Markov Models and Hidden Markov Modeling in Software Printing barcode pdf417 in Software Markov Models and Hidden Markov Modeling

How to generate, print barcode using .NET, Java sdk library control with example project source code free download:
16.3 Markov Models and Hidden Markov Modeling generate, create none none in none projects ANSI/AIM Code 93 hmm.h void HMM::f orwardbackward() { HMM forward-backward algorithm. Using the stored a, b, and obs matrices, the matrices alpha, beta, and pstate are calculated. The latter is the state estimation of the model, given the data.

Int i,j,t; Doub sum,asum,bsum; for (i=0; i<mstat; i++) alpha[0][i] = b[i][obs[0]]; arnrm[0] = 0; for (t=1; t<nobs; t++) { Forward pass. asum = 0; for (j=0; j<mstat; j++) { sum = 0.; for (i=0; i<mstat; i++) sum += alpha[t-1][i]*a[i][j]*b[j][obs[t]]; alpha[t][j] = sum; asum += sum; } arnrm[t] = arnrm[t-1]; Renormalize the s as necessary to avoid underif (asum < BIGI) { ow, keeping track of how many renormal++arnrm[t]; izations for each .

for (j=0; j<mstat; j++) alpha[t][j] *= BIG; } } for (i=0; i<mstat; i++) beta[nobs-1][i] = 1.; brnrm[nobs-1] = 0; for (t=nobs-2; t>=0; t--) { Backward pass. bsum = 0.

; for (i=0; i<mstat; i++) { sum = 0.; for (j=0; j<mstat; j++) sum += a[i][j]*b[j][obs[t+1]]*beta[t+1][j]; beta[t][i] = sum; bsum += sum; } brnrm[t] = brnrm[t+1]; if (bsum < BIGI) { Similarly, renormalize the s as necessary. ++brnrm[t]; for (j=0; j<mstat; j++) beta[t][j] *= BIG; } } lhood = 0.

; Overall likelihood is lhood with lnorm renormalfor (i=0; i<mstat; i++) lhood += alpha[0][i]*beta[0][i]; izations. lrnrm = arnrm[0] + brnrm[0]; while (lhood < BIGI) {lhood *= BIG; lrnrm++;} for (t=0; t<nobs; t++) { Get state probabilities from s and s. sum = 0.

; for (i=0; i<mstat; i++) sum += (pstate[t][i] = alpha[t][i]*beta[t][i]); The next line is an equivalent calculation of sum. But we d rather have the normalization of the Pi .t/ s be more immune to roundo error.

Hence we do the above sum for each value of t. // sum = lhood*pow(BIGI, lrnrm - arnrm[t] - brnrm[t]); for (i=0; i<mstat; i++) pstate[t][i] /= sum; } fbdone = 1; Flag prevents misuse of baumwelch(), later. }.

You may be none for none wondering how well forwardbackward is able to do at predicting the hidden states of Teen Life, given just a long string of output symbols. If we take the prediction to be the state with the highest probability at each time, then this is correct about 78% of the time. Another 17% of the time, the correct state has the second-highest probability, often when the top two probabilities are nearly equal.

It is an important property of HMMs that the output is not only a prediction, but also a quantitative assessment of how sure the model is of that prediction.. 16. Classi cation and Inference 16.3.2 Some Variations on HMM HMM state e stimation with the forward-backward algorithm is a very exible formalism, and many variants are possible. For example, in decoding codes on a trellis, as we did above, the symbols 0 or 1 are emitted not by the states, but by the transitions between the states. If we want to use HMM for that problem (we will say more about this below), we must replace bi .

k/ with bij .k/, the probability of emitting symbol k in a transition from state i to state j . The forward and backward recurrences now become t C1 .

j / D. M 1 X i D0 t .i /Aij bij .y t C1 / (16.

3.19) Aij bij .y t / t .

j /. 1 .i /. M 1 X j D0 and we star none none t off the s with the special rule 0 .i / D 1, since (like the case of N 1 .i / previously) the probability of the data is 1 before there are any data.

Another variant case is where one or more intermediate states are known exactly. In that case, one or more of the sums over i0 ; i1 ; : : : ; i t 1 in equation (16.3.

13) is left out, and the corresponding index on an A and b gets replaced by the known state number. If you track through how this affects the recurrence equation (16.3.

14), you ll see that the new procedure is calculate the s forward to, and including, a known state; zero all the values at that time except for that of the known state; don t renormalize anything (though you feel tempted to do so); and continue forward with the s for the next timestep. Proceed similarly for the s. The opposite variant is where you have missing data, meaning that for some values of t there is no observation of the symbol y t .

In this case, all you need to do is to make a special case for the symbol probability, bi .y t / 1; .0 i < M / t 2 fmissingg (16.

3.20). meaning tha t, regardless of state i , the probability of observing the data (meaning no data) at time t is unity. Now proceed as usual to calculate the state probabilities. If you then want to reconstruct the missing data, you can calculate its posterior probabilities, P .

y t D k j y/ D.
Copyright © . All rights reserved.