponsesTwo tries. 1st try: 3 : 3, 7th is fake if equal; otherwise, 2nd try: 1:1 picked from the light triple in 1st try. the lighter one is fake if any, the third one fake otherwise.If the stones are made of the same material... they likely have the same density... therefor whichever one looks the smallest, will weigh the least... there... 1 step... just look for the one that is the smallest.

Senior Hardware Verification Engineer at NVIDIA was asked...

Jan 29, 2010
 Describe a circuit that implements the following truth table using only NAND gates. A B OUT 0 0 1 0 1 1 1 0 0 1 1 17 Answers((A NAND B) NAND C)out = (A NAND (B NAND B))out = ((A NAND A) NAND (A NAND A)) NAND (B NAND B)Show More ResponsesOUT = (A NAND (B NAND 1)) or out = (A NAND (B NAND B)) like what anonymous said.OUT = A' + B OUT = A' + B.1= A' + B.(A+A') = A'+ BA+ BA'= A'(1+B) + BA = A' +BA= (A.(BA)')' [Demorgans] = (A NAND(A NAND B))out = A' & B' + A' & B + A & B; => A' & B' + B => (A' +B) & (B' +B) => A' + B => (A NAND (B NAND B))(A NAND B) NAND A

Senior Hardware Verification Engineer at NVIDIA was asked...

Jan 29, 2010
 Describe a function to check if an integer is a power of 2.6 AnswersFor an integer n: If n is less than 1, return false. If the bitwise & of n and n-1 is 0, return true. Otherwise, return false.Write the number in binary and count the number of ones in that.If the number os ones is only 1 then it the number is indeed a power of 2first check if no is 1 then return false else write the number in binary and then check number of ones in that.if only one 1 is there then its a power of 2Show More ResponsesI think the main idea is to use recursion function, for the integer which is larger than 0, if it is 1 return true, else return function(n-1)See if the sum of all bits is 1. If that's the case then the number is a power of 2.function int power2(int i); if (i%2==1) return 0; else if (i/2==1) return 1; else return power2(i/2); endfunction

Verification Engineer at Cavium was asked...

Aug 7, 2009
 List an application (computer) where the properties of caching cannot be used.2 AnswersLive video streamingI/O operation.

ASIC Verification Engineer at Zoran was asked...

Sep 9, 2010
 What are the five stages of a classic RISC pipeline?2 AnswersFetch, Decode, Execute, Memory Access, WritebackInstruction Fetch; Instruction Decode; Memory Access; Execute; Write back

Sep 9, 2010