Integer_Transition_Systems 2019-03-29 01.54 pair #432275946

details

property	value
status	complete
benchmark	wrong_loop.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n084.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	10.3976 seconds
cpu usage	10.5923
user time	5.86049
system time	4.73181
max virtual memory	688152.0
max residence set size	9984.0

stage attributes

key	value
starexec-result	MAYBE

output

10.50/10.39	MAYBE
10.50/10.39	
10.50/10.39	DP problem for innermost termination.
10.50/10.39	P =
10.50/10.39	  f8#(x1, x2, x3, x4) -> f7#(x1, x2, x3, x4) 
10.50/10.39	  f7#(I0, I1, I2, I3) -> f3#(I0, I1, 0, I3) 
10.50/10.39	  f3#(I4, I5, I6, I7) -> f5#(I4, I5, I6, I7) 
10.50/10.39	  f5#(I8, I9, I10, I11) -> f4#(I8, I9, I10, I11) [1 + I10 <= I8]
10.50/10.39	  f4#(I16, I17, I18, I19) -> f2#(I16, I17, rnd3, I19) [rnd3 = rnd3 /\ I17 <= I18 /\ I18 <= I17]
10.50/10.39	  f4#(I20, I21, I22, I23) -> f1#(I20, I21, I22, I23) [1 + I22 <= I21]
10.50/10.39	  f4#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) [1 + I25 <= I26]
10.50/10.39	  f2#(I28, I29, I30, I31) -> f3#(I28, I29, 1 + I30, I31) 
10.50/10.39	  f1#(I32, I33, I34, I35) -> f2#(I32, I33, I34, I34) 
10.50/10.39	R = 
10.50/10.39	  f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) 
10.50/10.39	  f7(I0, I1, I2, I3) -> f3(I0, I1, 0, I3) 
10.50/10.39	  f3(I4, I5, I6, I7) -> f5(I4, I5, I6, I7) 
10.50/10.39	  f5(I8, I9, I10, I11) -> f4(I8, I9, I10, I11) [1 + I10 <= I8]
10.50/10.39	  f5(I12, I13, I14, I15) -> f6(I12, I13, I14, I15) [I12 <= I14]
10.50/10.39	  f4(I16, I17, I18, I19) -> f2(I16, I17, rnd3, I19) [rnd3 = rnd3 /\ I17 <= I18 /\ I18 <= I17]
10.50/10.39	  f4(I20, I21, I22, I23) -> f1(I20, I21, I22, I23) [1 + I22 <= I21]
10.50/10.39	  f4(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) [1 + I25 <= I26]
10.50/10.39	  f2(I28, I29, I30, I31) -> f3(I28, I29, 1 + I30, I31) 
10.50/10.39	  f1(I32, I33, I34, I35) -> f2(I32, I33, I34, I34) 
10.50/10.39	
10.50/10.39	The dependency graph for this problem is:
10.50/10.39	  0 -> 1
10.50/10.39	  1 -> 2
10.50/10.39	  2 -> 3
10.50/10.39	  3 -> 4, 5, 6
10.50/10.39	  4 -> 7
10.50/10.39	  5 -> 8
10.50/10.39	  6 -> 8
10.50/10.39	  7 -> 2
10.50/10.39	  8 -> 7
10.50/10.39	Where:
10.50/10.39	  0) f8#(x1, x2, x3, x4) -> f7#(x1, x2, x3, x4) 
10.50/10.39	  1) f7#(I0, I1, I2, I3) -> f3#(I0, I1, 0, I3) 
10.50/10.39	  2) f3#(I4, I5, I6, I7) -> f5#(I4, I5, I6, I7) 
10.50/10.39	  3) f5#(I8, I9, I10, I11) -> f4#(I8, I9, I10, I11) [1 + I10 <= I8]
10.50/10.39	  4) f4#(I16, I17, I18, I19) -> f2#(I16, I17, rnd3, I19) [rnd3 = rnd3 /\ I17 <= I18 /\ I18 <= I17]
10.50/10.39	  5) f4#(I20, I21, I22, I23) -> f1#(I20, I21, I22, I23) [1 + I22 <= I21]
10.50/10.39	  6) f4#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) [1 + I25 <= I26]
10.50/10.39	  7) f2#(I28, I29, I30, I31) -> f3#(I28, I29, 1 + I30, I31) 
10.50/10.39	  8) f1#(I32, I33, I34, I35) -> f2#(I32, I33, I34, I34) 
10.50/10.39	
10.50/10.39	We have the following SCCs.
10.50/10.39	  { 2, 3, 4, 5, 6, 7, 8 }
10.50/10.39	
10.50/10.39	DP problem for innermost termination.
10.50/10.39	P =
10.50/10.39	  f3#(I4, I5, I6, I7) -> f5#(I4, I5, I6, I7) 
10.50/10.39	  f5#(I8, I9, I10, I11) -> f4#(I8, I9, I10, I11) [1 + I10 <= I8]
10.50/10.39	  f4#(I16, I17, I18, I19) -> f2#(I16, I17, rnd3, I19) [rnd3 = rnd3 /\ I17 <= I18 /\ I18 <= I17]
10.50/10.39	  f4#(I20, I21, I22, I23) -> f1#(I20, I21, I22, I23) [1 + I22 <= I21]
10.50/10.39	  f4#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) [1 + I25 <= I26]
10.50/10.39	  f2#(I28, I29, I30, I31) -> f3#(I28, I29, 1 + I30, I31) 
10.50/10.39	  f1#(I32, I33, I34, I35) -> f2#(I32, I33, I34, I34) 
10.50/10.39	R = 
10.50/10.39	  f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) 
10.50/10.39	  f7(I0, I1, I2, I3) -> f3(I0, I1, 0, I3) 
10.50/10.39	  f3(I4, I5, I6, I7) -> f5(I4, I5, I6, I7) 
10.50/10.39	  f5(I8, I9, I10, I11) -> f4(I8, I9, I10, I11) [1 + I10 <= I8]
10.50/10.39	  f5(I12, I13, I14, I15) -> f6(I12, I13, I14, I15) [I12 <= I14]
10.50/10.39	  f4(I16, I17, I18, I19) -> f2(I16, I17, rnd3, I19) [rnd3 = rnd3 /\ I17 <= I18 /\ I18 <= I17]
10.50/10.39	  f4(I20, I21, I22, I23) -> f1(I20, I21, I22, I23) [1 + I22 <= I21]
10.50/10.39	  f4(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) [1 + I25 <= I26]
10.50/10.39	  f2(I28, I29, I30, I31) -> f3(I28, I29, 1 + I30, I31) 
10.50/10.39	  f1(I32, I33, I34, I35) -> f2(I32, I33, I34, I34) 
10.50/10.39	
10.50/13.37	EOF

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