Integer_Transition_Systems 2019-03-29 01.54 pair #432274803

details

property	value
status	complete
benchmark	fibcall.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n077.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	15.913 seconds
cpu usage	16.1517
user time	8.23274
system time	7.91897
max virtual memory	726112.0
max residence set size	13248.0

stage attributes

key	value
starexec-result	YES

output

16.08/15.91	YES
16.08/15.91	
16.08/15.91	DP problem for innermost termination.
16.08/15.91	P =
16.08/15.91	  f5#(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) -> f4#(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) 
16.08/15.91	  f4#(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) -> f3#(1, 0, I2, I2, I4, 2, I2, I7, I8, I9) 
16.08/15.91	  f3#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f1#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) 
16.08/15.91	  f1#(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) -> f3#(I20 + I21, rnd2, I22, I23, I24, 1 + I25, I26, I27, I20, I29) [rnd2 = I20 /\ I25 <= I26]
16.08/15.91	R = 
16.08/15.91	  f5(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) -> f4(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) 
16.08/15.91	  f4(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) -> f3(1, 0, I2, I2, I4, 2, I2, I7, I8, I9) 
16.08/15.91	  f3(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f1(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) 
16.08/15.91	  f1(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) -> f3(I20 + I21, rnd2, I22, I23, I24, 1 + I25, I26, I27, I20, I29) [rnd2 = I20 /\ I25 <= I26]
16.08/15.91	  f1(I30, I31, I32, I33, I34, I35, I36, I37, I38, I39) -> f2(I30, I31, I32, I33, I30, I35, I36, I30, I38, I30) [1 + I36 <= I35]
16.08/15.91	
16.08/15.91	The dependency graph for this problem is:
16.08/15.91	  0 -> 1
16.08/15.91	  1 -> 2
16.08/15.91	  2 -> 3
16.08/15.91	  3 -> 2
16.08/15.91	Where:
16.08/15.91	  0) f5#(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) -> f4#(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) 
16.08/15.91	  1) f4#(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) -> f3#(1, 0, I2, I2, I4, 2, I2, I7, I8, I9) 
16.08/15.91	  2) f3#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f1#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) 
16.08/15.91	  3) f1#(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) -> f3#(I20 + I21, rnd2, I22, I23, I24, 1 + I25, I26, I27, I20, I29) [rnd2 = I20 /\ I25 <= I26]
16.08/15.91	
16.08/15.91	We have the following SCCs.
16.08/15.91	  { 2, 3 }
16.08/15.91	
16.08/15.91	DP problem for innermost termination.
16.08/15.91	P =
16.08/15.91	  f3#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f1#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) 
16.08/15.91	  f1#(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) -> f3#(I20 + I21, rnd2, I22, I23, I24, 1 + I25, I26, I27, I20, I29) [rnd2 = I20 /\ I25 <= I26]
16.08/15.91	R = 
16.08/15.91	  f5(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) -> f4(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) 
16.08/15.91	  f4(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) -> f3(1, 0, I2, I2, I4, 2, I2, I7, I8, I9) 
16.08/15.91	  f3(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f1(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) 
16.08/15.91	  f1(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) -> f3(I20 + I21, rnd2, I22, I23, I24, 1 + I25, I26, I27, I20, I29) [rnd2 = I20 /\ I25 <= I26]
16.08/15.91	  f1(I30, I31, I32, I33, I34, I35, I36, I37, I38, I39) -> f2(I30, I31, I32, I33, I30, I35, I36, I30, I38, I30) [1 + I36 <= I35]
16.08/15.91	
16.08/15.91	We use the reverse value criterion with the projection function NU:
16.08/15.91	  NU[f1#(z1,z2,z3,z4,z5,z6,z7,z8,z9,z10)] = z7 + -1 * z6
16.08/15.91	  NU[f3#(z1,z2,z3,z4,z5,z6,z7,z8,z9,z10)] = z7 + -1 * z6
16.08/15.91	
16.08/15.91	This gives the following inequalities:
16.08/15.91	    ==>  I16 + -1 * I15 >= I16 + -1 * I15
16.08/15.91	  rnd2 = I20 /\ I25 <= I26  ==>  I26 + -1 * I25 > I26 + -1 * (1 + I25)  with  I26 + -1 * I25 >= 0
16.08/15.91	
16.08/15.91	We remove all the strictly oriented dependency pairs.
16.08/15.91	
16.08/15.91	DP problem for innermost termination.
16.08/15.91	P =
16.08/15.91	  f3#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f1#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) 
16.08/15.91	R = 
16.08/15.91	  f5(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) -> f4(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) 
16.08/15.91	  f4(I0, I1, I2, I3, I4, I5, I6, I7, I8, I9) -> f3(1, 0, I2, I2, I4, 2, I2, I7, I8, I9) 
16.08/15.91	  f3(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f1(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) 
16.08/15.91	  f1(I20, I21, I22, I23, I24, I25, I26, I27, I28, I29) -> f3(I20 + I21, rnd2, I22, I23, I24, 1 + I25, I26, I27, I20, I29) [rnd2 = I20 /\ I25 <= I26]
16.08/15.91	  f1(I30, I31, I32, I33, I34, I35, I36, I37, I38, I39) -> f2(I30, I31, I32, I33, I30, I35, I36, I30, I38, I30) [1 + I36 <= I35]
16.08/15.91	
16.08/15.91	The dependency graph for this problem is:
16.08/15.91	  2 -> 
16.08/15.91	Where:
16.08/15.91	  2) f3#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) -> f1#(I10, I11, I12, I13, I14, I15, I16, I17, I18, I19) 
16.08/15.91	
16.08/15.91	We have the following SCCs.
16.08/15.91	  
16.08/18.88	EOF

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