details

property	value
status	complete
benchmark	fibcall.t2_fixed.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n141.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	12.8172 seconds
cpu usage	13.0106
user time	6.74939
system time	6.26122
max virtual memory	597540.0
max residence set size	12152.0

stage attributes

key	value
starexec-result	YES

output

YES DP problem for innermost termination. P = f5#(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f4#(x1, x2, x3, x4, x5, x6, x7, x8, x9) f4#(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f3#(1, 0, 30, I3, 2, 30, I6, I7, I8) f3#(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f1#(I9, I10, I11, I12, I13, I14, I15, I16, I17) f1#(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f3#(I18 + I19, rnd2, I20, I21, 1 + I22, I23, I24, I18, I26) [rnd2 = I18 /\ I22 <= I23] R = f5(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f4(x1, x2, x3, x4, x5, x6, x7, x8, x9) f4(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f3(1, 0, 30, I3, 2, 30, I6, I7, I8) f3(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f1(I9, I10, I11, I12, I13, I14, I15, I16, I17) f1(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f3(I18 + I19, rnd2, I20, I21, 1 + I22, I23, I24, I18, I26) [rnd2 = I18 /\ I22 <= I23] f1(I27, I28, I29, I30, I31, I32, I33, I34, I35) -> f2(I27, I28, I29, I27, I31, I32, I27, I34, I27) [1 + I32 <= I31] The dependency graph for this problem is: 0 -> 1 1 -> 2 2 -> 3 3 -> 2 Where: 0) f5#(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f4#(x1, x2, x3, x4, x5, x6, x7, x8, x9) 1) f4#(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f3#(1, 0, 30, I3, 2, 30, I6, I7, I8) 2) f3#(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f1#(I9, I10, I11, I12, I13, I14, I15, I16, I17) 3) f1#(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f3#(I18 + I19, rnd2, I20, I21, 1 + I22, I23, I24, I18, I26) [rnd2 = I18 /\ I22 <= I23] We have the following SCCs. { 2, 3 } DP problem for innermost termination. P = f3#(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f1#(I9, I10, I11, I12, I13, I14, I15, I16, I17) f1#(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f3#(I18 + I19, rnd2, I20, I21, 1 + I22, I23, I24, I18, I26) [rnd2 = I18 /\ I22 <= I23] R = f5(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f4(x1, x2, x3, x4, x5, x6, x7, x8, x9) f4(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f3(1, 0, 30, I3, 2, 30, I6, I7, I8) f3(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f1(I9, I10, I11, I12, I13, I14, I15, I16, I17) f1(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f3(I18 + I19, rnd2, I20, I21, 1 + I22, I23, I24, I18, I26) [rnd2 = I18 /\ I22 <= I23] f1(I27, I28, I29, I30, I31, I32, I33, I34, I35) -> f2(I27, I28, I29, I27, I31, I32, I27, I34, I27) [1 + I32 <= I31] We use the reverse value criterion with the projection function NU: NU[f1#(z1,z2,z3,z4,z5,z6,z7,z8,z9)] = z6 + -1 * z5 NU[f3#(z1,z2,z3,z4,z5,z6,z7,z8,z9)] = z6 + -1 * z5 This gives the following inequalities: ==> I14 + -1 * I13 >= I14 + -1 * I13 rnd2 = I18 /\ I22 <= I23 ==> I23 + -1 * I22 > I23 + -1 * (1 + I22) with I23 + -1 * I22 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f1#(I9, I10, I11, I12, I13, I14, I15, I16, I17) R = f5(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f4(x1, x2, x3, x4, x5, x6, x7, x8, x9) f4(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f3(1, 0, 30, I3, 2, 30, I6, I7, I8) f3(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f1(I9, I10, I11, I12, I13, I14, I15, I16, I17) f1(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f3(I18 + I19, rnd2, I20, I21, 1 + I22, I23, I24, I18, I26) [rnd2 = I18 /\ I22 <= I23] f1(I27, I28, I29, I30, I31, I32, I33, I34, I35) -> f2(I27, I28, I29, I27, I31, I32, I27, I34, I27) [1 + I32 <= I31] The dependency graph for this problem is: 2 -> Where: 2) f3#(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f1#(I9, I10, I11, I12, I13, I14, I15, I16, I17) We have the following SCCs. popout

output may be truncated. 'popout' for the full output.

job log

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actions

all output return to ITS